(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a rotation and a shear. The rotation is represented by \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\), which is an anticlockwise rotation about the origin through an angle \(\theta\). The shear is represented by \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), which fixes the x-axis and maps \((0,1)\) to \((2,1)\).

(b) The matrix \(\mathbf{M}\) is given by:

\(\mathbf{M} = \begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\)

Applying \(\mathbf{M}\) to a point \((x, y)\) gives:

\(\begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}\)

\(For a line of invariant points, we require:\)

\(x \sin \theta + y \cos \theta = y\) and \(x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta = x\).

Solving these equations gives:

\(x \cos \theta + 2x \sin \theta + \frac{x \sin \theta}{1 - \cos \theta} (2 \cos \theta - \sin \theta) = x\)

Simplifying, we find:

\((\cos \theta + 2 \sin \theta)(1 - \cos \theta) + 2 \sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos \theta - \cos^2 \theta + 2 \sin^2 \theta - \sin^2 \theta = 1 - \cos \theta \Rightarrow \sin \theta + \cos \theta = 1\)

Thus, \(\theta = \frac{1}{2} \pi\).

(a) Find direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

Calculate the normal to the plane ABC using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}\)

Substitute point \(A(1, 0, -2)\) into \(3x + 2y - z = d\) to find \(d = 5\).

(b) Find normal to plane ABD:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = (2t+4)\mathbf{i}\)

Use dot product to find angle:

\(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2t+4 \\ 0 \\ 0 \end{pmatrix} = \sqrt{14} \cos \theta\)

\(\theta = 36.7^\circ\) when \(t \neq -2\).

(c) Find common perpendicular:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ -1 & 1 & t+1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ 1 \end{pmatrix}\)

Use formula for perpendicular distance:

\(\frac{l + 2}{\sqrt{t^2 - 2t + 6}} = \sqrt{2}\)

Solve \(t^2 - 8t + 8 = 0\) to find \(t = 6.83, \; t = 1.17\).

(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\) represents a rotation followed by a shear. The rotation is anticlockwise, centered at the origin, through an angle \(\theta\), with the x-axis fixed and \((0,1)\) mapped to \((2,1)\).

(b) The combined matrix is \(\mathbf{M} = \begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\).

\(To find the line of invariant points, solve:\)

\(\begin{pmatrix} \cos \theta + 2\sin \theta & 2\cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)

This gives the equations:

\(x\sin \theta + y\cos \theta = y\)

\(x\cos \theta + 2x\sin \theta - y\sin \theta + 2y\cos \theta = x\)

Rearranging and simplifying, we find:

\(x\cos \theta + 2x\sin \theta + \frac{x\sin \theta}{1 - \cos \theta}(2\cos \theta - \sin \theta) = x\)

\((\cos \theta + 2\sin \theta)(1 - \cos \theta) + 2\sin \theta\cos \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos^2 \theta - \cos^2 \theta + 2\sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\Rightarrow \sin \theta + \cos \theta = 1\)

Solving gives \(\theta = \frac{1}{2} \pi\).

(a) To find the equation of the plane ABC, we first determine the direction vectors \(\overrightarrow{AB} = 2\mathbf{j} + 4\mathbf{k}\) and \(\overrightarrow{AC} = \mathbf{i} - \mathbf{j} + \mathbf{k}\).

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \begin{pmatrix} 6 \\ 4 \\ -2 \end{pmatrix}\)

Substituting point \(A(1, 0, -2)\) into the plane equation \(3x + 2y - z = d\), we find \(d = 5\).

(b) The normal to the plane ABD is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 4 \\ 0 & 0 & t+2 \end{vmatrix} = \begin{pmatrix} 2t + 4 \\ 0 \\ 0 \end{pmatrix}\)

The angle \(\theta\) between the planes is given by:

\(\cos \theta = \frac{\left( \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right)}{\sqrt{14} \cdot 1}\)

\(\theta = 36.7^\circ\).

(c) To find \(t\) such that the shortest distance between lines AB and CD is \(\sqrt{2}\), we use the formula for the perpendicular distance between skew lines:

\(\frac{1}{\sqrt{t^2 - 2t + 6}} \left| \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} \right| = \frac{t + 2}{\sqrt{t^2 - 2t + 6}}\)

Setting this equal to \(\sqrt{2}\) and solving gives:

\(t^2 - 8t + 8 = 0 \Rightarrow t = 2(2 \pm \sqrt{2})\)

Thus, \(t = 6.83\) and \(t = 1.17\).

(a) The stretch parallel to the x-axis with scale factor 14 is represented by the matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\).

The rotation anticlockwise about the origin through angle \(\frac{1}{3} \pi\) is represented by \(\begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

Thus, \(\mathbf{M} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & -\frac{\sqrt{3}}{2} \\ 7\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

Therefore, \(2\mathbf{M} = \begin{pmatrix} 14 & -\sqrt{3} \\ 14\sqrt{3} & 1 \end{pmatrix}\).

(b) The transformation \(\mathbf{M}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 7x - \frac{\sqrt{3}}{2}y \\ 7\sqrt{3}x + \frac{1}{2}y \end{pmatrix}\).

For invariant lines, \(y = mx\) and \(Y = mX\).

\(7\sqrt{3}x + \frac{1}{2}mx = m(7x - \frac{\sqrt{3}}{2}mx)\).

\(7\sqrt{3} + \frac{1}{2}m = 7m - \frac{\sqrt{3}}{2}m^2\).

\(\sqrt{3}m^2 - 13m + 14\sqrt{3} = 0\).

Solving gives \(m = 2\sqrt{3}\) and \(m = \frac{7}{\sqrt{3}}\).

Thus, the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{1}{\sqrt{3}}x\).

(c) The inverse of \(2\mathbf{M}\) is \(\mathbf{M}^{-1} = \frac{1}{14} \begin{pmatrix} 1 & \sqrt{3} \\ -14\sqrt{3} & 7 \end{pmatrix}\).

(a) To find the Cartesian equation of the plane \(\Pi\), we need a normal vector. The direction vectors are \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). The normal vector is the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 3 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

Using the point \((2, 3, -2)\) on the plane, substitute into \(6x - y + 8z = d\):

\(6(2) - 3 + 8(-2) = -7\)

Thus, the equation is \(6x - y + 8z = -7\).

(b) The position vector of the foot of the perpendicular \(\mathbf{F}\) is given by:

\(\mathbf{OF} = \mathbf{OP} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\mathbf{OF} = \begin{pmatrix} 4 \\ 2 \\ 9 \end{pmatrix} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} 4 + 6t \\ 2 - t \\ 9 + 8t \end{pmatrix}\)

Substitute into the plane equation:

\(6(4 + 6t) - (2 - t) + 8(9 + 8t) = -7\)

\(101t + 94 = -7\)

\(t = -1\)

\(\mathbf{OF} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)

(c) The acute angle \(\alpha\) between \(l\) and \(\Pi\) is found using:

\(\cos \alpha = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \|\mathbf{n}\|}\)

\(\mathbf{d} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}, \mathbf{n} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\cos \alpha = \frac{5}{\sqrt{35} \sqrt{101}}\)

\(\alpha = 4.8^\circ\)

(a) To find the equation of the plane, we first determine the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (-5\mathbf{i} + 3\mathbf{j} + \mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = -5\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + 2\mathbf{k}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 1 & -2 \\ 1 & 0 & 2 \end{vmatrix} = 2\mathbf{i} + 8\mathbf{j} - \mathbf{k}\)

Using point \(A(0, 2, 3)\), the equation of the plane is:

\(2(0) + 8(2) - 1(3) = 13 \Rightarrow 2x + 8y - z = 13\)

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|2(0) + 8(0) - 1(0) - 13|}{\sqrt{2^2 + 8^2 + (-1)^2}} = \frac{13}{\sqrt{69}} = 1.57\)

(c) The acute angle \(\theta\) between the line \(OA\) and the plane is found using the dot product:

\(\cos \theta = \frac{\mathbf{n} \cdot \overrightarrow{OA}}{\|\mathbf{n}\| \|\overrightarrow{OA}\|}\)

\(\overrightarrow{OA} = 2\mathbf{j} + 3\mathbf{k}\)

\(\mathbf{n} \cdot \overrightarrow{OA} = 2(0) + 8(2) - 1(3) = 13\)

\(\|\mathbf{n}\| = \sqrt{69}, \quad \|\overrightarrow{OA}\| = \sqrt{13}\)

\(\cos \theta = \frac{13}{\sqrt{69} \sqrt{13}}\)

\(\theta = \cos^{-1}\left(\frac{13}{\sqrt{69} \sqrt{13}}\right) = 0.449\) radians

(a) The transformation matrix for a stretch in the x-direction by a factor of 3 is \(\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\). The reflection in the line y = x is represented by \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Therefore, \(\mathbf{M} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 0 \end{pmatrix}\).

(b) The inverse of \(\mathbf{M}\) is \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix}\). This represents a reflection in the line y = x followed by a stretch in the x-direction by a factor of \(\frac{1}{3}\).

