Question

As in Problem 24, find the equations of the line intersections of the planes in Problem 23. Find the distance from the point (1,0,0) to the line.

Short answer

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

Step-by-step solution

Given

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

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

Find point on both planes

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 $\left(\frac{22}{15},\frac{1}{15},0\right)$

Find equations of line of intersections of planes

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

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

Find the point and the line

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,

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

Find the distance

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

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

So,

$$\begin{gathered} \left|\overrightarrow{PQ}\times u\right|=\frac{\sqrt{50}}{\sqrt{425}} \\ =\sqrt{\frac{2}{17}} \end{gathered}$$

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