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
$=\left|6a-b\right|$
$=\sqrt{36a^2+b^2-2\times 6\left|a\right|.\left|b\right|.\cos 45^{\circ }}$
$=\sqrt{36\times 8+9-12\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}$
$=\sqrt{288+9-12\times 6}=\sqrt{225}=15$
Other diagonal is $4a+5b$. Its length is
$=\sqrt{{\left(4a\right)}^2+\left(5b^2\right)+2\times \left|4a\right|\left|5b\right|\cos 45^{\circ }}$
$=\sqrt{16\times 8+25\times 9+40\times 6}=\sqrt{593}$