(c) Given \(\mathbf{MN} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}\), we find \(\mathbf{N}\) by multiplying by \(\mathbf{M}^{-1}\) on the left: \(\mathbf{N} = \mathbf{M}^{-1} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{3} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 1 & 2 \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ 3x + 2y \end{pmatrix}\). Setting \(y = mx\), we have \(3x + 2mx = m(x + 2y)\). This leads to the quadratic \(2m^2 - m - 3 = 0\), which factors to \((2m + 3)(m - 1) = 0\). Thus, \(m = \frac{3}{2}\) or \(m = -1\), giving the lines \(y = \frac{3}{2}x\) and \(y = -x\).

(a) The transformation matrix for a stretch parallel to the x-axis with scale factor k is \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\). The shear transformation with the x-axis fixed and (0, 1) mapped to (k, 1) is \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). The combined transformation matrix M is the product:

\(\mathbf{M} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix}\)

(b) The transformation \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx + ky \\ y \end{pmatrix}\) implies invariant points satisfy \(kx + ky = x\). Solving gives \(y = \frac{1-k}{k}x\).

(c) The inverse matrix \(\mathbf{M}^{-1}\) is found by inverting \(\mathbf{M}\):

\(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) The area of P is given by the determinant of M, \(|k| = 3k^2\). Solving \(|k| = 3k^2\) gives \(k = \pm \frac{1}{3}\).

(a) To find the equation of \(\Pi_1\), we need a normal vector to the plane. Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), the normal vector can be found using the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -4 & 2 \end{vmatrix} = \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\)

Substitute the point \((1, 3, -2)\) from \(l_1\) into the plane equation \(6x - 3y - 9z = d\) to find \(d\):

\(6(1) - 3(3) - 9(-2) = d\)

\(6 - 9 + 18 = d\)

\(d = 15\)

Thus, the equation of \(\Pi_1\) is \(2x - y - 3z = 5\).

(b) The normal to \(\Pi_2\) is the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\). The angle \(\theta\) between \(\Pi_1\) and \(\Pi_2\) is given by:

\(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}\)

\(\cos \theta = \frac{\begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}}{\sqrt{6^2 + (-3)^2 + (-9)^2} \sqrt{1^2 + (-4)^2 + 2^2}}\)

\(\cos \theta = \frac{-7}{\sqrt{14 \times 77}}\)

\(\theta = 77.7^\circ\)

(c) To find \(\mathbf{PQ}\), use the points \(\mathbf{P}\) and \(\mathbf{Q}\) on \(l_1\) and \(l_2\) respectively:

\(\mathbf{OP} = \begin{pmatrix} 1 + 2\lambda \\ 3 + \lambda \\ -2 + \lambda \end{pmatrix}\)

\(\mathbf{OQ} = \begin{pmatrix} 1 + \mu \\ -2 - 4\mu \\ 9 + 2\mu \end{pmatrix}\)

\(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix}\)

Set \(\mathbf{PQ}\) perpendicular to both direction vectors:

\(\begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = 0\)

\(\begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = 0\)

Solve these equations to find \(\lambda\) and \(\mu\), then substitute back to find \(\mathbf{PQ}\):

\(\mathbf{r} = \begin{pmatrix} -1 \\ 6 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\)

(a) To find the equation of the plane \(\Pi\), we need a normal vector to the plane. The direction vector of \(l_1\) is \(\mathbf{i} - \mathbf{j} - 4\mathbf{k}\), and the plane is parallel to \(2\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}\). The normal vector is the cross product of these two vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \mathbf{i}(16 - 20) - \mathbf{j}(-4 - (-8)) + \mathbf{k}(5 + 2) = -4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\)

Using the point \((1, 3, -1)\) on \(l_1\), the equation of the plane is:

\(24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5\).

(b) To find the acute angle between \(l_2\) and \(\Pi\), we use the dot product of the direction vector of \(l_2\), \(5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\), and the normal vector of \(\Pi\), \(24\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\):

\((24, -4, 7) \cdot (5, -5, -2) = 24 \times 5 + (-4) \times (-5) + 7 \times (-2) = 120 + 20 - 14 = 126\).

The magnitudes are \(\sqrt{24^2 + (-4)^2 + 7^2} = \sqrt{641}\) and \(\sqrt{5^2 + (-5)^2 + (-2)^2} = \sqrt{54}\).

\(\cos \alpha = \frac{126}{\sqrt{641} \times \sqrt{54}}\).

The angle \(\alpha\) is the angle between the line and the normal, so the acute angle between \(l_2\) and \(\Pi\) is \(90^\circ - \alpha = 42.6^\circ\).

(a) To find \(CAB\), first calculate \(AB\):

\(AB = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix}\)

Then calculate \(CAB\):

\(CAB = \begin{pmatrix} -2 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\):

\(-9x + 3mx = m(3x - 7y)\)

\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)

\(m = \pm \frac{3}{7}\)

Thus, \(y = \frac{3}{7}x\) and \(y = -\frac{3}{7}x\).

(c) The matrix \(M\) represents a stretch parallel to the x-axis with scale factor 3.

(d) Find \(N\) such that \(NM = CAB\):

\(M^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\)

\(N = CAB \cdot M^{-1} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -7 \\ -3 & 3 \end{pmatrix}\)

(a) The transformation is a stretch parallel to the x-axis with scale factor k, represented by \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\), followed by a shear with (0, 1) mapped to (k, 1), represented by \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). The combined transformation matrix M is:

\(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix}\)

(b) The line of invariant points satisfies \(\begin{pmatrix} k & k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This gives:

\(kx + ky = x\) and \(y = y\)

Solving \(kx + ky = x\) gives \(y = \frac{1-k}{k}x\).

(c) The inverse of M is:

\(\text{If } \text{det}(\textbf{M}) = k, \text{ then } \textbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & -k \\ 0 & k \end{pmatrix}\)

(d) Given \(|k| = 3k^2\), solving gives \(k = \pm \frac{1}{3}\).

(a) To find the equation of \(\Pi_1\), we need a normal vector to the plane. Since \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\), is parallel to \(\Pi_1\). The direction vector of \(l_1\) is \(\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\). The normal to \(\Pi_1\) is the cross product of these two vectors:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 1 & -4 & 2 \end{vmatrix} = \begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix}\)

Substitute the point on \(l_1\), \((1, 3, -2)\), into the plane equation \(6x - 3y + 3z = d\) to find \(d\):

\(6(1) - 3(3) + 3(-2) = -5\)

Thus, the equation of \(\Pi_1\) is \(2x - y - 3z = 5\).

(b) The normal to \(\Pi_2\) is the direction vector of \(l_2\), \(\begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}\). The angle \(\theta\) between \(\Pi_1\) and \(\Pi_2\) is given by:

\(\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|} = \frac{\begin{pmatrix} 6 \\ -3 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix}}{\sqrt{14} \sqrt{77}} = \frac{-7}{\sqrt{14 \times 77}}\)

\(\theta = 77.7^\circ\).

(c) To find \(\mathbf{PQ}\), we need points \(P\) and \(Q\) such that \(\mathbf{PQ}\) is perpendicular to both lines. Let \(\mathbf{OP} = \begin{pmatrix} 1 + 2\lambda \\ 3 + \lambda \\ -2 + \lambda \end{pmatrix}\) and \(\mathbf{OQ} = \begin{pmatrix} 1 + \mu \\ -2 - 4\mu \\ 9 + 2\mu \end{pmatrix}\).

\(\mathbf{PQ} = \mathbf{OQ} - \mathbf{OP} = \begin{pmatrix} \mu - 2\lambda \\ -5 - 4\mu - \lambda \\ 11 + 2\mu - \lambda \end{pmatrix}\).

Set \(\mathbf{PQ} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = 0\) and \(\mathbf{PQ} \cdot \begin{pmatrix} 1 \\ -4 \\ 2 \end{pmatrix} = 0\) to find \(\lambda\) and \(\mu\):

\(-6\lambda + 6 = 0 \rightarrow \lambda = 1\)

\(21\mu + 42 = 0 \rightarrow \mu = -2\)

Substitute back to find \(\mathbf{PQ}\):

\(\mathbf{r} = \begin{pmatrix} -1 \\ 6 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 6 \\ -3 \\ -9 \end{pmatrix}\).

(a) The matrix \(\mathbf{M}\) represents a stretch followed by a rotation. The stretch is parallel to the x-axis with a scale factor of 14. The rotation is \(\frac{\pi}{3}\) anticlockwise about the origin.

(b) \(\mathbf{M}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\)

(c) The equations of the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{\sqrt{3}}{3}x\).

(d) Given \(28 = 14 \times |\triangle ABC|\), the area of triangle \(ABC\) is \(2 \text{ cm}^2\).

(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

Simplifying gives \(-5x + 5y + 2z = 8\).

(b) The perpendicular distance from point D to the plane is given by:

\(\frac{|8 - (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (-5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} = 0.953\)

(c) The direction vector \(\overrightarrow{CD}\) is:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

Using the formula for the shortest distance between two skew lines:

\(\frac{|\overrightarrow{AB} \cdot (\overrightarrow{CD} \times \overrightarrow{AC})|}{|\overrightarrow{CD} \times \overrightarrow{AC}|} = \frac{35}{\sqrt{1049}} = 1.08\)

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a stretch and a rotation. The first matrix \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis with a scale factor of 14. The second matrix \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\) represents a rotation by \(\frac{\pi}{3}\) anticlockwise about the origin.

