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western50 · Jan 27, 2011, 12:37 AM

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?

galactus · Jan 27, 2011, 03:55 AM

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:

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

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

Unknown008 · Jan 27, 2011, 07:37 AM

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 $$

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