Q260 / 270 · NDA 2016 (Phase 1) · Mathematics

What is a vector of unit length orthogonal to both the vectors $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ ?

  • A
    $\frac{-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{26}}$
  • Correct answer
    B
    $\frac{-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{26}}$
  • C
    $\frac{-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{14}}$
  • D
    $\frac{-3 \hat{i}+2 \hat{j}+\hat{k}}{\sqrt{14}}$

Solution

$\begin{aligned} & \vec{A}=\hat{i}+\hat{j}+\hat{k} \\ & \vec{B}=2 \hat{i}+3 \hat{j}-\hat{k} \\ & \vec{A} \times \vec{B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 3 & -1 \end{array}\right| \\ & =\hat{i}(-1-3)-\hat{j}(-1-2)+\hat{k}(3-2) \\ & =-4 \hat{i}+3 \hat{j}+\hat{k} \end{aligned}$ Vector of unit length orthogonal to both the vectors $\vec{A}$ and $\vec{B}$ $=\frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\frac{-4 i+3 j+k}{\sqrt{16+9+1}}=\frac{-4 \hat{i}+3 \hat{j}+\hat{k}}{\sqrt{26}}$

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

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