Q74 / 180 · KCET 2015 · Mathematics

Let \( \vec{a}=i-2 j+3 k \) if \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b}=|\vec{b}|^{2} \) and \( |\vec{a}-\vec{b}|=\sqrt{7} \), then \( |\vec{b}|= \)

  • A
    \( 07 \)
  • B
    \( 14 \)
  • Correct answer
    C
    \( \sqrt{7} \)
  • D
    \( 21 \)

Solution

Given that, $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \rightarrow(1)$ $\vec{a} \cdot \vec{b}=|\vec{b}|^{2} \rightarrow(2)$ and $|\vec{a}-\vec{b}|=\sqrt{7} \rightarrow(3)$ Now, $|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}$ Squaring both sides of Eq. (3), we get $|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \vec{b} \mid \cdot \cos \theta=7$ $\Rightarrow 14+|\vec{b}|^{2}-2|\vec{b}|^{2}=7$ $\Rightarrow 7=|\vec{b}|^{2} \Rightarrow|\vec{b}|=\sqrt{7}$

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

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