Q96 / 180 · COMEDK 2014 · Mathematics

Find $\sin \theta$, if $\theta$ is the angle between the vectors $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\hat{\mathbf{j}}}+2 \hat{\mathbf{k}}$.

  • Correct answer
    A
    $\sqrt{\frac{5}{21}}$
  • B
    $\frac{5}{\sqrt{21}}$
  • C
    $\frac{4}{\sqrt{21}}$
  • D
    $\sqrt{\frac{3}{21}}$

Solution

Let $\mathbf{a}=3 \mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{b}=\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ $$ \begin{gathered} \mathbf{a} \times \mathbf{b}=\left|\begin{array}{lll} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ 1 & 1 & 2 \end{array}\right| \\ =\mathfrak{i}(2-2)-\mathfrak{j}(6-2)+\mathbb{k}(3-1)=-4 \mathfrak{j}+2 \mathbf{k} \end{gathered} $$ $$ |\mathbf{a} \times \mathbf{b}|=\sqrt{(-4)^{2}+(2)^{2}}=\sqrt{16+4}=\sqrt{20} $$ Since, $|\mathbf{a} \times \mathbf{b}|=|\mathbf{a}||\mathbf{b}| \sin \theta$, If $\theta$ is angle between $\mathbf{a}$ and $\mathbf{b}$. $$ \begin{aligned} \sqrt{20} &=\sqrt{9+1+4} \sqrt{1+1+4} \sin \theta \\ \Rightarrow \sin \theta &=\frac{\sqrt{20}}{\sqrt{6} \sqrt{14}} \\ \Rightarrow \sin \theta &=\sqrt{\frac{20}{6 \times 14}}=\sqrt{\frac{5}{21}} \end{aligned} $$

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

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