Question

If $|\overrightarrow{C}{|}^2=60$ and $\overrightarrow{C}\times (\hat{i}+2\hat{j}+5\hat{k})=\overrightarrow{0}$, then a value of $\overrightarrow{C}\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$ is :

• A. $4\sqrt{2}$
• B. $24$
• C. $12\sqrt{2}$
• D. $12$

Answer 1

Answer: (C)

$12\sqrt{2}$

$\overrightarrow{C}$ is parallel to $\hat{i}+2\hat{j}+5\hat{k}$.
$\Rightarrow \overrightarrow{C}=t(\hat{i}+2\hat{j}+5\hat{k})$, where '$t$' is a scalar.
$|\overrightarrow{C}{|}^2=60$
$\Rightarrow t^2(\hat{i}.\hat{i}+4\hat{j}.\hat{j}+25\hat{k}.\hat{k})=60$
$\Rightarrow 60=t^2\times 30$
$\Rightarrow t=\pm \sqrt{2}$

Now
$\overrightarrow{C}.(-7i+2\hat{j}+3\hat{k})=\pm \sqrt{2}(i+2\hat{j}+5\hat{k})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$
$=\pm \sqrt{2}[-7+4+15]$
$=\pm 12\sqrt{2}$