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

<p>$\because$ $\overrightarrow a$ and $\overrightarrow b$ be the vectors along the diagonals of a parallelogram having area 2$\sqrt2$.</p> <p>$\therefore$ ${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2$</p> <p>$|\overrightarrow a ||\overrightarrow b |\sin \theta = 4\sqrt 2$</p> <p>$\Rightarrow |\overrightarrow b |\sin \theta = 4\sqrt 2$ ..... (i)</p> <p>and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$$</p> <p>$$|\overrightarrow a ||\overrightarrow b |\cos \theta = |\overrightarrow a ||\overrightarrow b |\sin \theta $$</p> <p>$\Rightarrow \tan \theta = 1$</p> <p>$\therefore$ $\theta = {\pi \over 4}$</p> <p>By (i) $|\overrightarrow b | = 8$</p> <p>Now $$\overrightarrow c = 2\sqrt 2 (\overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b $$</p> <p>$$ \Rightarrow \overrightarrow c \,.\,\overrightarrow b = - 2|\overrightarrow b {|^2} = - 128$$ ...... (ii)</p> <p>and $$\overrightarrow c \,.\,\overrightarrow c = 8|\overrightarrow a \times \overrightarrow b {|^2} + 4|\overrightarrow b {|^2}$$</p> <p>$\Rightarrow |\overrightarrow c {|^2} = 8.32 + 4.64$</p> <p>$\Rightarrow |\overrightarrow c | = 16\sqrt 2$ ..... (iii)</p> <p>From (ii) and (iii)</p> <p>$|\overrightarrow c ||\overrightarrow b |\cos \alpha = - 128$</p> <p>$\Rightarrow \cos \alpha = {{ - 1} \over {\sqrt 2 }}$</p> <p>$\alpha = {{3\pi } \over 4}$</p>