Q72 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{c}$ is a vector perpendicular to $\mathbf{b}$, then $\left\{\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|^2}\right\}(\mathbf{b} \times \mathbf{c})+\left\{\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right\} \mathbf{b}+\left\{\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{c}|^2}\right\} \mid=$

  • Correct answer
    A
    $\sqrt{14}$
  • B
    14
  • C
    13
  • D
    $\sqrt{17}$

Solution

Any vector $\mathbf{r}$ can be written in linear combination of two non-parallel vector $\mathbf{b}$ and $\mathbf{c}$, as $$ \mathbf{r}=\left(\frac{\mathbf{r} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right) \mathbf{b}+\left(\frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|^2}\right) \mathbf{c}+\left(\frac{\mathbf{r} \cdot(\mathbf{b} \times \mathbf{c})}{|\mathbf{b} \times \mathbf{c}|^2}\right)(\mathbf{b} \times \mathbf{c}) $$ So, $\mathbf{r}=\mathbf{a} \Rightarrow|\mathbf{r}|=|\mathbf{a}|=\sqrt{1+4+9}=\sqrt{14}$.

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

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