Question

Find 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)

$\sqrt{593}$

Length of the two diagonals will be

$\Rightarrow$ $d_1=|(5a+2b)+(a-3b)|$

$\Rightarrow$ $d_2=|(5a+2b)-(a-3b)|$

$\Rightarrow$ $d_1=|6a-b|$ and $\Rightarrow$ $d_2=|4a+5b|$

So,

$d_1=\sqrt{|6a{|}^2+|-b{|}^2+2|6a||-b|cos(\pi -\frac{\pi }{4})}$

$d_1=\sqrt{36(2\sqrt{2}{)}^2+3^2+12(2\sqrt{2}\left)3\right(-\frac{1}{\sqrt{2}})}$

$d_1=15$

$d_2=\sqrt{|4a{|}^2+|5b{|}^2+2|4a||5b|cos\left(\frac{\pi }{4}\right)}$

$d_2=\sqrt{16(2\sqrt{2}{)}^2+5\times 3^2+40\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}$

$d_2=\sqrt{593}>15$

$\therefore d_2>d_1$

Hence, length of longer diagonal is $\sqrt{593}$.