Question

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

• A. $2\sqrt{16-(\overrightarrow{a}.\overrightarrow{b}{)}^2}$
• B. $2\sqrt{4-(\overrightarrow{a}.\overrightarrow{b}{)}^2}$
• C. $\sqrt{16-(\overrightarrow{a}.\overrightarrow{b}{)}^2}$
• D. $\sqrt{4}-(\overrightarrow{a}.\overrightarrow{b}{)}^2$

Answer 1

The correct option is A $2\sqrt{16-(\overrightarrow{a}.\overrightarrow{b}{)}^2}$

For vectors, $a,b$

${|\overrightarrow{a}\times \overrightarrow{b}|}^2={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{|\overrightarrow{a}.\overrightarrow{b}|}^2$

Now, let angle btw a and b be $\theta$

$|\overrightarrow{u}.\overrightarrow{v}|=|(\overrightarrow{a}-\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})|={\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=0$

${\left|\overrightarrow{u}\right|}^2=8-8cos\theta$

${\left|\overrightarrow{v}\right|}^2=8+8cos\theta$

${|\overrightarrow{u}\times \overrightarrow{v}|}^2={\left|\overrightarrow{u}\right|}^2{\left|\overrightarrow{v}\right|}^2-{|\overrightarrow{u}.\overrightarrow{v}|}^2$

$|\overrightarrow{u}\times \overrightarrow{v}|=\sqrt{(8-8cos\theta )(8+8cos\theta )-0}=\sqrt{64-64co{s}^2\theta }=2\sqrt{16-(\overrightarrow{a}.\overrightarrow{b}{)}^2}$

Hence, option A is correct.