Question

The length of 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 $\overrightarrow{b}$ is $\frac{\pi }{4}$ is?

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

Answer 1

The correct option is C

$\sqrt{593}$

Explanation for correct option:

Given that,

$\left|\overrightarrow{a}\right|=2\sqrt{2}$

$\left|\overrightarrow{b}\right|=3$

As per the given condition, the length of the two diagonals are

$$\begin{gathered} D_1=\left|\left(5\overrightarrow{a}+2\overrightarrow{b}\right)+\left(\overrightarrow{a}-3\overrightarrow{b}\right)\right|=\left|6\overrightarrow{a}-\overrightarrow{b}\right| \\ {D}_2=\left|\left(5\overrightarrow{a}+2\overrightarrow{b}\right)-\left(\overrightarrow{a}-3\overrightarrow{b}\right)\right|=\left|4\overrightarrow{a}+5\overrightarrow{b}\right| \end{gathered}$$

Length of $D_1$ IS

$$\begin{gathered} D_1=\left|6\overrightarrow{a}-\overrightarrow{b}\right| \\ =\sqrt{\left(6\overrightarrow{a}-\overrightarrow{b}\right)\left(6\overrightarrow{a}-\overrightarrow{b}\right)} \\ =\sqrt{36{\left|a\right|}^2+{\left|\overrightarrow{b}\right|}^2-2·\left|a\right|·\left|\overrightarrow{b}\right|\cos \left(\frac{\pi }{4}\right)} \\ =\sqrt{36·8+9-2·2\sqrt{2}·3·\frac{1}{\sqrt{2}}} \\ =\sqrt{225} \\ =15 \\ {D}_2=\left|4\overrightarrow{a}+5\overrightarrow{b}\right| \\ =\sqrt{\left(4\overrightarrow{a}+5\overrightarrow{b}\right)\left(4\overrightarrow{a}+5\overrightarrow{b}\right)} \\ =\sqrt{16{\left|a\right|}^2+25{\left|\overrightarrow{b}\right|}^2+2·\left|a\right|·\left|\overrightarrow{b}\right|\cos \left(\frac{\pi }{4}\right)} \\ =\sqrt{16·8+25·3+2·2\sqrt{2}·3·\frac{1}{\sqrt{2}}} \\ =\sqrt{593} \end{gathered}$$

So the length of the longer diagonal is $\sqrt{593}$

Hence, the correct option is option C.