Topic: Engineering Mathematics
A unit vector perpendicular to the vectors $\vec{a}=2i-3j+k$ and $\vec{b}=i+j-2k$, 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: B. $\frac{1}{\sqrt{3}}(i+j+k)$
Explanation: A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is their cross product $\vec{c} = \vec{a} \times \vec{b}$. $\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 1 & -2 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(-4-1) + \hat{k}(2-(-3)) = 5\hat{i} + 5\hat{j} + 5\hat{k}$. Its magnitude is $|\vec{c}| = \sqrt{5^{2} + 5^{2} + 5^{2}} = \sqrt{75} = 5\sqrt{3}$. The unit vector is $\frac{\vec{c}}{|\vec{c}|} = \frac{5(\hat{i} + \hat{j} + \hat{k})}{5\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.