(b) To find \(\mathbf{M}^{-1}\), we take the inverse of each matrix in the product: \(\begin{pmatrix} 14 & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix}\) and \(\begin{pmatrix} \frac{1}{2} & -\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}^{-1} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\).

(c) The invariant lines are found by solving \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives the equations \(y = 2\sqrt{3}x\) and \(y = \frac{2}{3}\sqrt{3}x\).

(d) The area of triangle \(ABC\) is found using the determinant of \(\mathbf{M}\). Given \(|\mathbf{DEF}| = |\det \mathbf{M}| \times |\mathbf{ABC}|\), we have \(28 = 14 \times |\mathbf{ABC}|\), so \(|\mathbf{ABC}| = 2\) cm².

(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = (-5(2) - 5(-5))\mathbf{i} + (0(-5) - (-5)(-5))\mathbf{j} + (0(-5) - (-5)(2))\mathbf{k}\)

\(= -10\mathbf{i} + 25\mathbf{j} + 10\mathbf{k}\)

\(= -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

\(-5x + 5y + 2z = 8\)

(b) The perpendicular distance from point \(D(2, 1, 3)\) to the plane is given by:

\(\frac{|(-5)(2) + 5(1) + 2(3) - 8|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} \approx 0.953\)

(c) The shortest distance between the lines AB and CD is found using the vector \(\overrightarrow{CD}\) and the cross product:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = -3\mathbf{j} + 4\mathbf{k}\)

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ 5 & 4 & -1 \end{vmatrix} = (18)\mathbf{i} + (-25)\mathbf{j} + (-10)\mathbf{k}\)

The shortest distance is:

\(\frac{|\overrightarrow{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{35}{\sqrt{1049}} \approx 1.08\)

(a) The matrix \(\mathbf{M} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch followed by a shear. The first matrix \(\begin{pmatrix} 7 & 0 \\ 0 & 1 \end{pmatrix}\) is a stretch parallel to the \(x\)-axis with a scale factor of 7. The second matrix \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) is a shear with the \(x\)-axis fixed, mapping \((0,1)\) to \((2,1)\).

(b) To find the invariant lines, consider \(\mathbf{M} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7x + 2y \\ y \end{pmatrix}\). For a line \(y = mx\) to be invariant, \(mx = m(7x + 2mx)\). This simplifies to \(2m^2 + 6m = 0\), giving solutions \(m = 0\) and \(m = -3\). Thus, the invariant lines are \(y = 0\) and \(y = -3x\).

(c) The area of \(PQR\) is \(35 \text{ cm}^2\). The area of \(DEF\) is related by the determinant of the transformation matrix \(\mathbf{M}\), which is \(|7| = 7\). Therefore, \(\text{Area of } DEF = \frac{35}{7} = 5 \text{ cm}^2\).

(a) The direction vectors of \(l_1\) and \(l_2\) are \(\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) respectively. The vector between points on \(l_1\) and \(l_2\) is \(\begin{pmatrix} -4 \\ 0 \\ 1 \end{pmatrix}\). The cross product of the direction vectors is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). The shortest distance is given by \(\frac{1}{\sqrt{30}} \begin{vmatrix} -4 & 0 & 1 \\ -2 & -2 & -1 \end{vmatrix} = \frac{21}{\sqrt{30}}\).

(b) The normal to \(\Pi_1\) is the cross product of the direction vector of \(l_1\) and the direction vector of \(l_2\), which is \(\begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). Using the point \((1, 4, -1)\) on \(l_1\), the equation of \(\Pi_1\) is \(5x - 2y - z = -2\).

(c) The direction vector of the line of intersection is \(\begin{pmatrix} 1 \\ 4 \\ -3 \end{pmatrix}\). The normal to \(\Pi_2\) is \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\). Using the point \((1, 1, 1)\), the equation of \(\Pi_2\) is \(x + 2y + 3z = 6\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vector from one line to another: \(\begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ -3 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

Next, find the common perpendicular by taking the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 18 \\ 14 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

Use the formula for shortest distance:

\(\frac{1}{\sqrt{139}} \left| \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix} \right| = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) To find the equation of the plane \(\Pi\), find a vector perpendicular to the plane using the cross product of the direction vector of \(l_1\) and the vector from the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) to a point on \(l_1\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -3 \\ -1 \end{pmatrix} \sim \begin{pmatrix} 9 \\ -3 \\ -1 \end{pmatrix}\).

Use the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) in the plane equation:

\(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0\)

Solving gives \(k = 2\).

Calculate \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 & 1 \\ 1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix}\)

\(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

\(-9x + 3mx = m(3x - 7mx)\)

\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)

\(m = \pm \frac{3}{\sqrt{7}}\)

Invariant lines: \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF}\), set \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}\), \(\mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}\), \(\mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}\)

\(\alpha = 3, \beta = \frac{7}{9}\)

\(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\)

(a) The matrix \(\mathbf{M}\) is the product of two matrices: \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\) represents a shear parallel to the x-axis, and \(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\) represents a stretch parallel to the x-axis by a factor of \(k\). The shear is applied first, followed by the stretch.

(b) The area of the parallelogram \(OPQR\) is given by the absolute value of the determinant of \(\mathbf{M}\). Calculating the determinant: \(\det(\mathbf{M}) = \det\left(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\right) = k\). Thus, \(|OPQR| = |k|\).

The inverse matrix \(\mathbf{M}^{-1}\) is calculated as follows: \(\mathbf{M}^{-1} = \frac{1}{k} \begin{pmatrix} 1 & 0 \\ -1 & k \end{pmatrix}\).

(c) To show the line is invariant, consider the transformation of a point \(\begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\) under \(\mathbf{M}\):

\(\begin{pmatrix} k & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ x + \frac{1}{k-1}x \end{pmatrix} = \begin{pmatrix} kx \\ \frac{k}{k-1}x \end{pmatrix} = k \begin{pmatrix} x \\ \frac{1}{k-1}x \end{pmatrix}\).

This shows that the line \(y = \frac{1}{k-1}x\) is invariant under the transformation.

(a) The normal vector to \(\Pi_1\) can be found using the cross product of the direction vectors \(\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\) and \(3\mathbf{i} - \mathbf{k}\). This gives \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -3 \\ 3 & 0 & -1 \end{vmatrix} = \mathbf{i}(2 - 0) - \mathbf{j}(-1 - (-9)) + \mathbf{k}(0 - (-6)) = 2\mathbf{i} - 8\mathbf{j} + 6\mathbf{k}\).

Using the point \((1, -1, -2)\), the equation is \(1(x - 1) - 4(y + 1) + 3(z + 2) = 0\), simplifying to \(x - 4y + 3z = -1\).

(b) The direction vector of \(l\) is \(\mathbf{i} + \mathbf{j} + \mathbf{k}\). The dot product with the normal vector \(1\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}\) is \(1(1) - 4(1) + 3(1) = 0\), showing \(l\) is parallel to \(\Pi_1\).

(c) The distance from a point on \(l\) to \(\Pi_1\) is \(\frac{|(-3)(1) + 0(-4) + 1(3) + 1|}{\sqrt{1^2 + (-4)^2 + 3^2}} = \frac{1}{\sqrt{26}}\).

(d) A point common to both planes is found by solving \(x - 4y + 3z = -1\) and \(3x + 3y + 2z = 1\). Solving these gives a point \(\begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix}\). The direction vector of the line of intersection is the cross product of the normals \(\begin{pmatrix} 1 \\ -4 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}\), which is \(\begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\). Thus, the line is \(\mathbf{r} = \begin{pmatrix} \frac{5}{7} \\ 0 \\ -\frac{4}{7} \end{pmatrix} + \lambda \begin{pmatrix} -17 \\ 7 \\ 15 \end{pmatrix}\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vectors of the lines: \(\mathbf{d_1} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix}\) and \(\mathbf{d_2} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\).

The vector between a point on \(l_1\) and a point on \(l_2\) is \(\mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

The cross product \(\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

The shortest distance is given by \(\frac{|\mathbf{b} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) The plane \(\Pi\) contains \(l_1\) and the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\). The normal vector to the plane is \(\mathbf{n} = \mathbf{d_1} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix}\).

The equation of the plane is \(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0.\)

This simplifies to \(-1 - k + 3 = 0 \Rightarrow k = 2.\)

Substitute \(k = 2\) into \(\mathbf{A}\) and compute \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 \\ 1 & 1 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \\ 1 & 5 & 12 \end{pmatrix}.\)

Then, \(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}.\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}.\)

This gives \(-9x + 3mx = m(3x - 7mx)\) and \(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9.\)

Thus, \(m = \pm \frac{3}{\sqrt{7}}\), giving lines \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Assume \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF},\) solve:

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}.\)

Equating gives \(\alpha = 3\) and \(\beta = \frac{7}{9}.\)

Thus, \(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

(a) For a matrix to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we equate \(b^2 = -c^2\), which is impossible for real \(b\) and \(c\) with \(b \neq 0\).

