Question

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonal of a parallelogram having area $2\sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. $|\vec{a}|=1$ and $|\vec{a}\cdot \vec{b}|=|\vec{a}\times \vec{b}|$. If $\vec{c}=2\sqrt{2}(\vec{a}\times \vec{b})-2\vec{b}$, then an angle between $\vec{b}$ and $\vec{c}$ is :

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{4}$
• C. $\frac{5\pi }{6}$
• D. $\frac{3\pi }{4}$

Answer 1

Answer: (D)

Area $=\frac{1}{2}|\vec{a}\times \vec{b}|=2\sqrt{2}$
$\Rightarrow |\vec{a}\times \vec{b}|=4\sqrt{2}$
$|\vec{a}|=1\text{ and }|\vec{a}\cdot \vec{b}|=|\vec{a}\times \overrightarrow{b}|$
$\Rightarrow \cos \theta =\sin \theta$
$\Rightarrow \theta =\frac{\pi }{4}$
$\therefore |\vec{a}\times \vec{b}|=4\sqrt{2}$
$\Rightarrow |\vec{a}||\vec{b}|\sin \frac{\pi }{4}=4\sqrt{2}$
$\Rightarrow |\vec{b}|=8$
Now, $\vec{c}=2\sqrt{2}(\vec{a}\times \vec{b})-2\vec{b}$
$|\vec{c}|=\sqrt{(2\sqrt{2}{)}^2|\vec{a}\times \vec{b}{|}^2+(2|\vec{b}|{)}^2}=16\sqrt{2}$
Now, $\vec{b}\cdot \vec{c}=-2|\vec{b}{|}^2$
$\Rightarrow 8\times 16\sqrt{2}\times \cos \alpha =-2.64$
$\Rightarrow \cos \alpha =-\frac{1}{\sqrt{2}}$
$\Rightarrow \alpha =\frac{3\pi }{4}$