Question

Let $\overrightarrow a$ and $\overrightarrow b$ be the vectors along the diagonals of a parallelogram having area $2\sqrt2$. Let the angle between $\overrightarrow a$ and $\overrightarrow b$ be acute,$|\overrightarrow a|=1$, and $|\overrightarrow a⋅\overrightarrow b|=|\overrightarrow a×\overrightarrow b|$ If $\overrightarrow c=2\sqrt2(\overrightarrow a×\overrightarrow b→)−2\overrightarrow b$ then an angle between$\overrightarrow b$ and $\overrightarrow c$ is

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

Answer 1

Answer: (D)

$\frac{3π}{4}$

Answer 2

$∵ \overrightarrow a$ and $\overrightarrow b$ be the vectors along the diagonals of a parallelogram having area $2\sqrt2$.

$∴\frac{1}{2}|\overrightarrow a×\overrightarrow b|=2\sqrt2$

$|\overrightarrow a||\overrightarrow b|sin⁡θ=4\sqrt2$

$⇒|\overrightarrow b|sin⁡θ=4\sqrt2…(i)$

and

$|\overrightarrow a⋅\overrightarrow b|=|\overrightarrow a×\overrightarrow b|$

$|\overrightarrow a||\overrightarrow b|cos⁡θ=|\overrightarrow a||\overrightarrow b|sin⁡θ$

$⇒tan⁡θ=1\;∴θ=\frac{π}{4}$

By (i)

$|\overrightarrow b|=8$

Now

$\overrightarrow c=2\sqrt2(\overrightarrow a×\overrightarrow b)−2\overrightarrow b$

$⇒\overrightarrow c⋅\overrightarrow b=−2|\overrightarrow b|2=−128…(ii)$

and

$\overrightarrow c⋅\overrightarrow c=8|\overrightarrow a×\overrightarrow b|2+4|\overrightarrow b|2$

$⇒|\overrightarrow c|2=8.32+4.64$

$⇒|\overrightarrow c|=16\sqrt2…(iii)$

From (ii) and (iii)

$|\overrightarrow c||\overrightarrow b|cos⁡α=−128$

$⇒cos⁡α=\frac{−1}{\sqrt2}$

$α=\frac{3π}{4}$