(b) Consider the transformation \(\begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines, \(y = mx\) and \(Y = mX\). Solving \(c^2x + amx = m(ax + b^2mx)\) gives \(c^2 + am = ma + b^2m^2 \Rightarrow c^2 = b^2m^2\). Thus, \(y = \frac{c}{b}x\) and \(y = -\frac{c}{b}x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). The combined transformation is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 25 \\ 0 & 5 \end{pmatrix}\).

(d) The area of triangle PQR is \(12 \times \det(\mathbf{M}) = 12 \times 25 = 300 \text{ cm}^2\).

(a) The equation of the plane \(\Pi_1\) can be found by setting up the system of equations from the given vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\)

Solving gives \(y + z = -7\).

(b) To find the line of intersection, find a point common to both planes, e.g., \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\), and the direction vector by solving:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -5 & 3 & 5 \end{vmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\)

The vector equation is \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) The perpendicular distance from \(A\) to \(\Pi_1\) is given by:

\(\frac{|a + a - 7 + 7|}{\sqrt{2}} = \sqrt{2}\)

Solving gives \(a = 1\).

(d) The obtuse angle condition gives:

\(\frac{|b + b|}{\sqrt{2((1-b)^2 + 2b^2)}} = \frac{1}{2}\sqrt{2}\)

Solving \(2b = \sqrt{(1-b)^2 + 2b^2}\) gives \(b = -1 + \sqrt{2}\).

(a) For \(\mathbf{M}\) to represent a rotation, it must be of the form \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\). Comparing with \(\mathbf{M} = \begin{pmatrix} a & b^2 \\ c^2 & a \end{pmatrix}\), we have \(b^2 = -c^2\), which is impossible since \(b\) and \(c\) are real and \(b \neq 0\).

(b) The transformation of \(\begin{pmatrix} x \\ y \end{pmatrix}\) by \(\mathbf{M}\) gives \(\begin{pmatrix} ax + b^2y \\ c^2x + ay \end{pmatrix}\). For invariant lines through the origin, \(y = mx\) gives \(c^2x + amx = m(ax + b^2y)\). Simplifying, \(c^2 + am = ma + b^2m^2\) leads to \(c^2 = b^2m^2\), giving \(y = \frac{c}{b} x\) and \(y = -\frac{c}{b} x\).

(c) The enlargement matrix is \(\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}\) and the shear matrix is \(\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\). Thus, \(\mathbf{M} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 5 & 5^2 \\ 0 & 5 \end{pmatrix}\).

(d) The area of \(PQR\) is \(12 \times \det(\mathbf{M})\). \(\det(\mathbf{M}) = 5 \times 5 - 0 \times 25 = 25\). Therefore, the area is \(12 \times 25 = 300 \text{ cm}^2\).

(a) The direction vectors of \(\Pi_1\) are \(\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\). The normal vector is the cross product:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}\)

Substitute \(\mathbf{r} = \begin{pmatrix} 0 \\ -4 \\ -3 \end{pmatrix}\) into \(\mathbf{n} \cdot \mathbf{r} = d\):

\(0 \cdot 0 + 2(-4) + 2(-3) = d\)

\(d = -8 - 6 = -14\)

Equation: \(0x + 2y + 2z = -14\) simplifies to \(y + z = -7\).

(b) A point on both planes is \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\). The direction vector is the cross product of normals:

\(\mathbf{n_1} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}, \mathbf{n_2} = \begin{pmatrix} -5 \\ 3 \\ 5 \end{pmatrix}\)

\(\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 2 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\)

Vector equation: \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\).

(c) Distance from \(A\) to \(\Pi_1\):

\(\frac{|a + a - 7 + 7|}{\sqrt{2}} = \sqrt{2}\)

\(|2a| = 2 \Rightarrow a = 1\)

(d) Angle between \(l\) and \(\Pi_1\):

\(\frac{|b + b|}{\sqrt{2((1-b)^2 + 2b^2)}} = \frac{1}{2} \sqrt{2}\)

\(2b = \sqrt{(1-b)^2 + 2b^2} \Rightarrow b^2 + 2b - 1 = 0\)

\(b = -1 + \sqrt{2}\)

(a) The direction vectors of the lines in the plane are:

\(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}, \ \overrightarrow{AC} = \begin{pmatrix} -3 \\ 0 \\ 2 \end{pmatrix}\)

The normal vector to the plane is found using the cross product:

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 4 \\ -3 & 0 & 2 \end{vmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}\)

The equation of the plane is \(x - 2y + z = d\). Substituting point \(A(1, 1, 0)\) gives \(d = -1\).

(b) The perpendicular distance from \(O\) to the plane is:

\(\frac{1}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{1}{\sqrt{6}} \approx 0.408\)

(c) Let \(\overrightarrow{OP} = \begin{pmatrix} -2\lambda \\ \lambda \\ 3\lambda \end{pmatrix}\) and \(\overrightarrow{OQ} = \begin{pmatrix} 1 - 2\mu \\ 1 + \mu \\ 4\mu \end{pmatrix}\).

\(\overrightarrow{PQ} = \begin{pmatrix} 1 + \mu - \lambda \\ 4\mu - 3\lambda \end{pmatrix}\).

Using the dot product with the line direction \(\begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}\):

\((1 - 2\mu + 2\lambda) \cdot (-2) + (1 + \mu - \lambda) \cdot 1 + (4\mu - 3\lambda) \cdot 3 = 0\)

Solving gives \(\lambda = \frac{4}{5}\).

The vector equation is \(\mathbf{r} = \frac{-4}{5} \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}\).

(a) To find the equation of the plane, we first determine the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (-4 - 4)\mathbf{i} + (3 + 4)\mathbf{j} + (-4 - 1)\mathbf{k} = -8\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (4 - 4)\mathbf{i} + (-1 + 4)\mathbf{j} + (-2 - 1)\mathbf{k} = 0\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 7 & -5 \\ 0 & 3 & -3 \end{vmatrix} = \mathbf{i}(7(-3) - (-5)(3)) - \mathbf{j}(-8(-3) - (-5)(0)) + \mathbf{k}(-8(3) - 7(0))\)

\(= \mathbf{i}(-21 + 15) - \mathbf{j}(24) + \mathbf{k}(-24) = -6\mathbf{i} - 24\mathbf{j} - 24\mathbf{k}\)

Normalizing gives \(\mathbf{n} = \mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\).

Using point \(A(4, -4, 1)\), the equation is:

\(x + 4y + 4z = -8\).

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|0 + 4(0) + 4(0) + 8|}{\sqrt{1^2 + 4^2 + 4^2}} = \frac{8}{\sqrt{33}} \approx 1.39\).

(c) The line \(OD\) has the equation \(\mathbf{r} = t(2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\).

Substitute into the plane equation:

\(2t + 12t - 12t = -8\)

\(2t = -8\)

\(t = -4\)

Coordinates of intersection: \((-8, -12, 12)\).

(a) To find when \(A\) is non-singular, set \(\det(A) \neq 0\). Calculate \(\det(A) = 1(k \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - k \cdot 7)\).

Simplifying, \(\det(A) = -12k + 60\). Set \(-12k + 60 \neq 0\), giving \(k \neq 5\).

(b) The inverse of \(A\) is \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\). The top row of \(\text{adj}(A)\) is \(\begin{pmatrix} k \cdot 9 - 6 \cdot 8 & -(4 \cdot 9 - 6 \cdot 7) & 4 \cdot 8 - k \cdot 7 \end{pmatrix}\).

Thus, the top row of \(A^{-1}\) is \(\begin{pmatrix} \frac{1}{60 - 12k} & \frac{6}{60 - 12k} & \frac{3k - 16}{4(5-k)} \end{pmatrix}\).

(c) Calculate \(BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \\ 7 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 4 & k & 6 \end{pmatrix}\).

Set \(C = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}\) such that \(BAC = \begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\).

Choose \(C = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\).

(d) The transformation matrix is \(\begin{pmatrix} 2 & 1 \\ k & 4 \end{pmatrix}\). The characteristic equation is \(\lambda^2 - 6\lambda + (8-k) = 0\).

For two distinct invariant lines, the discriminant must be positive: \(6^2 - 4(8-k) > 0\).

Simplifying, \(36 - 32 + 4k > 0\) gives \(4k > -4\), so \(k > 1\). Also, \(k \neq 8\) to avoid repeated roots.

(a) The direction vector of line AB is \(\mathbf{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\).

The direction vector of line CD is \(\mathbf{CD} = \mathbf{j} + (\lambda - 3)\mathbf{k}\).

Find the common perpendicular using the cross product and the formula for perpendicular distance:

\(\frac{1}{\sqrt{(\lambda - 2)^2 + (4\lambda - 12)^2 + 16}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda - 3 \end{vmatrix} = 3\)

Solving gives \(\lambda^2 - 5\lambda + 4 = 0\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 7 \end{vmatrix} = -8\mathbf{i} - 25\mathbf{j} + 7\mathbf{k}\)

Substitute a point to find \(-8x - 25y + 7z = -163\).

