Question

If $\overline{u}=\overline{a}+\overline{b},\overline{v}=\overline{a}-\overline{b},|\overrightarrow{a}|=|\overline{b}|=2$, then $|\overline{u}\times \overline{v}|=$

• A. $\sqrt{16-(\overline{a}\overline{b}{)}^2}$
• B. $2\sqrt{16-(\overline{a}\overline{b}{)}^2}$
• C. $3\sqrt{16-(\overline{a}\overline{b}{)}^2}$
• D. $0$

Answer 1

Answer: (B)

$2\sqrt{16-(\overline{a}\overline{b}{)}^2}$

$\overrightarrow{U}\times \overrightarrow{V}=(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})$
$=-\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{a}$
$=2\overrightarrow{b}\times \overrightarrow{a}$
$|\overrightarrow{U}\times \overrightarrow{V}|=2|\overrightarrow{b}\times \overrightarrow{a}|({|\overrightarrow{a}\times \overrightarrow{b}|}^2+(\overrightarrow{a}-\overrightarrow{b}{)}^2=\overrightarrow{a}\overrightarrow{b})$
$=2\sqrt{{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-(\overrightarrow{a}-\overrightarrow{b}{)}^2}$
$=2\sqrt{16-(\overrightarrow{a}-\overrightarrow{b}{)}^2}$