Q68 / 180 · KCET 2018 · Mathematics

If \( |\bar{a} \times \bar{b}|^{2}+|\bar{a} \cdot \bar{b}|^{2}=144 \) and \( |\bar{a}|=4 \), then the value of \( |\bar{b}| \) is

  • A
    \( 11 \)
  • B
    \( 12 \)
  • Correct answer
    C
    \( 03 \)
  • D
    \( 04 \)

Solution

Given that, $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$ and $|\vec{a}|=4$ We know, $|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2} \cdot|\vec{b}|^{2} \cos ^{2} \theta$ $=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|\vec{a}|^{2}|\vec{b}|^{2}$ $\Rightarrow|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}$ So, $(144)=16|\vec{b}|^{2}$ $\Rightarrow 9=|\vec{b}|^{2} \Rightarrow|\vec{b}|=3$

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

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