Finding perpendicular distance of a point from line l (using vector equation).

1. May 14, 2012

Alshia

1. The problem statement, all variables and given/known data

Find a vector equation of the line l containing the points (1,3,1) and (1,-3,-1). Find the perpendicular distance of the point with coordinates (2,-1,1) from l.

2. The attempt at a solution

Let (1,3,1) = a, let (2,-1,1) = q. Let N be the point where ANQ = 90 degrees.

r = (1,3,1) + t(1-1,-3-3,-1-1) = (1,3,1) + t(0,-6,-2)

|p| = sqrt (6^2 + 2^2) = 2 sqrt 10

Unit vector, u = (-3 sqrt 10/10)i + (- sqrt 10/10)j

Vector AQ = q - a = i - 4j

AN = (Vector AQ).u = 6 sqrt 10/5

Hence, perpendicular distance NQ = sqrt (|Vector AQ|^2 - AN^2) = 17 - (6 sqrt 10/5)^2 = sqrt 65/5.

3. Relevant equations

The book gives the answer AN = 21 sqrt 15/5. Can anyone please check my answer for any mistakes? Thanks.

2. May 14, 2012

HallsofIvy

Staff Emeritus
What is "ANQ"?

Yes, this is the equation of the line l.

You didn't say, but I guess "p" is the vector from (1, 3, 1) to (1, -3, 1).

This is a unit vector in the direction of vector -6i- 2j but where did that vector come from? Your line is in the direction of the vector -6j- 2k.

You keep using notation you haven't defined! What are A and Q? Presumably it is the vector from point a to point q but what is point a?

I really can't tell what you are doing. What I would do is find the equation of the plane perpendicular to line, l, through (1, 3, 1) and (1, -3, -1), and containing the point (2, -1, 1). Then find the point where line l crosses that plane and find the distance from that point to (2, -1, 1).

Last edited: May 15, 2012
3. May 15, 2012

Joffan

Your numerical answer is good but, as HallsofIvy says, your working needs more clarity. The book answer as given is wrong for the problem as given, assuming I understand your notation (function brackets would be nice for sqrt(15) eg.).

Can you calculate the coordinates of N, which I take to be the nearest point on the line l to (2,-1,1)?

4. May 18, 2012

Alshia

REVISION

NOTE: Small alphabets in bold indicate vectors. If A is a point, a represents its position vector.

Let (1,3,1), (1,-3,-1) and (2,-1,1) be the points A, B and Q respectively. Let N be the foot of the perpendicular from Q to the line.

r = a + tp
r = (1,3,1) + t(1-1,-3-3,-1-1) = (1,3,1) + t(0,-6,-2)

Length AB = |p| = √(6^2 + 2^2) = 2√10

Unit vector, u = (-3√10/10)j + (-√10/10)k

q - a = i-2j-2k

Length AN = u.(q-a) = (4√10)/5

∴ Perpendicular distance:

Length NQ = √(AQ^2-AN^2) = √65/5

How is this? Yes, I made a mistake with the unit vector components.

@Joffan:

I can. Since length AN is (4√10)/5, the coordinates of N would be (1,3,1)+((4√10)/5)(0,-6,-2). I won't simplify here because the exact answer is in rational form.

By the way, is there a way to write vectors as column vectors here? I don't like writing in component form. Too much work :S.

5. May 18, 2012

HallsofIvy

Staff Emeritus
In LaTeX,
[ tex ]\ begin{bmatrix}a \\ b \\ c\ end{bmatrix}[ /tex ]
(without the spaces) gives
$$\begin{bmatrix}a \\ b \\ c\end{bmatrix}$$

6. May 18, 2012

Joffan

Your position of N is wrong, because t is not in coordinate units.

The coordinates of N can be specified without surds.