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, $\vec{ b } \cdot \vec{ c }=-2|\vec{ b }|^{2}$
$\Rightarrow 8 \times 16 \sqrt{2} \times \cos \alpha=-2.64$
$\Rightarrow \cos \alpha=-\frac{1}{\sqrt{2}}$
$\Rightarrow \alpha=\frac{3 \pi}{4}$