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

Concept: Diagonal of parallelogram: \[ \vec{d} = \vec{a} + \vec{b} \]

Step 1: Add vectors.
\[ \vec{d} = (3\vec{\alpha}-\vec{\beta}) + (\vec{\alpha}+3\vec{\beta}) = 4\vec{\alpha} + 2\vec{\beta} \]

Step 2: Find magnitude.
\[ |\vec{d}|^2 = |4\vec{\alpha} + 2\vec{\beta}|^2 \] \[ =16|\vec{\alpha}|^2 +4|\vec{\beta}|^2 +16(\vec{\alpha}\cdot\vec{\beta}) \] \[ =16(4) +4(4) +16(2\cdot2\cdot\cos\frac{\pi}{3}) \] \[ =64+16+32 =112 \] \[ |\vec{d}|=\sqrt{112}=4\sqrt{7} \]