Answer to Question #245551 in Analytic Geometry for moe

Question

You are given the two vectors $\mathbf{a}=(1,2,1)$ and $\mathbf{b}=(3,-4,2)$ . Suppose that the pair of unit vectors perpendicular to both a and b are given by

$\pm \frac{1}{\sqrt{p}}(q,1,r),$

where p, q and r are some constants.

Determine the values of p, q and r.

Answer 1

$c=\bold{a}\times\bold{b}=\begin{vmatrix} i & j & k \\ 1 & 2 & 1\\ 3 & -4 & 2 \end{vmatrix}\\ =i(2\cdot2-(-4)\cdot1)-j(1\cdot2-1\cdot3)+k(1\cdot(-4)-3\cdot2)=8i+j-10k\\ =\dfrac{1}{\sqrt{p}}(q,1,r)\\ 8=\dfrac{q}{\sqrt{p}}\\ 1=\dfrac{1}{\sqrt{p}}\\ \therefore p=1\\ q=8\sqrt{1}=8\\ -10=\dfrac{r}{\sqrt{p}}\\ r=-10\sqrt{1}=-10$