Question

If $\vec{u}=\vec{a}-\vec{b},\vec{v}=\vec{a}+\vec{b},|\vec{a}|=|\vec{b}|=2$ then $|\vec{u}\times \vec{v}|$ is equal to

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

Answer 1

Answer: (A)

We have, $\vec{u}=\vec{a}-\vec{b},\vec{v}=\vec{a}+\vec{b}$
$\Rightarrow \vec{u}\times \vec{v}=(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})$
$=0-\vec{b}\times \vec{a}+\vec{a}\times \vec{b}-0$
$=2\vec{a}\times \vec{b}$
$\Rightarrow |\vec{u}\times \vec{v}|=2|\vec{a}\times \vec{b}|$
$=2\sqrt{|\vec{a}\times \vec{b}{|}^2}$
$=2\sqrt{|\vec{a}{|}^2|\vec{b}{|}^2}{\sin }^2\theta |\hat{n}{|}^2$
$\{\because \hat{n}=$ unit vector $|\hat{n}|=1\}$
$=2\sqrt{4\cdot 4\cdot {\sin }^2\theta \cdot 1}$
$=2\sqrt{16\left(1-{\cos }^2\theta \right)}$
$=2\sqrt{16-16{\cos }^2\theta }$
$=2\sqrt{16-16{\left(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)}^2}$
$\because \{\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta \}$
$=2\sqrt{16-16\frac{(\vec{a}\cdot \vec{b}{)}^2}{|\vec{a}{|}^2|\vec{b}{|}^2}}$
$=2\sqrt{16-16\frac{(\vec{a}\cdot \vec{b}{)}^2}{4\cdot 4}}$
$\Rightarrow 2\sqrt{16-(\vec{a}\cdot \vec{b}{)}^2}$