91Ó°ÊÓ

As per Problem 23, the planes are$2x+y-2z=3$ and $3x-6y-2z=4$.

The given point is $\left(1,0,0\right)$

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

Then equations of the planes become$2x+y=3$ and $3x-6y=4$.

Solving the two equations using elimination method, it implies

$x=\frac{22}{15}$and $y=\frac{1}{15}$

Thus, the common point is $\left(\frac{22}{15},\frac{1}{15},0\right)$

For the given planes, normal vectors are:$2\hat{i}+\hat{j}-2\hat{k}$

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

The cross product of the two normal vectors:

$$\begin{gathered} N=\left[\begin{array}{ccc}i& j& k\\ 2& 1& -2\\ 3& -6& -2\end{array}\right] \\ =\left(-2-12\right)\hat{i}-\left(-4+6\right)\hat{j}+\left(-12-3\right)\hat{k} \\ =-14\hat{i}-2\hat{j}-15\hat{k} \end{gathered}$$

Now, the equation of line of intersection of the planes through$\left(\frac{22}{15},\frac{1}{15},0\right)$ and parallel to$-14\hat{i}-2\hat{j}-15\hat{k}$ is

$\frac{x-\frac{22}{15}}{-14}=\frac{y-\frac{1}{15}}{-2}=\frac{z-0}{-15}$

So the equations are

$\frac{x-\frac{22}{15}}{-14}=\frac{y-\frac{1}{15}}{-2}=\frac{z}{-15}$

And parameteric form of equations is

$r=\left(\frac{22}{15}\right)\hat{i}+\left(\frac{1}{15}\right)\hat{j}+\left(-14\hat{i}-2\hat{j}-15\hat{k}\right)t$

The point is$P=\left(1,0,0\right)$

The line is$r=\left(\frac{22}{15}\right)\hat{i}+\left(\frac{1}{15}\right)\hat{j}+\left(-14\hat{i}-2\hat{j}-15\hat{k}\right)t$

Here,

$$\begin{gathered} Q=\frac{22}{15}\hat{i}+\frac{1}{15}\hat{j} \\ =\left(\frac{22}{15},\frac{1}{15},0\right) \end{gathered}$$

And $A=-14\hat{i}-2\hat{j}-15\hat{k}$

So

$$\begin{gathered} \left|A\right|=\sqrt{{\left(-14\right)}^2+{\left(-2\right)}^2+{\left(-15\right)}^2} \\ =\sqrt{425} \end{gathered}$$

Unit vector along the line is $u=\frac{-14\hat{i}-2\hat{j}-15\hat{k}}{\sqrt{425}}$

Also,

$$\begin{gathered} \stackrel{→}{PQ}=\left(\frac{22}{15},\frac{1}{15},0\right)-\left(1,0,0\right) \\ =\left(\frac{7}{15},\frac{1}{15},0\right) \\ =\frac{7}{15}\hat{i}\frac{1}{15}\hat{j} \end{gathered}$$

The distance of the point$P=\left(1,0,0\right)$ to the line is

$$\begin{gathered} \left|\stackrel{→}{PQ}×u\right|=\left|\left(\frac{7}{15}\hat{i}+\frac{1}{15}\hat{j}\right)×\frac{-14\hat{i}-2\hat{j}-15\hat{k}}{\sqrt{425}}\right| \\ =\frac{1}{\sqrt{425}}\left|\left(\frac{7}{15}\hat{i}+\frac{1}{15}\hat{j}\right)×\left(-14\hat{i}-2\hat{j}-15\hat{k}\right)\right| \\ =\frac{1}{\sqrt{425}}\left|-\hat{i}+7\hat{j}\right| \\ =\frac{1}{\sqrt{425}}\left(\sqrt{{\left(-1\right)}^2+7^2}\right) \end{gathered}$$

So,

$$\begin{gathered} \left|\stackrel{→}{PQ}×u\right|=\frac{\sqrt{50}}{\sqrt{425}} \\ =\sqrt{\frac{2}{17}} \end{gathered}$$

Hence, the distance from the point (1,0,0) to the line is$\sqrt{\frac{2}{17}}$ units.