Consider the line through the origin in three dimensional space parametrized by t*v, where v=i+2j+3k. What is the distance from the point (-3,2,5) to this line? By the distance between a given point P and a line l, we mean the minimal distance between P and a point of l.

I was thinking that to get the angle between two vector (one is to get from the origin to the given point), and get the projection of vector a on vector b, so we could calculate the length of two side, and use the relationship to get the distance?

The line equation can be written as:

x=t, \;\ y=2t, \;\ z=3t

The distance from the point (-3,2,5) and the line:

Since (0,0,0) is on the line, we have

u=-3i+2j+5k

v=1+2j+3k

Cross product of u and v:
u\times v=-4i+14j-8k

The distance is found by using the 'norms' of uxv and v:

d=\frac{||u\times v||}{||v||}

\frac{\sqrt{(-4)^{2}+(14)^{2}+(-8)^{2}}}{\sqrt{1^{2}+2^{2}+3^{2}}}=\frac{\sqrt{966 }}{7}\approx 4.44

If you don't remember the formula, you can still do it using the basics.

Make a quick sketch.

The line v has equation:

\vec{v}= t\(1 \\ 2 \\ 3 \)

http://p1cture.me/images/42210695227422283694.png

The black line is the line v.

The length of the blue line is what you are looking for. Let the red line be l and the blue line be x.

Use the dot product of the line v and the red line, with vector <-3, 2, 5>:

\vec{v}.\vec{l} = |\vec{v}||\vec{l}|\cos\theta

\(1\\2\\3\).\(-3\\2\\5\) = \sqrt{1^2 + 2^2 + 3^2} \sqrt{-3^2 + 2^2 + 5^2}\cos\theta

Find theta.

You get the length of the red line from this formula itself, as you have to calculate the length of l.

Then use a simple trigonometric ratio:

\sin\theta = \frac{x}{l}

And you get 4.44 units