Question

If $\overrightarrow{a}=2\hat{i}+2\hat{j}+\hat{k}$, $\overrightarrow{a}.\overrightarrow{b}=14$, $\overrightarrow{a}\times \overrightarrow{b}=3\hat{i}+\hat{j}-8\hat{k}$, then $b=$

• A. $5\hat{i}-\hat{j}+2\hat{k}$
• B. $5\hat{i}+\hat{j}-\hat{k}$
• C. $5\hat{i}+\hat{j}+2\hat{k}$
• D. $5\hat{i}+\hat{j}-2\hat{k}$

Answer 1

Answer: (C)

$5\hat{i}+\hat{j}+2\hat{k}$

$(\overrightarrow{a}\times \overrightarrow{b}{)}^2+(\overrightarrow{a}-\overrightarrow{b}{)}^2=a^2{b}^2$
$74+196=9\times b^2$
$\sqrt{30}=|b|$
Let $\hat{b}=a\hat{i}+b\hat{j}+c\hat{k}$
$\hat{a}-\hat{b}=14$
$2a+2b+c=14$ ------(1)
$a^2+b^2+c^2=30$ -------(2)
$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& 1\\ a& b& c\end{array}\right|=(2c-b)\hat{i}+(a-2c)\hat{j}+(2b-2a)\hat{k}$
$2c-b=3;$ $a-2c=1;$ $b-a=-4$ -------(3)
solve $\left(1\right),\left(2\right)$ & $\left(3\right)$
$a=5;$ $b=1;$ $c=2$