Question

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

• A. 15
• B. $\sqrt{113}$
• C. $\sqrt{593}$
• D. $\cos\theta = \frac{3}{19}$

Answer 1

Answer: (C)

$\sqrt{593}$

Answer 2

Length of the two diagonals will be

$d_{1} = |(5\mathbf{a} + 2\mathbf{b}) + (\mathbf{a} - 3\mathbf{b})|$ and $d_{2} = |(5\mathbf{a} + 2\mathbf{b}) - (\mathbf{a} - 3\mathbf{b})|$

⇒ $d_{1} = |6\mathbf{a} - \mathbf{b}|$, $d_{2} = |4\mathbf{a} + 5\mathbf{b}|$

Thus, $d_{1} = \sqrt{|6\mathbf{a}|^{2} + | - \mathbf{b}|^{2} + 2|6\mathbf{a}|| - \mathbf{b}|\cos(\pi - \pi/4)}$

= $\sqrt{36(2\sqrt{2})^{2} + 9 + 12.2\sqrt{2}.3.\left( - \frac{1}{\sqrt{2}} \right)}$ = 15.

$d_{2} = \sqrt{|4\mathbf{a}|^{2} + |5\mathbf{b}|^{2} + 2|4\mathbf{a}||5\mathbf{b}|\cos\frac{\pi}{4}}$

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

$\therefore$ Length of the longer diagonal = $\sqrt{593}$