Q73 / 180 · KCET 2012 · Mathematics

If $(\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144$ and $|\mathbf{a}|=4$, then $|\mathbf{b}|$ is equal to

  • A
    16
  • B
    8
  • Correct answer
    C
    3
  • D
    12

Solution

Given, $(\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144$ $\Rightarrow \quad\left(a^{2} b^{2} \cdot 1 \cdot \sin ^{2} \theta\right)+a^{2} b^{2} \cos ^{2} \theta=144$ $\Rightarrow \quad \mathrm{a}^{2} \mathrm{~b}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144$ $\Rightarrow \quad a^{2} b^{2}=144$ $\Rightarrow \quad 16 \mathrm{~b}^{2}=144 \quad(\because|\mathrm{a}|=4)$ $\Rightarrow \quad \mathrm{b}^{2}=9$ $\Rightarrow \quad \mathrm{b}=3$ or $|\mathbf{b}|=3$

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

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