# Thread: finding a unit vector orthogonal to two other vectors U and V

1. ## finding a unit vector orthogonal to two other vectors U and V

The exercise: find a unit vector orthogonal to both U= (1,1,0) and V = (0,1,1).

My question is why is it true that because (U dot X) = x_1 + x_3 and (V dot X) is equal to x_2 + x_3 that both U and V are orthogonal to X?

2. ## Re: finding a unit vector orthogonal to two other vectors U and V

Originally Posted by kingsolomonsgrave
The exercise: find a unit vector orthogonal to both U= (1,1,0) and V = (0,1,1).
My question is why is it true that because (U dot X) = x_1 + x_3 and (V dot X) is equal to x_2 + x_3 that both U and V are orthogonal to X?
Do you understand how to find the dot product of vectors?

Do you understand why $U\cdot V=1~?$

3. ## Re: finding a unit vector orthogonal to two other vectors U and V

yup, I know that (U dot V) is = 1 because U dot V = (u1 * v1 + u2 * v2 +...+uN * vN)

I think it see the problem more clearly now: Is it that we create a vector x and dot multiply it by both vectors U and V and then pick values that will give the dot product between X and V and X and V = 0 for both cases?

In this case x_1 =x_2 = -x_3 works to make both dot products zero so we can create X = (1, 1, -1).

If this is the correct procedure, is there a general way of creating an orthogonal vector, or is it done by inspection?

4. ## Re: finding a unit vector orthogonal to two other vectors U and V

Hello, kingsolomonsgrave!

$\text{Find a unit vector orthogonal to both }\vec u= \langle1,1,0\rangle\text{ and }\vec v = \langle0,1,1\rangle.$

Why isn't this basic theorem being used?

A vector $\vec n$ orthogonal to $\vec u$ and $\vec v$ is given by: . $\vec u \times \vec v$

Hence: . $\vec n \;=\;\begin{vmatrix} i&j&k \\ 1&1&0 \\ 0&1&1\end{vmatrix} \;=\;i(1-0) - j(1-0) + k(1-0) \;=\;i - j + k \;=\;\langle1,\text{-}1,1\rangle$

A unit vector is: . $\vec U_1 \:=\:\left\langle \tfrac{1}{\sqrt{3}},\:\tfrac{-1}{\sqrt{3}},\:\tfrac{1}{\sqrt{3}}\right\rangle$

Another is its negative: . $\vec U_2 \:=\:\left\langle \tfrac{-1}{\sqrt{3}},\:\tfrac{1}{\sqrt{3}},\:\tfrac{-1}{\sqrt{3}}\right\rangle$