3D vectors finding perpindicular vector help!!!

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Mar 2nd 2008, 03:42 PM
dmitri

3D vectors finding perpindicular vector help!!!

hello everybody, first of all i would like to thank you for even trying to help me with my problem, I am doing my 3d vectors homework and I came upon a problem that I am having difficulties with, the problem is as follows:

Determine a vector perpindicular to both the y-axis and the vector AB = [2,-1,2].

any help will be appreciated.

thanks
Mar 2nd 2008, 04:45 PM
topsquark

Quote:

Originally Posted by dmitri

hello everybody, first of all i would like to thank you for even trying to help me with my problem, I am doing my 3d vectors homework and I came upon a problem that I am having difficulties with, the problem is as follows:

Determine a vector perpindicular to both the y-axis and the vector AB = [2,-1,2].

any help will be appreciated.

thanks

The y-axis has a unit vector of <0, 1, 0> and you want a vector mutually perpendicular to this and <2, -1, 2>. Take the cross product of these two vectors.

-Dan
Mar 2nd 2008, 04:52 PM
Soroban

Hello, dmitri!

Quote:

Determine a vector perpendicular to both the y-axis and the vector AB = [2,-1,2].

We have two vectors: . $\begin{array}{ccc}\vec{u} &=&\langle 2,1,2\rangle \\ \vec{v} &=& \langle0,1,0\rangle \end{array}$

A vector perpendicular to both is their cross product.

$\vec{w} \;=\;\vec{u} \times \vec{v} \;=\;\left|\begin{array}{ccc}\bold i&\bold j&\bold k \\ 2&1&2\\0&1&0\end{array}\right| \;=\;\bold i(0-2) -\bold j(0-0) + \bold k(2-0) \;=\;-2\bold i + 2\bold k$

. . Therefore: . $\vec{w} \;=\;\langle -2,0,2\rangle$
Mar 2nd 2008, 05:00 PM
dmitri

thanks guys! i appreciate it(Yes)