Question

If $\left|\overline{a}\right|=\left|\overline{b}\right|=2$, then ${(\overline{a}\times \overline{b})}^2+{(\overline{a}.\overline{b})}^2=$

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

Answer: (D)

$16$

$|a|=|b|=2$

$(\overline{a}\times \overline{b}{)}^2+(\overline{a}\cdot \overline{b}{)}^2$

$|a{|}^2|b{|}^2{\sin }^2\theta +|a{|}^2|b{|}^2{\cos }^2\theta$

$|a{|}^2|b{|}^2({\sin }^2\theta +{\cos }^2\theta )$

$|a{|}^2|b{|}^2=2^2\cdot 2^2=2^4=16$.