[Math] Perpendicular unit vector

Question

$$x = 4i + 2j − 3k$$
I need to find a unit vector perpendicular to $x$.

I calculated and got the answer $\sqrt{1/5}*i – 2\sqrt{1/5}*j$

Is that correct?
I assumed $k = 0$, so I could solve equation.

Answer 1

You want to find a unit vector $ Y= ai + bj + ck $ such that $X\cdot Y = 0.$ That is; $ 4a + 2b -3c = 0 . $ Clearly we can make some arbitrary choices here since many combinations of $a,b,c$ satisfy that requirement. It appears you chose $c=0$ (not the same as $k=0$, $k$ is a predefined vector remember), so then $a=1,b=-2.$

Thus you've found that $ i - 2j $ is a vector perpendicular to $X.$ Now you want to make it a unit vector, so divide by the length of the vector, which is $ \sqrt{ 1^2 + (-2)^2} = \sqrt{5}.$ Thus the required vector is $ Y = \frac{i}{\sqrt{5}} - \frac{2}{\sqrt{5}} j $ as you found.