Question

The length of the longer diagonal of the parallelogram constructed on $5\overrightarrow{a}+2\overrightarrow{b}$ and $\overrightarrow{a}-3\overrightarrow{b}$, if it is given that $\left|\overrightarrow{a}\right|=2\sqrt{2},\left|\overrightarrow{b}\right|=3$ and the angle between $\overrightarrow{a}$ and $\overrightarrow{b}=\frac{\pi }{4}$ is

• A. $15$
• B. $\sqrt{113}$
• C. $\sqrt{593}$
• D. $\sqrt{369}$

Answer 1

Answer: (C)

$\sqrt{593}$

The diagonals of the parallelogram are

$\overrightarrow{\alpha }=5\overrightarrow{a}+2\overrightarrow{b}+\overrightarrow{a}-3\overrightarrow{b}=6\overrightarrow{a}-\overrightarrow{b}$

$\overrightarrow{\beta }=\pm (4\overrightarrow{a}+5\overrightarrow{b})$

Now $\left|\overrightarrow{\alpha }\right|=|6\overrightarrow{a}-\overrightarrow{b}|=\sqrt{36{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-12(\overrightarrow{a}.\overrightarrow{b})}$

$=\sqrt{36\times 8+9-12\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}=15$

and $\left|\overrightarrow{\beta }\right|=|4\overrightarrow{a}+5\overrightarrow{b}|=\sqrt{16{\left|\overrightarrow{a}\right|}^2+25{\left|\overrightarrow{b}\right|}^2+40(\overrightarrow{a}.\overrightarrow{b})}$

$=\sqrt{16\times 8+25\times 9+40\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}=\sqrt{593}$