A unit vector perpendicular to the vectors $a=2i-3j+k$ and $b=i+j-2k$, is
- A. $\frac{1}{\sqrt{3}}(-i+j+k)$
- B. $\frac{1}{\sqrt{3}}(i+j-k)$
- C. $\frac{1}{\sqrt{3}}(i+j+k)$
- D. $(i+j+k)$
Answer: C. $\frac{1}{\sqrt{3}}(i+j+k)$
Explanation: $\vec{a} \times \vec{b} = \begin{vmatrix}i&j&k\\2&-3&1\\1&1&-2\end{vmatrix} = i(6-1) - j(-4-1) + k(2+3) = 5i + 5j + 5k$. Unit vector perpendicular to both is $\frac{5(i+j+k)}{\sqrt{5^2+5^2+5^2}} = \frac{1}{\sqrt{3}}(i+j+k)$.