Q93 / 160 · TS EAMCET 2011 · Mathematics

If $|\mathbf{a}|=1,|\mathbf{b}|=2$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $120^{\circ}$, then $\{(\mathbf{a}+3 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})\}^2$ is equal to

  • A
    425
  • B
    375
  • C
    325
  • Correct answer
    D
    300

Solution

$\begin{aligned} & \therefore\{(\mathbf{a}+3 \mathbf{b}) \times(3 \mathbf{a}-\mathbf{b})\}^2 \\ & =\{3 \mathbf{a} \times \mathbf{a}-\mathbf{a} \times \mathbf{b}+9 \mathbf{b} \times \mathbf{a}-3 \mathbf{b} \times \mathbf{b}\}^2 \\ & =\{-10 \mathbf{a} \times \mathbf{b}\}^2 \\ & =100\left[|\mathbf{a}||\mathbf{b}| \hat{\mathbf{n}} \sin 120^{\circ}\right]^2 \\ & =100\left[1 \times 2 \times \hat{\mathbf{n}} \times \frac{\sqrt{3}}{2}\right]^2=100 \times 3=300\end{aligned}$

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

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