Question

find a vector whose magnitude is $7$ units and which is perpendicular to each of the vector $\overrightarrow{a}=2\hat{i}+3\hat{j}+6\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$

• A. $\frac{\hat{i}-8\hat{j}-5\hat{k}}{\sqrt{2}}$
• B. $\frac{3\hat{i}-\hat{j}-5\hat{k}}{\sqrt{2}}$
• C. $\frac{3\hat{i}-8\hat{j}-\hat{k}}{\sqrt{2}}$
• D. $\frac{3\hat{i}-8\hat{j}-5\hat{k}}{\sqrt{2}}$

Answer 1

Answer: (D)

$\frac{3\hat{i}-8\hat{j}-5\hat{k}}{\sqrt{2}}$

Given $\overrightarrow{A}=2\hat{i}-3\hat{j}+6\hat{k}\overrightarrow{B}=\hat{i}+\hat{j}-\hat{k}\overrightarrow{A}\times \overrightarrow{B}=-3\hat{i}+8\hat{j}+5\hat{k}|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{(-3{)}^2+(8{)}^2+(5{)}^2}=\sqrt{98}=7\sqrt{2}$

Therefore required vectors$\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}\times 7=\frac{-3\hat{i}+8\hat{j}+5\hat{k}}{7\sqrt{2}}\times 7=\frac{-3\hat{i}+8\hat{j}+5\hat{k}}{\sqrt{2}}$

If multiply this vector by $-1$ it would have same magnitude at some time and perpendicular to A and B. Hence

$\frac{3\hat{i}-8\hat{j}-5\hat{k}}{\sqrt{2}}$ is correct.