Q52 / 170 · Manipal (MET) 2018 · Mathematics

A parallelogram is constructed on the vectors $\mathbf{a}=3 \alpha-\beta, \mathbf{b}=\alpha+3 \beta$. If $|\alpha|=|\beta|=2$ and the angle between $\alpha$ and $\beta$ is $\frac{\pi}{3}$, then length of a diagonal of the parallelogram is

  • A
    $4 \sqrt{3}$
  • B
    $4 \sqrt{5}$
  • Correct answer
    C
    $4 \sqrt{7}$
  • D
    None of these

Solution

$\begin{aligned} & \text {Given, } \overrightarrow{\mathbf{a}}=3 \vec{\alpha}-\vec{\beta}, \overrightarrow{\mathbf{b}}=\vec{\alpha}+3 \vec{\beta} \text { and } \\ & |\vec{\alpha}|=|\vec{\beta}|=2 \\ & \therefore|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^2=|4 \vec{\alpha}+2 \vec{\beta}|^2 \\ & =16|\vec{\alpha}|^2+4|\vec{\beta}|^2+16|\vec{\alpha}||\vec{\beta}| \cos \frac{\pi}{3} \\ & =16(4)+4(4)+16(2 \times 2) \times \frac{1}{2} \\ & =64+16+32=112 \\ & \Rightarrow \quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=4 \sqrt{7}\end{aligned}$

Hence, the correct answer is option: 'C'.

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