Finding an angle between two vectors

Question

I am trying to answer part $d)$ by using my answer to part $c)$. From what I can see, the only possible way to do this is to find the lenght of $AB$ and $OB$, and, using the angle in part $c)$, apply the sine rule to find angle $OAB$. Are there are any quicker ways to find angle $OAB$ by using my answer to part $c)$?

Answer 1

You don't need to use the answer to part C. The vector from $A$ to $B$ is $$\mathbf{u}=\langle6-2,4-4,2-6\rangle=\langle4,0,-4\rangle$$ and the vector from $A$ to $O$ is $$\mathbf{v}=\langle-2,-4,-6\rangle$$ The angle between these is $$\theta=\arccos \left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\right)=\arccos \left(\frac{16}{4\sqrt{2}\times2\sqrt{14}}\right)=\arccos\left(\frac{1}{\sqrt{7}}\right)\approx67.8^\circ$$

Answer 2

The questioner is right to suspect that it must be quite easy:

Length $OA$ = length $OB$.

Triangle $AOB$ is isosceles.

Now use the angle you have. So it is "hence"