Answer:
- (a) The shortest distance between the lines is \(\displaystyle \frac{15}{\sqrt{77}}=\frac{15\sqrt{77}}{77}\), approximately \(1.71\).
- (b) The plane is \(4x+3y-4z=0\).
- (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
\(\displaystyle \theta=\sin^{-1}\!\left(\frac{9}{\sqrt{2501}}\right)\approx 10.4^\circ\).
