# 3D Vector Problem

• June 3rd 2008, 05:41 AM
thermalwarrior
3D Vector Problem
I have been give the following problem: the vector perpendicular to OA and OB where O, B, A are points with respective coordinates of (0,0,0), (-4,2,1) and (1,2,3) is of the form 4i+yj+zk. What are the values of y and z?

How do i go about working this out???
• June 3rd 2008, 07:22 AM
topsquark
Quote:

Originally Posted by thermalwarrior
I have been give the following problem: the vector perpendicular to OA and OB where O, B, A are points with respective coordinates of (0,0,0), (-4,2,1) and (1,2,3) is of the form 4i+yj+zk. What are the values of y and z?

How do i go about working this out???

A vector perpendicular to two other vectors is given by the cross product:
$(-4, 2, 1) \times (1, 2, 3) = det \left ( \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right )$

This will either be parallel to your vector or antiparallel.

-Dan
• June 3rd 2008, 07:47 AM
thermalwarrior
how does that lead to the values of y and z in the final form 4i + yj + zk?
• June 3rd 2008, 07:54 AM
topsquark
Quote:

Originally Posted by thermalwarrior
how does that lead to the values of y and z in the final form 4i + yj + zk?

Quote:

Originally Posted by topsquark
A vector perpendicular to two other vectors is given by the cross product:
$(-4, 2, 1) \times (1, 2, 3) = det \left ( \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right )$

This will either be parallel to your vector or antiparallel.

-Dan

The result of the cross product will be of the form
$a(4i + yj + zk)$
where a is some constant. ie. they are parallel.

So find the determinant, and set the x component equal to 4a. Solve for a. Then the y component will be ay, etc.

-Dan
• June 3rd 2008, 07:58 AM
thermalwarrior
sorry im a bit confused - if i do that then a can only be 1 because the answer given in the questions has the x component as 4i. or have i misunderstood what you have wrote?
• June 3rd 2008, 08:12 AM
earboth
Quote:

Originally Posted by thermalwarrior
sorry im a bit confused - if i do that then a can only be 1 because the answer given in the questions has the x component as 4i. or have i misunderstood what you have wrote?

You are right. With your problem a = 1

(Btw: I've got y = 13 and z = -10)
• June 3rd 2008, 08:16 AM
thermalwarrior
can you show me how? that fits the question!
• June 3rd 2008, 10:02 AM
o_O
As suggested take the cross product of OA and OB which will yield a vector perpendicular to both vectors. You are told that there is a vector of the form (4,y,z) that is also perpendicular to both OA and OB. So, (4,y,z) is some scalar multiple of the cross product between OA and OB:

$(-4, 2, 1) \times (1, 2, 3) = \left( \left| \begin{array}{cc} 2 & 1 \\ 2 & 3 \end{array} \right|, -\left| \begin{array}{cc} -4 & 1 \\ 1 & 3 \end{array} \right|, \left| \begin{array}{cc} -4 & 2 \\ 1 & 2 \end{array} \right| \right) = (a, b, c)$ (leaving it up to you to find the cross product i.e. what a, b, and c are)

So, (a, b, c) = k(4, y, z).

Looking at the x component, you can see that a = 4k. Solve for k, then you should be able to find y and z (b = ky; c = kz).