Question

What is the length of the longer diagonal of the parallelogram constructed on \[5 \vec{a} + 2 \vec{b} \text{ and } \vec{a} - 3 \vec{b}\] if it is given that \[\left| \vec{a} \right| = 2\sqrt{2}, \left| \vec{b} \right| = 3\] and the angle between \[\vec{a} \text{ and } \vec{b}\] is π/4? 

पर्याय

• 15 

•  \[\sqrt{113}\] 

•  \[\sqrt{593}\] 

•  \[\sqrt{369}\] 

Answer 1

 \[\sqrt{593}\] 

\[\text{ Let } \text{ ABCD }\text{ be a parallelogram in which } \]
\[\text{ side } \vec{AB} = \vec{DC} = 5 \vec{a} + 2 \vec{b} \]
\[\text{ and } \vec{AD} = \vec{BC} = \vec{a} - 3 \vec{b} \]
\[\text{ and diagonals are AC and BD } . \]
\[\text{ Now }, \vec{AC} = \vec{AB} + \vec{BC} \]
\[ = \left( 5 \vec{a} + 2 \vec{b} \right) + \left( \vec{a} - 3 \vec{b} \right)\]
\[ = 6 \vec{a} - \vec{b} \]
\[ \therefore \left| \vec{AC} \right| = \left| 6 \vec{a} - \vec{b} \right|\]
\[ = \sqrt{\left| 6 \vec{a} \right|^2 + \left| \vec{b} \right|^2 - 2 \times \left| 6 \vec{a} \right| \times \left| \vec{b} \right|\cos\theta}\]
\[ = \sqrt{36 \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - 12 \times \left| \vec{a} \right| \times \left| \vec{b} \right|\cos\frac{\pi}{4}}\]
\[ = \sqrt{36 \left| 2\sqrt{2} \right|^2 + \left| 3 \right|^2 - 12 \times \left| 2\sqrt{2} \right| \times \left| 3 \right| \times \frac{1}{\sqrt{2}}}\]
\[ = \sqrt{288 + 9 - 72}\]
\[ = \sqrt{225} = 15 \text{ units }\]
\[ \vec{BD} = \vec{BA} + \vec{BD} \]
\[ = - \vec{AB} + \vec{BD} \]
\[ = - \left( 5 \vec{a} + 2 \vec{b} \right) + \left( \vec{a} - 3 \vec{b} \right)\]
\[ = - 4 \vec{a} - 5 \vec{b} \]
\[ \therefore \left| \vec{BD} \right| = \left| - 4 \vec{a} - 5 \vec{b} \right|\]
\[ = \left| 4 \vec{a} + 5 \vec{b} \right|\]
\[ = \sqrt{\left| 4 \vec{a} \right|^2 + \left| 5 \vec{b} \right|^2 + 2 \times \left| 4 \vec{a} \right| \times \left| 5 \vec{b} \right|\cos\theta}\]
\[ = \sqrt{16 \left| \vec{a} \right|^2 + 25 \left| \vec{b} \right|^2 + 40 \times \left| \vec{a} \right| \times \left| \vec{b} \right|\cos\frac{\pi}{4}}\]
\[ = \sqrt{16 \left| 2\sqrt{2} \right|^2 + 25 \left| 3 \right|^2 + 40 \times \left| 2\sqrt{2} \right| \times \left| 3 \right| \times \frac{1}{\sqrt{2}}}\]
\[ = \sqrt{128 + 225 + 240}\]
\[ = \sqrt{593} \text{ units }\]
\[ \text{ Therefore, the larger diagonal } = \sqrt{593}\]