(c) The angle between \(\Pi_1\) and \(\Pi_2\) is found using the dot product of normals:

\(\cos \theta = \frac{414}{\sqrt{243 \times 738}}\)

\(\theta = 12.1^\circ\).

(a) To find the Cartesian equation of the plane \(\Pi\), we first find a normal vector to the plane by taking the cross product of the direction vectors of the lines:

Direction vectors: \(\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\).

Cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 1 \\ 3 & 2 & -1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ -8 \end{pmatrix}\).

Normal vector: \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Using point \((3, -2, 1)\) on the plane, substitute into \(2x - y + 4z = d\):

\(2(3) - (-2) + 4(1) = 12\).

Thus, the equation is \(2x - y + 4z = 12\).

(b) The line \(l\) is parallel to \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). The normal to the plane is \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Dot product: \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \sqrt{2} \sqrt{21} \cos \alpha\).

\(\cos \alpha = \frac{3}{\sqrt{42}}\).

Acute angle: \(90^\circ - \alpha = 27.6^\circ\).

(c) The position vector of the foot of the perpendicular \(\overrightarrow{OF}\) is given by:

\(\overrightarrow{OF} = \overrightarrow{OP} + t \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 + 2t \\ 3 - t \\ 1 + 4t \end{pmatrix}\).

Substitute into the plane equation: \(2(2 + 2t) - (3 - t) + 4(1 + 4t) = 12\).

Solve for \(t\): \(2t + 5 = 12\), \(t = \frac{7}{2}\).

\(\overrightarrow{OF} = \frac{1}{3} \begin{pmatrix} 8 \\ 8 \\ 7 \end{pmatrix}\).

(a) The matrix \(M\) is a product of two matrices: \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\) represents a shear in the x-direction, and \(\begin{pmatrix} \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{2} \end{pmatrix}\) represents a rotation anticlockwise about the origin through 45°.

(b) To find \(M^{-1}\), we calculate the inverse of the product of the two matrices: \(M^{-1} = \begin{pmatrix} \frac{k+1}{\sqrt{2}} & \frac{k-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}\).

(c) For no invariant lines through the origin, the characteristic equation of \(M\) must have no real roots. The characteristic equation is derived from \(\det(M - \lambda I) = 0\). Solving gives \(k^2 + 4k - 4 < 0\), leading to the solution \(2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})\).

(a) The matrix \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\) represents a shear in the y-direction. The matrix \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\) represents a stretch parallel to the y-axis with scale factor \(k\). The transformations are applied in the order of shear followed by stretch.

(b) To find \(\mathbf{M}^{-1}\), we first find the inverse of each matrix: \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\). Thus, \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & 0 \\ -k & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{k} \end{pmatrix}\).

(c) The transformation \(\mathbf{M} = \begin{pmatrix} 1 & 0 \\ k^2 & k \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\). Setting \(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ k^2x + ky \end{pmatrix}\), we get \(k^2x + ky = y\), which simplifies to \(y = \frac{k^2}{1-k}x\).

(d) The area of triangle DEF is equal to the area of triangle ABC when \(|\det(\mathbf{M})| = 1\). The determinant \(\det(\mathbf{M}) = 1 \times k - 0 \times k^2 = k\). Thus, \(|k| = 1\), giving \(k = -1\).

(a) To find the length \(PQ\), we first find a vector joining any point on \(l_1\) to any point on \(l_2\). The vector is \(\begin{pmatrix} 0 \\ 2 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix}\).

Next, find the common perpendicular using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix}\).

The length \(PQ\) is given by \(\frac{1}{\sqrt{29}} \begin{pmatrix} -2 \\ 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ -3 \end{pmatrix} = \frac{27}{\sqrt{29}} = 5.01\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 2\mathbf{i} + \mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(2\mathbf{i} - 4\mathbf{j} - 3\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ 2 & -4 & -3 \end{vmatrix} = \begin{pmatrix} -14 \\ -1 \\ -8 \end{pmatrix}\).

Substitute a point to find the equation: \(14(0) + (2) + 8(6) = 50 \Rightarrow 14x + y + 8z = 50\).

(c) The normals to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 11 \\ 7 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} 14 \\ 1 \\ 8 \end{pmatrix}\) respectively.

Use the dot product to find the angle: \(\cos \theta = \frac{145}{\sqrt{174} \sqrt{261}}\).

The acute angle is \(47.1^\circ\).

(a) The rotation matrix for 60° anticlockwise is \(\begin{pmatrix} \cos 60^{\circ} & -\sin 60^{\circ} \\ \sin 60^{\circ} & \cos 60^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

The stretch matrix is \(\begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix}\).

Thus, \(\textbf{M} = \begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{d}{2} & -\frac{d\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

(b) The area of the parallelogram is given by the determinant of \(\textbf{M}\), which is \(\frac{1}{2}d^2\). Therefore, \(\frac{1}{2}d^2 = \frac{1}{2} \Rightarrow d^2 = 1 \Rightarrow d = 2\) since \(d \neq 0\).

(c) To find \(\textbf{N}\), we use \(\textbf{MN} = \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\).

First, find \(\textbf{M}^{-1}\):

\(\textbf{M}^{-1} = \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & 1 \end{pmatrix}\).

Then, \(\textbf{N} = \textbf{M}^{-1} \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 1 + \sqrt{3} & 1 + \sqrt{3} \\ 1 - \sqrt{3} & 1 - \sqrt{3} \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

This gives \(\frac{1}{2}x + \frac{1}{2}y = mx\) and \(\frac{1}{2}x + \frac{1}{2}y = my\).

Solving, we find \(y = \frac{1}{2}x\) and \(y = -x\).

(a) Let line \(l_1\) pass through \(\mathbf{a} = (t,1,0)\) with direction \(\mathbf{b} = (-2,-1,0)\), and line \(l_2\) pass through \(\mathbf{c} = (0,1,t)\) with direction \(\mathbf{d} = (0,-2,1)\).

The shortest distance between two skew lines is \[ \text{distance} = \frac{\big|(\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a})\big|} {\big|\mathbf{b} \times \mathbf{d}\big|}. \] Here \(\mathbf{b} \times \mathbf{d} = (-1,\,2,\,4)\), so \(\big|\mathbf{b} \times \mathbf{d}\big| = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}.\)

Also \(\mathbf{c} - \mathbf{a} = (0-t,\,1-1,\,t-0) = (-t,\,0,\,t)\), so \[ (\mathbf{b} \times \mathbf{d}) \cdot (\mathbf{c}-\mathbf{a}) = (-1,2,4)\cdot(-t,0,t) = t + 4t = 5t. \] Therefore \[ \text{distance} = \frac{|5t|}{\sqrt{21}}. \]

Given the shortest distance is \(\sqrt{21}\), \[ \frac{|5t|}{\sqrt{21}} = \sqrt{21} \;\Rightarrow\; |5t| = 21 \;\Rightarrow\; |t| = \frac{21}{5}. \] Since \(t\) is positive, \(t = \dfrac{21}{5}.\)

(b) The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\), so it contains the direction vectors of both lines: \((-2,-1,0)\) and \((0,-2,1)\), and passes through \((t,1,0) = \left(\dfrac{21}{5},1,0\right)\).

Hence an equation of \(\Pi_1\) is \[ \mathbf{r} = \frac{21}{5}\,\mathbf{i} + \mathbf{j} + \lambda(-2\mathbf{i} - \mathbf{j}) + \mu(-2\mathbf{j} + \mathbf{k}), \] where \(\lambda,\mu \in \mathbb{R}.\)

(c) The direction vector of \(l_2\) is \(\mathbf{d} = (0,-2,1)\) and a normal to \(\Pi_2\colon 5x - 6y + 7z = 0\) is \(\mathbf{n} = (5,-6,7)\).

The angle \(\alpha\) between a line and a plane satisfies \[ \sin\alpha = \frac{|\,\mathbf{d}\cdot\mathbf{n}\,|}{|\mathbf{d}|\,|\mathbf{n}|}. \] Here \(\mathbf{d}\cdot\mathbf{n} = 0\cdot5 + (-2)(-6) + 1\cdot7 = 19,\) \(|\mathbf{d}| = \sqrt{0^2+(-2)^2+1^2} = \sqrt{5},\) \(|\mathbf{n}| = \sqrt{5^2+(-6)^2+7^2} = \sqrt{110}.\)

So \[ \sin\alpha = \frac{19}{\sqrt{5}\,\sqrt{110}} = \frac{19}{\sqrt{550}}, \quad \alpha \approx 54.1^\circ. \]

(d) The acute angle between the planes \(\Pi_1\) and \(\Pi_2\) is the angle between their normals.

A normal to \(\Pi_1\) is \(\mathbf{n}_1 = \mathbf{b}\times\mathbf{d} = (-1,2,4)\), and a normal to \(\Pi_2\) is \(\mathbf{n}_2 = (5,-6,7)\).

