Question

If $\overline{a}=2\overrightarrow{i}+2\overrightarrow{j}+\overline{k},\overline{b}=5\overrightarrow{i}+\overline{j}+2\overline{k}$ then
$|\overline{a}\times \overline{b}{|}^2+(\overline{a}\cdot \overline{b}{)}^2=$

• A. $270$
• B. $120$
• C. $170$
• D. $110$

Answer 1

The correct option is A $270$

$\overrightarrow{a}=2\hat{i}+2\hat{j}+\hat{k}$ and

$\overrightarrow{b}=5\hat{i}+\hat{j}+2\hat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& 1\\ 5& 1& 2\end{array}\right|$

$=\hat{i}(4-1)-\hat{j}(4-5)+\hat{k}(2-10)$

$=3\hat{i}+\hat{j}-8\hat{k}$

$|\overrightarrow{a}\times \overrightarrow{b}{|}^2=3^2+1^2+(-8{)}^2$

$=9+1+64=74$

$(\overrightarrow{a}\cdot \overrightarrow{b})=(2\hat{i}+2\hat{j}+\hat{k})\cdot (5\hat{i}+\hat{j}+2\hat{k})$

$=\left(2.5\right)+\left(2.1\right)+\left(1.2\right)$

$=10+2+2=14$

$(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2={14}^2=196$

$|\overrightarrow{a}\times \overrightarrow{b}{|}^2+(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2=74+196=270$.