Finding a vector that's perpendicular to another given vector?

Question

So I have the vector

$\vec{u}=\left<0.82,6.45,-1.07\right>$

and I need to find a vector $\vec{v}$ that's perpendicular to this one. The given hint is that I need to use "projection onto $\vec{u}$", which I presume is referring to the vector projection, to find $\vec{v}$. I don't see how this hint, or any other way, could help me find such a vector.

Answer 1

For any non-zero vector $\vec u=\langle a,b,c \rangle$, at least two of $\langle 0,c,-b \rangle$, $\langle -c,0,a \rangle$ and $\langle b,-a,0 \rangle$ are also non-zero. (These are the cross products with the standard basis vectors.) You can easily verify that all three are orthogonal to $\vec u$.

However, the hint wants you to use projection onto $\vec u$, so you basically need to perform a step of the Gram-Schmidt process: pick any non-zero vector $\vec v$ that’s not a multiple of $\vec u$, compute its projection onto $\vec u$ and then subtract that from $\vec v$. What’s left—the orthogonal rejection of $\vec v$ from $\vec u$—is orthogonal to $\vec u$. I.e., compute $$\vec v - {\vec v\cdot \vec u \over \vec u\cdot\vec u}\vec u.$$

Answer 2

$\vec{v}(a,b,c)\perp\vec{u}$ says $0.82a+6.45b-1.07c=0.$

Now, take $c=0$, $b=-0.82$ and $a=6.45.$

Answer 3

Every 3D vector $\vec{v}=\left<a,b,c\right>$ which is peerpendicular to $\vec{u}=\left<0.82,6.45,-1.07\right>$ have the property $u.v=0$ so $$0.82a+6.45b-1.07c=0$$ this is the equation of a plane and also $c=\frac{0.82a+6.45b}{1.07}$, then with arbitrary $a$ and $b$ you can find $c$ and hence $v$.