Question

A vector of magnitude $12$ unit perpendicular to the plane containing the vectors $4\hat{i}+6\hat{j}-\hat{k}$ and $3\hat{i}+8\hat{j}+\hat{k}$ is

• A. $-8\hat{i}+4\hat{j}+8\hat{k}$
• B. $8\hat{i}+4\hat{j}+8\hat{k}$
• C. $8\hat{i}-4\hat{j}+8\hat{k}$
• D. $8\hat{i}-4\hat{j}-8\hat{k}$
• E. $4i-8\hat{j}-8\hat{k}$

Answer 1

Answer: (C)

Let $\overrightarrow{a}=4\hat{i}+6\hat{j}-\hat{k}$ and $\overrightarrow{b}=3\hat{i}+8\hat{j}+\hat{k}$
$\therefore$ $\overrightarrow{c}=|\overrightarrow{a}\times \overrightarrow{b}|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6& -1\\ 3& 8& 1\end{array}\right|$
$=14\hat{i}-7\hat{j}+14\hat{k}$
$\Rightarrow$ $\hat{c}=\frac{14\hat{i}-7\hat{j}+14\hat{k}}{21}$
$\Rightarrow$ $\overrightarrow{d}=\hat{c}\times 12$
$=12.\frac{(14\hat{i}-7\hat{j}+14\hat{k})}{21}$
$=8\hat{i}-4\hat{j}+8\hat{k}$