Q44 / 120 · VITEEE 2014 · Mathematics
The length of longer diagonal of the parallelogram constructed on $5 a+2 b$ and $a-3 b$, if it is given that $|\mathrm{a}|=2 \sqrt{2},|\mathrm{~b}|=3$ and the angle between $a$ and $b$ is $\frac{\pi}{4}$, is
- A15
- B$\sqrt{113}$
- Correct answerC$\sqrt{593}$
- D$\sqrt{369}$
Solution
Given : $|\mathbf{a}|=2 \sqrt{2},|\mathbf{b}|=3$
One diagonal is $5 \mathbf{a}+2 \mathbf{b}+\mathbf{a}-3 \mathbf{b}=6 \mathbf{a}-\mathbf{b}$
Length of one diagonal
$\begin{array}{l}
=|6 \mathbf{a}-\mathbf{b}| \\
=\sqrt{36 \mathbf{a}^{2}+\mathbf{b}^{2}-2 \times 6|\mathbf{a}| \cdot|\mathbf{b}| \cdot \cos 45^{\circ}} \\
=\sqrt{36 \times 8+9-12 \times 2 \sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}} \\
=\sqrt{288+9-12 \times 6}=\sqrt{225}=15
\end{array}$
other diagonal is $(5 \mathbf{a}+2 \mathbf{b})-(\mathbf{a}-3 \mathbf{b})$ $=4 \mathbf{a}+5 \mathbf{b}$
Its length is
$\begin{array}{l}
=\sqrt{(4 a)^{2}+\left(5 b^{2}\right)+2 \times|4 a||5 b| \cos 45^{\circ}} \\
=\sqrt{16 \times 8+25 \times 9+40 \times 6}=\sqrt{593}
\end{array}$
Hence, the correct answer is option: 'C'.