# Math Help - Another linear Algebra vector problem.

1. ## Another linear Algebra vector problem.

Find the distance from the point P (-3, 2, -7) to the line that passes through the points Q (-4, 3, 0) and R (-2, 1, -2).

I started it via getting the direction vector between R and Q, (-4, 3, 0)-(-2, 1, 2) = (-2, 2, 2)

Then got vector v from R to P (-2, 1, -2)-(-3, 2, -7) = (1, -1, 5)

Proj of v on d = (v*d/||d||^2)*d then gave me -1/2 (-2, 2, 2) ==> (1, -1, -1)

Now this is where I am a little confused, the projection is the vector or the point at the right angle? I am guessing the vector because from here I did the distance between (-3, 2, -7) and (1, -1, -1), getting sqrt (61) and that is wrong.

So I am not sure how to get from here to an actual distance between the right angle on the line passing through those two points with what I have so far.

2. Welp, I used the point (1, -1, -1) doing the sqrt (1^2+(-1)^2+(-1)^2) and getting a length of sqrt (3), it came back as wrong as well.

I am quite lost and have about 1.5 hours left for this last and only question I cannot get, can anyone figure out where I am amiss?

3. Originally Posted by LostInCalculus
Find the distance from the point P (-3, 2, -7) to the line that passes through the points Q (-4, 3, 0) and R (-2, 1, -2).

I started it via getting the direction vector between R and Q, (-4, 3, 0)-(-2, 1, 2) = (-2, 2, 2)

Then got vector v from R to P (-2, 1, -2)-(-3, 2, -7) = (1, -1, 5)

Proj of v on d = (v*d/||d||^2)*d then gave me -1/2 (-2, 2, 2) ==> (1, -1, -1)

...
I can only show you how I would solve this problem:

1. Calculate the equation of the line QR:

$QR: \vec r=\left(\begin{array}{c}-4\\3\\0\end{array}\right) + t \cdot \left(\begin{array}{c}2\\-2\\-2\end{array}\right)$

2. Calculate the equation of an auxiliar plane a which contains the point P and which is perpendicular to QR that means the direction vector of the line is the normal vector of the plane a:

$\left(\begin{array}{c}2\\-2\\-2\end{array}\right) \left( \left(\begin{array}{c}x\\y\\z\end{array}\right) - \left(\begin{array}{c}-3\\2\\-7\end{array}\right)\right) = 0 ~\implies~ -2x -2y - 2z -4 =0$

3. Calculate the coordinates of the point of intersection I between the plane a and the line QR. I've got I(-1, 0, -3)

4. Calculate the distance between I and P:

$\overrightarrow{IP}=\left(\begin{array}{c}-3\\2\\-7\end{array}\right) - \left(\begin{array}{c}-1\\0\\-3\end{array}\right) = \left(\begin{array}{c}-2\\2\\-4\end{array}\right)$ .... which has the length: $|\overrightarrow{IP}|= \left| \left(\begin{array}{c}-2\\2\\-4\end{array}\right) \right| = \sqrt{24} \approx 4.8989...$