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

In summary: 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.Can you calculate the coordinates of N, which I take to be the nearest point on the line l to (2,-1,1)?Yes, you can calculate the coordinates of N.
  • #1
Alshia
28
0

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.
 
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  • #2
Alshia said:

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).
 
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  • #3
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
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.
 
  • #5
In LaTeX,
[ tex ]\ begin{bmatrix}a \\ b \\ c\ end{bmatrix}[ /tex ]
(without the spaces) gives
[tex]\begin{bmatrix}a \\ b \\ c\end{bmatrix}[/tex]
 
  • #6
Your position of N is wrong, because t is not in coordinate units.

The coordinates of N can be specified without surds.
 

What is the formula for finding the perpendicular distance of a point from a line using the vector equation?

The formula for finding the perpendicular distance of a point (x1,y1) from a line represented by the vector equation r = r0 + tv, where r0 is a point on the line and v is a vector parallel to the line, is given by |(x1-x0) x v| / |v|, where x0 is a point on the line closest to (x1,y1) and |...| denotes the magnitude of the vector.

How do you find the point on a line closest to a given point using the vector equation?

To find the point on a line closest to a given point (x1,y1) using the vector equation r = r0 + tv, where r0 is a point on the line and v is a vector parallel to the line, you can set up a system of equations and solve for t. Once you have the value of t, you can plug it back into the vector equation to find the point on the line closest to (x1,y1).

Can the vector equation be used to find the perpendicular distance of a point from a line in 3-dimensional space?

Yes, the vector equation can be used to find the perpendicular distance of a point from a line in 3-dimensional space. The formula for finding the distance remains the same, but the vectors and points will have three components instead of two.

Is it necessary to know the equation of a line in order to find the perpendicular distance of a point from that line using the vector equation?

No, it is not necessary to know the equation of a line in order to find the perpendicular distance of a point from that line using the vector equation. As long as you have a point on the line and a vector parallel to the line, you can use the vector equation to find the distance from any given point.

Can the vector equation be used to find the perpendicular distance from a point to a curved line?

No, the vector equation can only be used to find the perpendicular distance from a point to a straight line. Curved lines require different methods, such as using calculus to find the shortest distance from the point to the curve.

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