Question

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

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

Answer 1

Answer: (C)

Given : $\left| a\right| = 2\sqrt{2}, \left| b\right| = 3 $
One diagonal is $5a + 2b + a - 3b = 6a - b$
Length of one diagonal
$= \left|6a - b\right| $
$ = \sqrt{36a^{2} + b^{2} - 2 \times6 \left| a\right|.\left| b\right| .\cos45^{\circ}} $
$ = \sqrt{36 \times8+9-12 \times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}$
$ = \sqrt{288 + 9 -12 \times 6 } = \sqrt{225} = 15 $
Other diagonal is $4 a +5 b$. Its length is
$ = \sqrt{\left(4a\right)^{2} + \left(5b^{2}\right) + 2 \times \left|4a\right|\left|5b\right|\cos 45^{\circ }} $
$ = \sqrt{16 \times8+ 25 \times9+40 \times6} = \sqrt{593 }$