Distance between a point and a line Question?

[Guest]

Hi everyone, i have a midterm exam coming up and i'm looking at problems in the section of the book to help myself prepare. Can anyone answer this question for me? I have attempted to answer this problem but without the answer i don't know if i'm right.

Find the distance between the point P(1,-2,4) and the line given by the parametric equations:

x= 2t
y= t-3
z= 2t+2

I tried using the techniques given in the book but it's not for a point and line. I graphed the line on the calculator but honestly have no idea how to do this mathematically.

Can someone do this one out for me? It's not a homework, i just need to see it done out step by step so i can follow it, along with the answer.

Thank you guys, really desperate...
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[galactus]

Some calc texts do not cover this. They just concentrate on the distance between a point and a PLANE. I will show you then you'll know how.

Anyway, the distance between a point and a line in space is given by

\(\displaystyle D=\frac{||PQ\times u||}{||u||}\)

Where u is the direction vector and P is a point on the line.

From the given equations, the direction vector is \(\displaystyle [2,1,2]\)

To find a point on the line, let t=0 and get:

\(\displaystyle P=(0,-3,2)\)

Therefore, \(\displaystyle PQ=[2-0,1-(-3),2-2]=[2,4,0]\)

Now, use the cross product between that and the direction vector:

\(\displaystyle PQ\times u=\begin{vmatrix}i&j&k\\2&4&0\\2&1&2\end{vmatrix}=8i-4j-6k=[8,-4,-6]\)

Now, use the formula from above:

\(\displaystyle ||PQ\times u||=2\sqrt{29}\)

\(\displaystyle ||u||=3\)

\(\displaystyle D=\frac{2\sqrt{29}}{3}\approx{3.59}\)

There, see how now?. I hope that helped.