Question

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

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

Answer 1

Answer: (D)

$\bar{C}\times \left(\hat{i}+2\hat{j}+5\hat{k}\right)=0$
$\bar{C}=\lambda \left(\hat{i}+2\hat{j}+5\hat{k}\right)$
$\left|C\right|=\lambda \sqrt{30}\Rightarrow {\lambda }^2\left(30\right)={\left|c\right|}^2=60$
$\lambda =\pm \sqrt{2}$
$\Rightarrow \bar{C}.\left(-7\hat{i}+2\hat{j}+3\hat{k}\right)$
$\Rightarrow \lambda \left(\hat{i}+2\hat{j}+5\hat{k}\right).\left(-7\hat{i}+2\hat{j}+3\hat{k}\right)$
$\Rightarrow \lambda \left(-7+4+15\right)=12\lambda$
$=12\sqrt{2}$ or $-12\sqrt{2}$