A parallelogram is constructed on the vectors $\vec{a}= 3\vec{\alpha }-\vec{\beta }, \vec{b}=\vec{\alpha } + 3\vec{\beta} .$ If $|\vec{\alpha}|=|\vec{\beta }|=2$ and the angle between $\vec{\alpha }$ and $\vec{\beta }$ is $\frac{\pi }{3}$, then length of the diagonal of the parallelogram is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $4\sqrt{7}$
Given
$\vec{a} = 3\vec{\alpha} - \vec{\beta}$, $\vec{b} = \vec{\alpha} + 3\vec{\beta}$,
$|\vec{\alpha}| = |\vec{\beta}| = 2$, and angle between $\vec{\alpha}$ and $\vec{\beta}$ is $60^\circ$.
Diagonal of a parallelogram $= |\vec{a} + \vec{b}|$
$\vec{a} + \vec{b} = (3\vec{\alpha} - \vec{\beta}) + (\vec{\alpha} + 3\vec{\beta}) = 4\vec{\alpha} + 2\vec{\beta}$
Now,
$|\vec{a} + \vec{b}|^2 = |4\vec{\alpha} + 2\vec{\beta}|^2 = 16|\vec{\alpha}|^2 + 4|\vec{\beta}|^2 + 16(\vec{\alpha} \cdot \vec{\beta})$
$|\vec{\alpha}|^2 = |\vec{\beta}|^2 = 4, \quad \vec{\alpha} \cdot \vec{\beta} = |\vec{\alpha}||\vec{\beta}| \cos 60^\circ = 2 \times 2 \times \frac{1}{2} = 2$
$|\vec{a} + \vec{b}|^2 = 16(4) + 4(4) + 16(2) = 64 + 16 + 32 = 112$
$|\vec{a} + \vec{b}| = \sqrt{112} = 4\sqrt{7}$