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Cross Product and Normal Vector Related Question

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Homework Statement


Find the distance from the point (2, 4, 4) to the line x = 0, y = 4 + 3t, z = 4 + 2t.


Homework Equations



The cross product and the dot product and d = |n * b|/|n|

The Attempt at a Solution



So the distance from the point to the line is the line directly perpendicular from the point to the line. To find that distance we can find the dot product (which produces a scalar projection of the distance from the point to the line onto the line's normal vector) of the vector from any point on the line (call it q) and the point (call it p) and the line itself. To get the line from p to q (call it pq) I simply do the following:
Set t = 3 and retrieve the following points <0,13,10> and we subtract that from p to get pq = <2,-9,-6>.
Next I need to find the normal vector, which is where I am having trouble with. I can find the direction numbers of the line which results in <0, 3, 2>, however my problem now is that I don't know what I should cross product it with to get the normal vector. If I cross product it with pq I'll get a line coming straight out of the board and for me it seems like that a normal vector that isn't pointing in the same direction as p will screw up the projection. Can you please explain?
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Find the distance from the point (2, 4, 4) to the line x = 0, y = 4 + 3t, z = 4 + 2t.


Homework Equations



The cross product and the dot product and d = |n * b|/|n|

The Attempt at a Solution



So the distance from the point to the line is the line directly perpendicular from the point to the line. To find that distance we can find the dot product (which produces a scalar projection of the distance from the point to the line onto the line's normal vector) of the vector from any point on the line (call it q) and the point (call it p) and the line itself. To get the line from p to q (call it pq) I simply do the following:
Set t = 3 and retrieve the following points <0,13,10> and we subtract that from p to get pq = <2,-9,-6>.
Next I need to find the normal vector, which is where I am having trouble with. I can find the direction numbers of the line which results in <0, 3, 2>, however my problem now is that I don't know what I should cross product it with to get the normal vector. If I cross product it with pq I'll get a line coming straight out of the board and for me it seems like that a normal vector that isn't pointing in the same direction as p will screw up the projection. Can you please explain?
I don't know what you mean by the "normal vector" to the line. A plane has a normal direction (and so normal vectors in that direction) but a line has a normal plane- any vector in that plane will be normal to the line.

The simplest way to find the shortest distance from a point in to a line is to find the normal plane to the line that contains the point. A plane, with normal vector <A, B, C> containing point [itex](x_0, y_0, z_0)[/itex] is given by [itex]A(x- x_0)+ B(y- y_0)+ C(z-z_0)= 0[/itex]. Here the vector perpendicular to the plane you want is <0, 3, 2>, the vector in the direction of the line, and the point is (2, 4, 4) so the plane is 0(x- 2)+3(y- 4)+ 2(z- 4)= 0 or 3y+ 2z= 20. Find the point where the given line crosses that plane and then find the distance from that point to (2, 4, 4).
 
  • #3
LCKurtz
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Find a point P on the line and a unit direction vector D for the line (both easy).
If Q is the point off the line, let V be the vector from P to Q (also easy). Then

distance = |V x D|

Draw a picture to see why this works.
 
  • #4
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In response to LCKurtz, sorry but I don't see how it works conceptually. What it looks like to me is that you are getting the magnitude of the cross product on the direction vector of the line and the line from the line to Q. From what I understand about cross products is that this will give us a vector which is perpendicular to both the line and the line V. I don't understand why the magnitude of this vector will give me the distance unless it is a property of the vector which results from the cross product to be equal to the perpendicular distance from point Q to the line.
Also can you check my work since I don't think I am doing this correctly.
Unit Direction Vector = <0, 3, 2> (I got this from isolating t from the line equation and getting their a, b, c).
Point of the line L(1) = <0,7,6>
V = <2,4,4> - <0,7,6> = <2,-3,-2>
Doing the cross product of <2,-3,-2> x <0,3,2> I get < <0,-4,6>.
Getting the magnitude of the vector will mean.
[tex]\sqrt{16+36}[/tex] = 7.21110255 which doesn't appear to be the answer. Am I doing something wrong?
 
  • #5
LCKurtz
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In response to LCKurtz, sorry but I don't see how it works conceptually. What it looks like to me is that you are getting the magnitude of the cross product on the direction vector of the line and the line from the line to Q. From what I understand about cross products is that this will give us a vector which is perpendicular to both the line and the line V. I don't understand why the magnitude of this vector will give me the distance unless it is a property of the vector which results from the cross product to be equal to the perpendicular distance from point Q to the line.
Also can you check my work since I don't think I am doing this correctly.
Unit Direction Vector = <0, 3, 2> (I got this from isolating t from the line equation and getting their a, b, c).
Point of the line L(1) = <0,7,6>
V = <2,4,4> - <0,7,6> = <2,-3,-2>
Doing the cross product of <2,-3,-2> x <0,3,2> I get < <0,-4,6>.
Getting the magnitude of the vector will mean.
[tex]\sqrt{16+36}[/tex] = 7.21110255 which doesn't appear to be the answer. Am I doing something wrong?
Yes. Your direction vector is not a unit vector (having length 1). You have to multiply it by 1/sqrt(13) to make it a unit vector. That makes your answer sqrt(52)/sqrt(13) = sqrt(4) = 2 which is the correct answer.

To see why it works draw a picture of the line with P = (0,7,6) on it, Q = (2,4,4,) off of it, the vector V from P to Q, and drop a perpendicular from Q to the nearest point W on the line. That perpendicular line's length h is what you want to calculate. Draw a little vector at P pointing towards W representing the unit vector. Call theta the angle between your unit vector and V. From the right triangle, h = |V| sin(theta).

But remember your geometric formula for the magnitude of a cross product:

|V x D| = |V||D| sin(theta). Since |D| = 1 it equals h.

The fact that V x D is perpendicular to V and D is irrelevant.
 

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