Question

Let $\overrightarrow{c}$ be a vector perpendicular to the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}.$ If $\overrightarrow{c}\cdot (\hat{i}+\hat{j}+3\hat{k}),$ then the value of $\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})$ is equal to

• A. 28.00
• B. 28
• C. 28.0

Answer 1

Answer: (A), (B), (C)

$\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})=\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]$
$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& 2& 1\end{array}\right|=(3,-2,1)$

Given, $\overrightarrow{c}\perp \overrightarrow{a}$ and $\overrightarrow{c}\perp \overrightarrow{b}$
$\Rightarrow c||\overrightarrow{a}\times \overrightarrow{b}$
$\Rightarrow \overrightarrow{c}=\lambda (\overrightarrow{a}\times \overrightarrow{b})$
$\Rightarrow \overrightarrow{c}=\lambda (3\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k})$

$\overrightarrow{c}\cdot (\hat{i}+\hat{j}+3\hat{k})=8$
$\Rightarrow 3\lambda -2\lambda +3\lambda =8$
$\Rightarrow 4\lambda =8\Rightarrow \lambda =2$
$\overrightarrow{c}=6\hat{i}-4\hat{j}+2\hat{k}$
$\overrightarrow{c}\cdot (\overrightarrow{a}\times \overrightarrow{b})=\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]$
$=\left|\begin{array}{ccc}6& -4& 2\\ 1& 1& -1\\ 1& 2& 1\end{array}\right|=18+8+2=28$