Then \[ \cos\phi = \frac{|\,\mathbf{n}_1\cdot\mathbf{n}_2\,|} {|\mathbf{n}_1|\,|\mathbf{n}_2|} = \frac{|(-1,2,4)\cdot(5,-6,7)|} {\sqrt{21}\,\sqrt{110}} = \frac{11}{\sqrt{21\cdot110}} = \frac{11}{\sqrt{2310}}, \] so \(\phi \approx 76.8^\circ.\)

(a) To find \(CAB\), first compute \(AB\):

\(AB = \begin{pmatrix} 2 & k & k \\ 5 & -1 & 3 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix}\)

Then compute \(CAB\):

\(CAB = \begin{pmatrix} 0 & 1 & 1 \\ -1 & 2 & 0 \end{pmatrix} \begin{pmatrix} 2+k & k \\ 8 & -1 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 10 & -1 \\ -k+14 & -k-2 \end{pmatrix}\)

(b) Since \(A\) is singular, its determinant is zero:

\(\det(A) = 2(-1 \cdot 1 - 3 \cdot 0) - k(5 \cdot 1 - 3 \cdot 1) + k(5 \cdot 0 - (-1) \cdot 1) = 0\)

\(-2 - 2k + k = 0\)

\(-2 - k = 0\)

\(k = -2\)

(c) Using \(k = -2\), the matrix \(CAB\) becomes:

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix}\)

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\):

\(\begin{pmatrix} 10 & -1 \\ 16 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10x - y \\ 16x \end{pmatrix}\)

Equating \(y = mx\) and \(Y = mX\):

\(10x - mx = X\)

\(16x = mX\)

\(16 = 10m - m^2\)

\(m^2 - 10m + 16 = 0\)

Solving gives \(m = 2\) and \(m = 8\), so the invariant lines are:

\(y = 2x\) and \(y = 8x\)

(a) The direction vector of \(l_1\) is \(2\mathbf{i} - 3\mathbf{j}\). The vector from \(P\) to a point on \(l_1\) is \(\mathbf{i} - 3\mathbf{k}\). Thus, the equation of \(\Pi_1\) is \(\mathbf{r} = -2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda (2\mathbf{i} - 3\mathbf{j}) + \mu (\mathbf{i} - 3\mathbf{k})\).

(b) The normal vector to \(\Pi_2\) is \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 0 \\ 3 & -1 & 3 \end{vmatrix} = -9\mathbf{i} - 6\mathbf{j} + 7\mathbf{k}\). Using the point \((3, 0, -2)\) on \(\Pi_2\), the equation is \(9x + 6y - 7z = 41\).

(c) The normal vectors to \(\Pi_1\) and \(\Pi_2\) are \(\begin{pmatrix} 9 \\ 6 \\ -7 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}\) respectively. The dot product is \(32\), and the magnitudes are \(\sqrt{166}\) and \(\sqrt{14}\). Thus, \(\cos \alpha = \frac{32}{\sqrt{14 \times 166}}\), giving \(\alpha = 48.4^\circ\) or \(0.845 \text{ rad}\).

(d) \(\overrightarrow{OQ} = -5\overrightarrow{OP} = \begin{pmatrix} 10 \\ 10 \\ -20 \end{pmatrix}\). The equation of \(\Pi_2\) is \(9x + 6y - 7z = 41\). Substituting \(\mathbf{OF} = \begin{pmatrix} 10 + 9t \\ 10 + 6t \\ -20 - 7t \end{pmatrix}\) into the plane equation gives \(t = -\frac{3}{2}\), leading to \(\mathbf{OF} = \left( -\frac{7}{2}, 1, -\frac{19}{2} \right)\).

(a) The normal vector to the plane \(\Pi\) is found using the cross product of the direction vectors \(\mathbf{i} + \mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j}\). The determinant is:

This gives the normal vector \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\).

Substitute the point \((-2, 3, 3)\) into the plane equation:

\(-3(-2) + 2(3) + 3(3) = 21\)

Thus, the Cartesian equation is \(-3x + 2y + 3z = 21\).

(b) The line \(l\) has the equation \(\begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}\). Substitute into the plane equation:

\(-3(2) + 2(-3) + 3(5 + t) = 21\)

This simplifies to \(t = 6\), giving the point \(\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}\).

(c) The direction vector of \(l\) is \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). The normal vector to \(\Pi\) is \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\). The dot product is:

\(0 \cdot (-3) + 0 \cdot 2 + 1 \cdot 3 = 3\)

The magnitudes are \(\sqrt{1}\) and \(\sqrt{22}\), so \(\cos \alpha = \frac{3}{\sqrt{22}}\).

The acute angle is \(90^\circ - \alpha = 39.8^\circ\).

(d) The direction vector from \(P\) to the plane is \(\begin{pmatrix} 0 \\ -3 \\ 7 \end{pmatrix}\). The dot product with the normal vector is:

\(\frac{1}{\sqrt{22}} \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84\).

(a) The matrix \(\begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\) represents an enlargement with a scale factor of 6.

(b) The determinant of \(\mathbf{A}\) is \(\det(\mathbf{A}) = 3 \times 2 - 4 \times 2 = 6 - 8 = -2\). The area of triangle PQR is \(2 \times 13 = 26 \text{ cm}^2\).

(c) To find \(\mathbf{B}\), we need \(\mathbf{AB} = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}\). First, find \(\mathbf{A}^{-1} = -\frac{1}{2} \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix}\). Then, \(\mathbf{B} = -3 \begin{pmatrix} 2 & -4 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -6 & 12 \\ 6 & -9 \end{pmatrix}\).

(d) For \(\mathbf{A} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\), we have \(\begin{pmatrix} 3x + 4y \\ 2x + 2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). This leads to the equations \(3x + 4y = x\) and \(2x + 2y = y\). Solving these gives \(y = -2x\) and \(-6x = 0\), leading to \(x = 0, y = 0\). Thus, the origin is the only invariant point.

(a) To find the equation of the plane OAB, we need a normal vector. The vectors \(\overrightarrow{OA} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\) are in the plane. The cross product \(\overrightarrow{OA} \times \overrightarrow{OB}\) gives the normal vector \(\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\). The equation of the plane is \(\mathbf{r} \cdot \mathbf{n} = 0\).

(b) The perpendicular distance from the origin to the plane \(\Pi\) is given by \(\frac{|1|}{\sqrt{1^2 + (-3)^2 + (-2)^2}} = \frac{1}{\sqrt{14}}\).

(c) The angle between the planes is found using the dot product of their normals. The normal to OAB is \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) and for \(\Pi\) is \(\begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}\). The cosine of the angle \(\alpha\) is \(\cos \alpha = \frac{\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -3 \\ -2 \end{pmatrix}}{\sqrt{3} \sqrt{14}} = \frac{6}{\sqrt{42}}\). The angle is \(22.2^\circ\).

(d) To find the common perpendicular, we first find the direction vector by solving the system of equations derived from the dot product conditions. The direction vector is \(\begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\). The position vector is \(\begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix}\). Thus, the equation is \(\mathbf{r} = \begin{pmatrix} -0.1 \\ 0.7 \\ 0.3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -3 \\ 1 \end{pmatrix}\).

Answer:

1. (a) The plane \(ABC\) is \(-2x-6y+7z=10\).
2. (b) The perpendicular distance from \(O\) to the plane is \(\dfrac{10}{\sqrt{89}}\).
3. (c) The acute angle between the planes \(OAB\) and \(ABC\) is \(36.2^\circ\) to 3 significant figures.

The position vectors give the coordinates \(A(-1,1,2)\), \(B(-2,-1,0)\) and \(C(2,0,2)\).

1. (a) Two direction vectors in the plane \(ABC\) are

   \(\overrightarrow{AB}=B-A=(-2,-1,0)-(-1,1,2)=(-1,-2,-2)\),

   \(\overrightarrow{AC}=C-A=(2,0,2)-(-1,1,2)=(3,-1,0)\).

   A normal vector to the plane is given by the cross product:

   \(\overrightarrow{AB}\times \overrightarrow{AC}=(-1,-2,-2)\times(3,-1,0)=(-2,-6,7)\).

   So the equation of the plane has the form \(-2x-6y+7z=d\). Substituting \(A(-1,1,2)\),

   \(-2(-1)-6(1)+7(2)=2-6+14=10\).

   Hence the equation of the plane is \(-2x-6y+7z=10\).

2. (b) The distance from the origin to the plane \(-2x-6y+7z=10\) is

   \(\dfrac{|0-10|}{\sqrt{(-2)^2+(-6)^2+7^2}}=\dfrac{10}{\sqrt{4+36+49}}=\dfrac{10}{\sqrt{89}}\).

3. (c) The plane \(OAB\) contains the vectors \(\overrightarrow{OA}=(-1,1,2)\) and \(\overrightarrow{OB}=(-2,-1,0)\). A normal vector to this plane is

   \(\overrightarrow{OA}\times \overrightarrow{OB}=(-1,1,2)\times(-2,-1,0)=(2,-4,3)\).

   A normal vector to the plane \(ABC\) is \((-2,-6,7)\). The angle between two planes is the angle between their normal vectors, taking the acute angle.

