Q52 / 170 · Manipal (MET) 2016 · Mathematics

If $\mathbf{a}$ and $\mathbf{b}$ be two perpendicular unit vectors such that $\mathbf{x}=\mathbf{b}+(\mathbf{a} \times \mathbf{x})$, then $|\mathbf{x}|$ is equal to

  • A
    $1$
  • B
    $\frac{1}{\sqrt{2}}$
  • Correct answer
    C
    $\sqrt{2}$
  • D
    $\sqrt{3}$

Solution

We have, $\mathbf{a} \cdot \mathbf{x}=0$ Also, $\quad \mathbf{x}=\mathbf{b}+(\mathbf{x} \times \mathbf{a})$ $\begin{array}{ll}\Rightarrow & \mathbf{x} \times \mathbf{a}=\mathbf{b} \times \mathbf{a}+(\mathbf{x} \times \mathbf{a}) \times \mathbf{a} \\ \Rightarrow & \mathbf{x} \times \mathbf{a}=\mathbf{b} \times \mathbf{a}+(\mathbf{x} \cdot \mathbf{a}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{x} \\ \Rightarrow & \mathbf{x - b}=\mathbf{b} \times \mathbf{a}-\mathbf{x}\end{array}$ $|2 \mathbf{x}|=|\mathbf{b}+\mathbf{b} \times \mathbf{a}|$ $\Rightarrow \quad|\mathbf{x}|=\frac{1}{\sqrt{2}}$

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

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