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

[Alshia]

Homework Statement

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.

 

[HallsofIvy]

Homework Statement

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.

What is "ANQ"?

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

Yes, this is the equation of the line l.

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

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

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

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.

Vector AQ = q - a = i - 4j

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?

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

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).

 

[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)?

 

[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) = √65/5How 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.

 

[HallsofIvy]

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

 

[Joffan]

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

The coordinates of N can be specified without surds.