Exam-Style Problem

9231 P13 - Jun 2010 - Q12 - 19 marks

6589

Answer only one of the following two alternatives.

EITHER
The line \(l_{1}\) passes through the point \(A\) whose position vector is \(3 \mathbf{i}+\mathbf{j}+2 \mathbf{k}\) and is parallel to the vector \(\mathbf{i}+\mathbf{j}\). The line \(l_{2}\) passes through the point \(B\) whose position vector is \(-\mathbf{i}-\mathbf{k}\) and is parallel to the vector \(\mathbf{j}+2 \mathbf{k}\). The point \(P\) is on \(l_{1}\) and the point \(Q\) is on \(l_{2}\) and \(P Q\) is perpendicular to both \(l_{1}\) and \(l_{2}\).
(i) Find the length of \(P Q\).

(ii) Find the position vector of \(Q\).

(iii) Show that the perpendicular distance from \(Q\) to the plane containing \(A B\) and the line \(l_{1}\) is \(\sqrt{ } 3\).

OR
The linear transformation \(\mathrm{T}: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) is represented by the matrix \(\mathbf{M}=\left(\begin{array}{rrrr}1 & 1 & 5 & 7 \\ 3 & 9 & 17 & 25 \\ 1 & 7 & 7 & 11 \\ 3 & 6 & 16 & 23\end{array}\right)\).
(i) In either order,
(a) show that the dimension of \(R\), the range space of T , is equal to 2 ,
(b) obtain a basis for \(R\).

(ii) Show that the vector \(\left(\begin{array}{r}1 \\ -15 \\ -17 \\ -6\end{array}\right)\) belongs to \(R\).

(iii) It is given that \(\left\{\mathbf{e}_{1}, \mathbf{e}_{2}\right\}\) is a basis for the null space of T , where \(\mathbf{e}_{1}=\left(\begin{array}{r}14 \\ 1 \\ -3 \\ 0\end{array}\right)\) and \(\mathbf{e}_{2}=\left(\begin{array}{r}19 \\ 2 \\ 0 \\ -3\end{array}\right)\). Show that, for all \(\lambda\) and \(\mu\),
\(\mathbf{x}=\left(\begin{array}{r} 4 \\ -3 \\ 0 \\ \end{array}\right)+\lambda \mathbf{e}_{1}+\mu \mathbf{e}_{2}\)
is a solution of
\(\mathbf{M x}=\left(\begin{array}{r} 1 \\ -15 \\ -17 \\ -6 \end{array}\right) .\)
(iv) Hence find a solution of \((*)\) of the form \(\left(\begin{array}{l}\alpha \\ 0 \\ \gamma \\ \delta\end{array}\right)\).

Solution

Answer: Either: \(P=(1,-1,2)\), \(Q=(-1,1,1)\), \(\overrightarrow{PQ}=(-2,2,-1)\), so \(|PQ|=3\).

The position vector of \(Q\) is \(-\mathbf i+\mathbf j+\mathbf k\).

The distance of \(Q\) from the plane through \(A,B\) and \(l_1\) is \(\sqrt3\).

Or: \(\operatorname{rank}(\mathbf M)=2\). A basis for the range is \(\{(1,3,1,3)^T,(1,9,7,6)^T\}\).

One solution in the form \((\alpha,0,\gamma,\delta)^T\) is \((46,0,-9,0)^T\).

Either

A general point on \(l_1\) is

\(P=(3,1,2)+\lambda(1,1,0)=(3+\lambda,1+\lambda,2).\)

A general point on \(l_2\) is

\(Q=(-1,0,-1)+\mu(0,1,2)=(-1,\mu,-1+2\mu).\)

For the shortest line segment between the two skew lines, \(\overrightarrow{PQ}\) must be perpendicular to both direction vectors \((1,1,0)\) and \((0,1,2)\).

Solving the two dot-product equations gives

\(\lambda=-2,\qquad \mu=1.\)

\(P=(1,-1,2),\qquad Q=(-1,1,1).\)

Hence

\(\overrightarrow{PQ}=(-2,2,-1),\qquad |PQ|=\sqrt{4+4+1}=3.\)

The position vector of \(Q\) is

\(-\mathbf i+\mathbf j+\mathbf k.\)

The plane through \(A,B\) and \(l_1\) has a normal vector

\((1,1,0)\times(-4,-1,-3)=(-3,3,3),\)

so an equivalent plane equation is

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

The distance of \(Q=(-1,1,1)\) from this plane is

\(\frac{|(-1)-1-1|}{\sqrt{1^2+(-1)^2+(-1)^2}}=\frac3{\sqrt3}=\sqrt3.\)

Let the columns of the matrix be

\(\mathbf c_1=\begin{pmatrix}1\\3\\1\\3\end{pmatrix},\quad \mathbf c_2=\begin{pmatrix}1\\9\\7\\6\end{pmatrix},\quad \mathbf c_3=\begin{pmatrix}5\\17\\7\\16\end{pmatrix},\quad \mathbf c_4=\begin{pmatrix}7\\25\\11\\23\end{pmatrix}.\)

The third and fourth columns are linear combinations of the first two:

\(\mathbf c_3=\frac{14}{3}\mathbf c_1+\frac13\mathbf c_2,\qquad \mathbf c_4=\frac{19}{3}\mathbf c_1+\frac23\mathbf c_2.\)

Since \(\mathbf c_1\) and \(\mathbf c_2\) are independent, the rank is

\(\operatorname{rank}(\mathbf M)=2.\)

A basis for the range is therefore

\(\left\{\begin{pmatrix}1\\3\\1\\3\end{pmatrix},\begin{pmatrix}1\\9\\7\\6\end{pmatrix}\right\}.\)

The vector

\(\begin{pmatrix}1\\-15\\-17\\-6\end{pmatrix}=4\mathbf c_1-3\mathbf c_2,\)

so one particular solution is \((4,-3,0,0)^T\).

The null space has basis

\(\begin{pmatrix}14\\1\\-3\\0\end{pmatrix},\qquad \begin{pmatrix}19\\2\\0\\-3\end{pmatrix}.\)

Thus

\(\mathbf x=\begin{pmatrix}4\\-3\\0\\0\end{pmatrix}+s\begin{pmatrix}14\\1\\-3\\0\end{pmatrix}+t\begin{pmatrix}19\\2\\0\\-3\end{pmatrix}.\)

For a solution with second component \(0\), choose \(s=3\), \(t=0\). This gives

\(\mathbf x=\begin{pmatrix}46\\0\\-9\\0\end{pmatrix}.\)