   Using the scalar product,

   \((-2,-6,7)\cdot(2,-4,3)=(-2)(2)+(-6)(-4)+7(3)=-4+24+21=41\).

   The magnitudes are \(|(-2,-6,7)|=\sqrt{89}\) and \(|(2,-4,3)|=\sqrt{29}\).

   Therefore \(\cos\theta=\dfrac{41}{\sqrt{89}\sqrt{29}}\), so

   \(\theta=\cos^{-1}\left(\dfrac{41}{\sqrt{89}\sqrt{29}}\right)=36.2^\circ\) to 3 significant figures.

Answer:

1. (a) The shortest distance between the lines is \(\displaystyle \frac{15}{\sqrt{77}}=\frac{15\sqrt{77}}{77}\), approximately \(1.71\).
2. (b) The plane is \(4x+3y-4z=0\).
3. (c) The acute angle between \(l_2\) and \(\Pi\) is \(\displaystyle \sin^{-1}\!\left(\frac{9}{\sqrt{2501}}\right)\), approximately \(10.4^\circ\).

Let \(A=(3,0,3)\) be a point on \(l_1\), with direction vector \(\mathbf d_1=(1,4,4)\). Let \(B=(3,-5,-6)\) be a point on \(l_2\), with direction vector \(\mathbf d_2=(0,5,6)\).

(a) For two skew lines, the shortest distance is

\(\displaystyle \frac{|\overrightarrow{AB}\cdot(\mathbf d_1\times\mathbf d_2)|}{|\mathbf d_1\times\mathbf d_2|}\).

Here \(\overrightarrow{AB}=B-A=(0,-5,-9)\), and

\(\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&4&4\\0&5&6\end{vmatrix}=(4,-6,5)\).

So

\(\overrightarrow{AB}\cdot(\mathbf d_1\times\mathbf d_2)=(0,-5,-9)\cdot(4,-6,5)=30-45=-15\),

and \(|\mathbf d_1\times\mathbf d_2|=\sqrt{4^2+(-6)^2+5^2}=\sqrt{77}\).

Therefore

\(\displaystyle \text{distance}=\frac{15}{\sqrt{77}}=\frac{15\sqrt{77}}{77}\approx 1.71\).

(b) The plane \(\Pi\) contains \(l_1\), so it contains the direction vector \((1,4,4)\). It is also parallel to \(\mathbf i+\mathbf k=(1,0,1)\).

A normal vector to the plane is

\(\mathbf n=(1,4,4)\times(1,0,1)=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&4&4\\1&0&1\end{vmatrix}=(4,3,-4)\).

Using the point \((3,0,3)\) on \(l_1\), the plane equation is

\(4(x-3)+3(y-0)-4(z-3)=0\).

Hence

\(4x+3y-4z=0\).

(c) The direction vector of \(l_2\) is \(\mathbf d_2=(0,5,6)\), and a normal to \(\Pi\) is \(\mathbf n=(4,3,-4)\).

The angle \(\alpha\) between \(\mathbf d_2\) and \(\mathbf n\) satisfies

\(\displaystyle \cos\alpha=\frac{\mathbf d_2\cdot\mathbf n}{|\mathbf d_2||\mathbf n|}=\frac{0\cdot4+5\cdot3+6(-4)}{\sqrt{61}\sqrt{41}}=-\frac{9}{\sqrt{2501}}\).

The acute angle \(\theta\) between the line and the plane is the complement of the acute angle between the line and the normal, so equivalently

\(\displaystyle \sin\theta=\frac{|\mathbf d_2\cdot\mathbf n|}{|\mathbf d_2||\mathbf n|}=\frac{9}{\sqrt{2501}}\).

Thus

\(\displaystyle \theta=\sin^{-1}\!\left(\frac{9}{\sqrt{2501}}\right)\approx 10.4^\circ\).

Answer: (i) The shortest distance between the lines is \(\sqrt6\).

(ii) The acute angle is \(\sin^{-1}\left(\dfrac{\sqrt{42}}{14}\right)\), approximately \(27.6^\circ\) or \(0.481\) radians.

For \(l_1\),

\(\overrightarrow{AB}=(-1+3,5-1,9-4)=(2,4,5)\).

For \(l_2\),

\(\overrightarrow{CD}=(-1+2,7-6,5-5)=(1,1,0)\).

(i) The direction of the common perpendicular is

\((2,4,5)\times(1,1,0)=(-5,5,-2)\).

Using \(\overrightarrow{AC}=(-2+3,6-1,5-4)=(1,5,1)\), the shortest distance is

\(\displaystyle \frac{|(1,5,1)\cdot(-5,5,-2)|}{\sqrt{(-5)^2+5^2+(-2)^2}}=\frac{18}{\sqrt{54}}=\sqrt6\).

(ii) The plane through \(A,B,D\) has direction vectors

\(\overrightarrow{AB}=(2,4,5)\), and \(\overrightarrow{AD}=(-1+3,7-1,5-4)=(2,6,1)\).

A normal vector to the plane is

\((2,4,5)\times(2,6,1)=(-26,8,4)\), so we can use \((-13,4,2)\).

The direction vector of \(l_2\) is \((1,1,0)\). If \(\theta\) is the acute angle between the line and the plane, then

\(\displaystyle \sin\theta=\frac{|(1,1,0)\cdot(-13,4,2)|}{\sqrt{1^2+1^2}\sqrt{(-13)^2+4^2+2^2}}\).

Thus

\(\displaystyle \sin\theta=\frac{9}{\sqrt2\sqrt{189}}=\frac{\sqrt{42}}{14}\).

Therefore

\(\displaystyle \theta=\sin^{-1}\left(\frac{\sqrt{42}}{14}\right)\approx27.6^\circ\).

Answer: (i) The shortest distance between the lines is \(\frac{6}{\sqrt{19}}\).

(ii) The required plane is \(19x+18y+13z=0\).

Let the position vectors of the points be

\(A=(1,-1,0),\quad B=(2,1,7),\quad C=(1,-1,1).\)

The line \(OC\) has direction vector

\(\mathbf{d}_1=(1,-1,1).\)

The line \(AB\) has direction vector

\(\mathbf{d}_2=B-A=(1,2,7).\)

Also, since \(O\) lies on \(OC\), a point on \(OC\) is \(O=(0,0,0)\), and a point on \(AB\) is \(A=(1,-1,0)\).

(i) Shortest distance between the lines

For two skew lines with direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\), the shortest distance is

\(\displaystyle \frac{|(\mathbf{a}-\mathbf{o})\cdot(\mathbf{d}_1\times \mathbf{d}_2)|}{|\mathbf{d}_1\times \mathbf{d}_2|}\),

where \(\mathbf{o}\) and \(\mathbf{a}\) are position vectors of one point on each line.

Here \(\mathbf{a}-\mathbf{o}=A-O=(1,-1,0)\).

Now

\(\mathbf{d}_1\times \mathbf{d}_2=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&-1&1\\ 1&2&7\end{vmatrix}=(-9,-6,3).\)

So

\(|\mathbf{d}_1\times \mathbf{d}_2|=\sqrt{(-9)^2+(-6)^2+3^2}=\sqrt{126}=3\sqrt{14}.\)

Also

\((1,-1,0)\cdot(-9,-6,3)=-9+6=-3,\)

so the absolute value is \(3\).

Hence the shortest distance is

\(\displaystyle \frac{3}{3\sqrt{14}}=\frac{1}{\sqrt{14}}.\)

(ii) Cartesian equation of the plane

The plane contains the line \(OC\), so it contains the vector \((1,-1,1)\). It also contains the common perpendicular of \(OC\) and \(AB\), so it must also contain a direction perpendicular to both lines. That common perpendicular has direction \(\mathbf{d}_1\times\mathbf{d}_2\), so the plane contains both \(\mathbf{d}_1\) and \(\mathbf{d}_1\times\mathbf{d}_2\).

Therefore a normal vector to the plane is perpendicular to both of these vectors. We can take

\(\mathbf{n}=\mathbf{d}_1\times(\mathbf{d}_1\times\mathbf{d}_2).\)

Using \(\mathbf{d}_1=(1,-1,1)\) and \(\mathbf{d}_1\times\mathbf{d}_2=(-9,-6,3)\),

\(\mathbf{n}=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&-1&1\\ -9&-6&3\end{vmatrix}=(3,-12,-15).\)

So a simpler normal vector is \((1,-4,-5)\).

The plane passes through \(O\), so its equation is

\(x-4y-5z=0.\)

But this is not the only possibility if we have chosen the wrong orientation? Let's verify by checking that the plane contains \(OC\): substituting \((1,-1,1)\) gives \(1+4-5=0\), so it does contain \(OC\). It also contains the common perpendicular direction \((-9,-6,3)\): \(-9+24-15=0\), so it is correct.

Answer: (i)(a) \(a=-\frac{6}{13}\).

(i)(b) The plane is \(-13x+5y-z=43\).

(ii) \(a=3\).

(i)(a) A point on \(l_1\) is

\((\lambda a,\,9-\lambda,\,2+\lambda)\),

and a point on \(l_2\) is

\((-6-\mu a,\,-5+2\mu,\,10+4\mu)\).

