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{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 4 & 6 & -1 \\ 3 & 8 & 1 \\ \end{matrix} \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} $