Write the formula for the distance and calculate it

Question

is there anyone who knows how to (can?) solve it. Any hints or directions would be helpful.

Write the formula for the distance from a point to a line L. Then calculate the distance of the point $P_1(1,2,4)$ to the line L which is the intersection of two planes $x-y + 2z = 1$ and $x + 3y-z = 4$ .

Answer 1

Say you have a point $P$ and some line $L$ with direction vector $\vec v$, and you know $P_2$ is on the line. We are interested in the distance from $P$ to $L$. Then we construct a vector going from $P_2$ to $P$, that is $\vec P-\vec P_2$.

The magnitude of cross product $|(\vec P-\vec P_2) \times \vec v|$ gives the area of the parallelogram formed by these two vectors. The distance we are seeking is the height of this parallelogram. The base is of length $|\vec v|$. And $\text{Area}=(\text{base}) (\text{height})$. So the distance is,

$$\text{height}=\frac{\text{Area}}{\text{base}}=\frac{|(\vec P-\vec P_2) \times \vec v|}{|\vec v|}$$

First let $z=0$ then from your two planes we have,

$$x-y=1$$

$$x+3y=4$$

So then $x=1.75$ and $y=0.75$. So one choice of $\vec P_2$ is $\langle 1.75,0.75,0 \rangle$. A direction vector of the line from the intersection of your two planes is the cross product of the normals of the two planes.

$$\vec v=\langle 1,-1,2 \rangle \times \langle 1,3,-1 \rangle=\langle -5,3,4 \rangle$$

So then the distance is,

$$d=\frac{|(\langle 1,2,4 \rangle-\langle 1.75,0.75,0 \rangle) \times \langle -5,3,4 \rangle|}{| \langle -5,3,4 \rangle|}$$

$$d=\frac{|\langle -0.75,1.25,4 \rangle \times \langle -5,3,4 \rangle|}{\sqrt{50}}$$

$$d=\frac{|\langle -7,-17,4 \rangle|}{\sqrt{50}}$$

$$=\sqrt{\frac{354}{50}}$$

$$=\frac{\sqrt{177}}{5}$$

Answer 2

Hint:

The projection of $P$ to $L$ (let $Q$) is the intersection of a plane perpendicular to $L$, through $P$. This plane is perpendicular to any plane that contains $L$ and in particular the two given planes.

Then the direction of the normal to the plane is the cross product of the normals, $$(1,-1,2)\times(1,3,-1)=(-5,3,4)$$

so that the equation of the plane is

$$-5(x-1)+3(y-2)+4(z-4)=0.$$

Now you solve the $3\times3$ system formed by the three plane equations,

$$\begin{cases}-5x+3y+4z=17,\\x-y + 2z = 1,\\x + 3y-z = 4,\end{cases}$$

giving $Q$, and you return the distance $PQ$.

Answer 3

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

perpendicular distance formula

$$d=$$ $$\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}$$