If the lines intersect, then

\((\lambda a,9-\lambda,2+\lambda)=(-6-\mu a,-5+2\mu,10+4\mu).\)

From the second and third components,

\(9-\lambda=-5+2\mu,\qquad 2+\lambda=10+4\mu.\)

These give \(\lambda=12\) and \(\mu=1\). Substituting in the first component gives

\(12a=-6-a\),

so

\(\boxed{a=-\frac{6}{13}}.\)

(i)(b) With \(a=-\frac{6}{13}\), the direction vectors are

\(\mathbf{d}_1=\left(-\frac{6}{13},-1,1\right),\qquad \mathbf{d}_2=\left(\frac{6}{13},2,4\right).\)

A normal vector to the plane is

\(\mathbf{d}_1\times\mathbf{d}_2=\left(-6,\frac{30}{13},-\frac{6}{13}\right),\)

which is proportional to \((-13,5,-1)\). Using the point \((0,9,2)\),

\(-13(x-0)+5(y-9)-(z-2)=0.\)

Therefore the plane is

\(\boxed{-13x+5y-z=43}.\)

(ii) For the general value of \(a\),

\(\mathbf{d}_1=(a,-1,1),\qquad \mathbf{d}_2=(-a,2,4),\)

so

\(\mathbf{d}_1\times\mathbf{d}_2=(-6,-5a,a).\)

Hence

\(|\mathbf{d}_1\times\mathbf{d}_2|=\sqrt{36+26a^2}.\)

The vector between the given points on the two lines is

\((-6,-5,10)-(0,9,2)=(-6,-14,8).\)

Therefore

\((-6,-14,8)\cdot(-6,-5a,a)=36+78a.\)

The distance between the skew lines is

\(\frac{|36+78a|}{\sqrt{36+26a^2}}.\)

Since the distance is \(3\sqrt{30}\),

\(3\sqrt{30}\sqrt{36+26a^2}=|36+78a|.\)

Squaring and simplifying gives

\(15(18+13a^2)=(6+13a)^2.\)

Thus

\(26(a-3)^2=0\),

so

\(\boxed{a=3}.\)

Answer: (i) \(a=2\).

(ii) The perpendicular distance from \(P\) to the plane is \(\sqrt{10}\).

(iii) The perpendicular distance from \(P\) to \(l_2\) is \(\sqrt{10}\).

A point on \(l_1\) is \((a+\lambda,9+2\lambda,13+3\lambda)\), and a point on \(l_2\) is \((-3-\mu,7+2\mu,-2-3\mu)\).

Since the lines intersect, compare coordinates. From the \(y\)-coordinates,

\(9+2\lambda=7+2\mu\), so \(\lambda=\mu-1\).

From the \(z\)-coordinates,

\(13+3\lambda=-2-3\mu\).

Substitute \(\lambda=\mu-1\):

\(13+3(\mu-1)=-2-3\mu\),

so \(10+3\mu=-2-3\mu\), hence \(\mu=-2\) and \(\lambda=-3\).

Using the \(x\)-coordinates,

\(a+\lambda=-3-\mu\),

so \(a-3=-1\), giving

\(a=2\).

The common point is obtained from \(l_2\) using \(\mu=-2\):

\(C=(-3,7,-2)+(-2)(-1,2,-3)=(-1,3,4).\)

The direction vectors of the two lines are

\(\mathbf d_1=(1,2,3),\qquad \mathbf d_2=(-1,2,-3).\)

A normal to the plane containing both lines is

\(\mathbf n=\mathbf d_1\times\mathbf d_2=(-12,0,4),\)

so we may use \(\mathbf n=(-3,0,1)\). The plane through \(C=(-1,3,4)\) is

\(-3(x+1)+(z-4)=0,\)

or

\(-3x+z-7=0.\)

For \(P=(3,1,6)\), the perpendicular distance to this plane is

\(\dfrac{| -3(3)+6-7 |}{\sqrt{(-3)^2+1^2}}=\dfrac{10}{\sqrt{10}}=\sqrt{10}.\)

For the distance from \(P\) to \(l_2\), take \(B=(-3,7,-2)\) on \(l_2\). Then

\(\overrightarrow{BP}=P-B=(6,-6,8).\)

The distance from a point to a line is

\(\dfrac{|\overrightarrow{BP}\times\mathbf d_2|}{|\mathbf d_2|}.\)

Now

\(\overrightarrow{BP}\times\mathbf d_2=(6,-6,8)\times(-1,2,-3)=(2,10,6),\)

so

\(|\overrightarrow{BP}\times\mathbf d_2|=\sqrt{2^2+10^2+6^2}=2\sqrt{35},\qquad |\mathbf d_2|=\sqrt{14}.\)

Therefore the distance is

\(\dfrac{2\sqrt{35}}{\sqrt{14}}=\sqrt{10}.\)

Answer: (a) \(5x-2y+z=-14\)

(b) \(\dfrac7{\sqrt{30}}\) \((\approx1.28)\)

(c) \(\mathbf r=\begin{pmatrix}0\\7\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}\)

(a)

The direction vectors in the plane are

\(\mathbf a=(0,1,2),\qquad \mathbf b=(1,3,1).\)

A normal vector is

\(\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&1&2\\1&3&1\end{vmatrix}=(-5,2,-1).\)

Use the equivalent normal vector \((5,-2,1)\). The point \((-3,-1,-1)\) lies on the plane. Hence

\(5(x+3)-2(y+1)+(z+1)=0.\)

So

\(5x-2y+z=-14.\)

(b)

The point is \((-1,0,-2)\), and the plane is \(5x-2y+z+14=0\). Therefore

\(d=\frac{|5(-1)-2(0)+(-2)+14|}{\sqrt{5^2+(-2)^2+1^2}}=\frac7{\sqrt{30}}.\)

(c)

A common point is found by taking \(x=0\), \(z=0\). Then \(3x+2y-z=14\) gives \(y=7\), so \((0,7,0)\) lies on both planes.

The direction vector of the intersection line is perpendicular to both normal vectors \((5,-2,1)\) and \((3,2,-1)\):

\(\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\5&-2&1\\3&2&-1\end{vmatrix}=(0,8,16)\sim(0,1,2).\)

Hence

\(\mathbf r=\begin{pmatrix}0\\7\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\2\end{pmatrix}.\)

Mark scheme

Answer: (a) \(\dfrac1{\sqrt{14}}\)

(b) \(7.5^\circ\) approximately.

(c) \(\mathbf r=\dfrac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}\)

(a)

The perpendicular distance from \((0,0,0)\) to \(x+3y+2z-1=0\) is

\(\frac{|0+0+0-1|}{\sqrt{1^2+3^2+2^2}}=\frac1{\sqrt{14}}.\)

(b)

Use \(\overrightarrow{OA}=(0,-1,2)\) and \(\overrightarrow{OB}=(2,0,-1)\). A normal to plane \(OAB\) is

\(\overrightarrow{OA}\times\overrightarrow{OB}=(1,4,2).\)

A normal to \(\Pi\) is

\((1,3,2).\)

Thus, if \(\alpha\) is the acute angle between the planes,

\(\cos\alpha=\frac{(1,4,2)\cdot(1,3,2)}{\sqrt{1^2+4^2+2^2}\sqrt{1^2+3^2+2^2}}=\frac{17}{\sqrt{21}\sqrt{14}}.\)

Therefore

\(\alpha\approx7.5^\circ.\)

(c)

The line \(OC\) has direction \((2,-1,-1)\), so a point \(P\) on \(OC\) has

\(\overrightarrow{OP}=\lambda\begin{pmatrix}2\\-1\\-1\end{pmatrix}.\)

The line \(AB\) has direction

\(\overrightarrow{AB}=(2,0,-1)-(0,-1,2)=\begin{pmatrix}2\\1\\-3\end{pmatrix}.\)

A point \(Q\) on \(AB\) has

\(\overrightarrow{OQ}=\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-3\end{pmatrix}=\begin{pmatrix}2\mu\\-1+\mu\\2-3\mu\end{pmatrix}.\)

Then

\(\overrightarrow{PQ}=\begin{pmatrix}2\mu-2\lambda\\-1+\mu+\lambda\\2-3\mu+\lambda\end{pmatrix}.\)

For the common perpendicular, \(\overrightarrow{PQ}\) is perpendicular to both line directions.

Using \(\overrightarrow{PQ}\cdot(2,-1,-1)=0\),

\(6\mu-6\lambda=1.\)

Using \(\overrightarrow{PQ}\cdot(2,1,-3)=0\),

\(14\mu-6\lambda=7.\)

Solving gives \(\mu=\frac34\). Hence

\(\overrightarrow{OQ}=\begin{pmatrix}\frac32\\-\frac14\\-\frac14\end{pmatrix}=\frac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}.\)

The common perpendicular direction is

\((2,-1,-1)\times(2,1,-3)=4(1,1,1).\)

Therefore an equation of the common perpendicular is

\(\mathbf r=\frac14\begin{pmatrix}6\\-1\\-1\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}.\)

Mark scheme