# find a vector perpendicular to both vector a and vector b

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• Jan 11th 2011, 10:59 AM
colorado
find a vector perpendicular to both vector a and vector b
I did this one the way I did a previous problem like this but came up with a completely incorrect answer

the question is:
vector a = (2,1,0) and vector b = (-1,6,1) find a vector perpendicular to both.

:)
• Jan 11th 2011, 11:04 AM
dwsmith
$\displaystyle ||a\times b||=\begin{vmatrix}i&j&k\\2&1&0\\-1&6&1\end{vmatrix}=\mbox{normal vector}$
• Jan 11th 2011, 11:06 AM
colorado
how about using dot product? haven't got to cross product yet
• Jan 11th 2011, 11:07 AM
dwsmith
The dot product produces a scalar not a vector.
• Jan 11th 2011, 11:08 AM
dwsmith
Quote:

Originally Posted by colorado
how about using dot product? haven't got to cross product yet

Have you ever taking a determinant of a matrix?
• Jan 11th 2011, 11:10 AM
Ackbeet
You could use the dot product by setting up a system of two equations in three unknowns. That is, your unknown vector is r=(x,y,z). Set the dot product of r with both your initial vectors equal to zero, and solve the resulting system.
• Jan 11th 2011, 11:10 AM
colorado
no, haven't gotten to that point
• Jan 11th 2011, 11:13 AM
colorado
Quote:

Originally Posted by Ackbeet
You could use the dot product by setting up a system of two equations in three unknowns. That is, your unknown vector is r=(x,y,z). Set the dot product of r with both your initial vectors equal to zero, and solve the resulting system.

yeah, I tried that, thats how I've done it before, didn't work out. perhaps its time for a math break.
• Jan 11th 2011, 11:17 AM
mr fantastic
Quote:

Originally Posted by colorado
yeah, I tried that, thats how I've done it before, didn't work out. perhaps its time for a math break.

Please show what you did.

To get a unique answer, you will need to impose a third condition on the vector you're trying to find (eg. you want it to be a unit vector).