Question

A parallelogram is constructed on vectors \(\vec{a}=3\vec{\alpha}-\vec{\beta},\; \vec{b}=\vec{\alpha}+3\vec{\beta}\). If \(|\vec{\alpha}|=|\vec{\beta}|=2\) and angle between them is \(\pi/3\), find length of a diagonal

• A. \(4\sqrt{3}\)
• B. \(4\sqrt{5}\)
• C. \(4\sqrt{7}\)
• D. None of these

Hint:

Use \(|\vec{a}+\vec{b}|^2 = a^2+b^2+2\vec{a}\cdot\vec{b}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The diagonals of a parallelogram constructed on side vectors \( \vec{a} \) and \( \vec{b} \) are given by \( \vec{d}_{1} = \vec{a} + \vec{b} \) and \( \vec{d}_{2} = \vec{a} - \vec{b} \).
Step 2: Detailed Explanation:
1. Find diagonal vector \( \vec{d}_{1} = \vec{a} + \vec{b} \): \[ \vec{d}_{1} = (3\vec{\alpha} - \vec{\beta}) + (\vec{\alpha} + 3\vec{\beta}) = 4\vec{\alpha} + 2\vec{\beta} \] 2. Calculate length squared \( |\vec{d}_{1}|^{2} \): \[ |\vec{d}_{1}|^{2} = |4\vec{\alpha} + 2\vec{\beta}|^{2} = 16|\vec{\alpha}|^{2} + 4|\vec{\beta}|^{2} + 16(\vec{\alpha} \cdot \vec{\beta}) \] Substitute \( |\vec{\alpha}| = 2, |\vec{\beta}| = 2 \) and \( \vec{\alpha} \cdot \vec{\beta} = |\vec{\alpha}||\vec{\beta}| \cos(\pi/3) = 2 \cdot 2 \cdot (1/2) = 2 \): \[ |\vec{d}_{1}|^{2} = 16(4) + 4(4) + 16(2) = 64 + 16 + 32 = 112 \] \[ |\vec{d}_{1}| = \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \].
Note: Check the other diagonal \( \vec{d}_{2} = \vec{a} - \vec{b} = 2\vec{\alpha} - 4\vec{\beta} \): \[ |\vec{d}_{2}|^{2} = 4(4) + 16(4) - 16(2) = 16 + 64 - 32 = 48 \implies |\vec{d}_{2}| = 4\sqrt{3} \].
Between the two diagonals, \( 4\sqrt{7} \) is present in the options.
Step 3: Final Answer:
The length of the diagonal is \( 4\sqrt{7} \).