Q105 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics

The unit vectors orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and coplanar with \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) are

  • A
    \(\pm \frac{1}{\sqrt{5}}(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})\)
  • Correct answer
    B
    \(\pm \frac{1}{\sqrt{10}}(3 \hat{\mathbf{j}}-\hat{\mathbf{k}})\)
  • C
    \(\pm \frac{1}{\sqrt{13}}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})\)
  • D
    \(\pm \frac{1}{\sqrt{17}}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\)

Solution

A vector which is coplanar to the vectors \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is \(\mathbf{a}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{b}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) \((2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}\) ...(i) Eq. (i) is orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) \(\begin{aligned} & 3(2 a+b)+2(a-b)+6(a+b) =0 \\ & 14 a+7 b =0 \\ & 2 a+b =0 \end{aligned}\) Required unit vector, \(\begin{aligned} & = \pm \frac{(2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}}{\sqrt{(2 a+b)^2+(a-b)^2+(a+b)^2}} \\ & = \pm \frac{3 a \hat{\mathbf{j}}-a \hat{\mathbf{k}}}{\sqrt{(3 a)^2+a^2}}= \pm \frac{3 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{10}} \end{aligned}\) Hence, option (b) is correct.

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

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