91Ó°ÊÓ

As per Problem 22, the planes are$2x-y-z=4$ and $3x-2y-6z=7$.

The given point is $\left(2,1,-1\right)$.

In order to find a point that lies on both the planes, put $z=0$.

Then equations of the planes become$2x-y=4$ and $3x-2y=7$.

Solving the two equations using elimination method, it implies

$x=1$and $y=-2$

Thus, the common point is $\left(1,-2,0\right)$

For the given planes, normal vectors are:

$2\hat{i}-\hat{j}-\hat{k}$and$3\hat{i}-2\hat{j}-6\hat{k}$ respectively.

The cross product of the two normal vectors:

$$\begin{gathered} N=\left|\begin{array}{ccc}i& j& k\\ 2& -1& -1\\ 3& -2& -6\end{array}\right| \\ =\left(6-2\right)\hat{i}-\left(-12+3\right)\hat{j}+\left(-4+3\right)\hat{k} \\ =4\hat{i}+9\hat{j}-\hat{k} \end{gathered}$$

Now, the equation of line of intersection of the planes through$\left(1,-2,0\right)$ and parallel to$4\hat{i}+9\hat{j}-\hat{k}$ is

$\frac{x-1}{4}=\frac{y-\left(-2\right)}{9}=\frac{z-0}{-1}$

So the equations are

$\frac{x-1}{4}=\frac{y+2}{9}=\frac{z}{-1}$

And parameteric form of equations is

$r=\hat{i}-2\hat{j}+\left(4\hat{i}+9\hat{j}-\hat{k}\right)t$

The point is $P=\left(2,1,-1\right)$

The line is$r=\hat{i}-2\hat{j}+\left(4\hat{i}+9\hat{j}-\hat{k}\right)t$

Here,

$$\begin{gathered} Q=\hat{i}-2\hat{j} \\ =\left(1,-2,0\right) \end{gathered}$$

And $A=4\hat{i}+9\hat{j}-\hat{k}$

So

$$\begin{gathered} \left|A\right|=\sqrt{{\left(4\right)}^2+{\left(9\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{98} \end{gathered}$$

Unit vector along the line is $u=\frac{4\hat{i}+9\hat{j}-\hat{k}}{\sqrt{98}}$

Also,

$$\begin{gathered} \stackrel{→}{PQ}=\left(1,-2,0\right)-\left(2,1,-1\right) \\ =\left(-1,-3,1\right) \\ =-\hat{i}-3\hat{j}+\hat{k} \end{gathered}$$

The distance of the point$P=\left(2,1,-1\right)$ to the line is

$$\begin{gathered} \left|\stackrel{→}{PQ}×u\right|=\left|\left(-\hat{i}-3\hat{j}+\hat{k}\right)×\frac{4\hat{i}+9\hat{j}-\hat{k}}{\sqrt{98}}\right| \\ =\frac{1}{\sqrt{98}}\left|\left(-\hat{i}-3\hat{j}+\hat{k}\right)×\left(4\hat{i}+9\hat{j}-\hat{k}\right)\right| \\ =\frac{1}{\sqrt{98}}\left|-6\hat{i}+3\hat{j}+3\hat{k}\right| \\ =\frac{1}{\sqrt{98}}\left(\sqrt{{\left(-6\right)}^2+3^2+3^2}\right) \end{gathered}$$

So,

$$\begin{gathered} \left|\stackrel{→}{PQ}×u\right|=\frac{\sqrt{54}}{\sqrt{98}} \\ =\frac{3\sqrt{3}}{7} \end{gathered}$$

Hence, the distance from the point (2,1,-1) to the line is$\frac{3\sqrt{3}}{7}$ units.