Find all the unit vectors in the plane determined by u=(3,0,1) and v=(1,-1,1) that are perpendicular to the vector w=(1,2,0).
Why can't I solve this using the following method (and what is the best method using the cross product)?
$ \mathbf{y} = a.\mathbf{u} + b.\mathbf{v} \qquad (a,b\in \Bbb{R}) $
y gives me any point on the plane.
$ c.\mathbf{w} = \mathbf{v} \times \mathbf{y} \qquad (c\in \Bbb{R}) $
Cross product gives me the perpendicular vector.
Find y such that a,b solve the above equation then divide by ||y|| to get $ \mathbf{\hat y} $. I have tried it 3 times and each time I arrive at the nonsense answer.
Note that $\vec v \times \vec y$ is perpendicular to the plane but $w$ is not thus your condition can’t have solution.
We don't need cross product but dot product.
Indeed note that the genric vector in the plane determined by $\vec u$ and $\vec v$ are given by
$$\vec x= a\vec u + b \vec v $$
with $a,b \in \mathbb{R}$.
By perpendicularity condition $\vec x \cdot \vec w=0$ we obtain
$$\vec x \cdot \vec w=0\vec x \cdot \vec w =a\vec u \cdot \vec w+b\vec v \cdot \vec w=0\iff3a-b=0 \iff b=3a$$
thus all vectors in the plane determined by $\vec u$ and $\vec v$ and perpendicular to $\vec w$ are
$$\vec x= a\vec u + 3a \vec v=a(6,-3,4)$$
and the corresponding unit vectors are
$$|\vec x|=a\sqrt{36+9+16}=a\sqrt{61}$$
$$\vec y=\pm \frac{\vec x}{|\vec x|}=\pm\frac{1}{\sqrt{61}}(6,-3,4)$$