Question
Let $\overrightarrow{a} = \sqrt{7}\widehat{i} + \widehat{j} - \widehat{k}$ and $\overrightarrow{b} = \widehat{j} + 2\widehat{k}$. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$, then $|3\overrightarrow{r}|^{2}$ is equal to :
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Solution
Sol. [Key Concept: Rewrite $(r - b) \times \ a = 0 \Rightarrow \ r = b + \lambda\ a$; impose $r \cdot \ a = 0$ to fix $\lambda$, then compute $|3{r|}^{2}$.]
$${\overrightarrow{r} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{a} = \overrightarrow{0} }{(\overrightarrow{r} - \overrightarrow{b}) \times \overrightarrow{a} = \overrightarrow{0} }{\overrightarrow{r} - \overrightarrow{b} = \lambda\overrightarrow{a} }{\overrightarrow{r} = \overrightarrow{b} + \lambda\overrightarrow{a} }{\overrightarrow{r} \cdot \overrightarrow{a} = 0 \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b} + \lambda|\overrightarrow{a}|^{2} = 0 }{\lambda = - \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}|^{2}} = - \frac{(1 - 2)}{9} = \frac{1}{9} }{\overrightarrow{r} = \overrightarrow{b} + \frac{\overrightarrow{a}}{9} }{|3\overrightarrow{r}|^{2} = 9|\overrightarrow{r}|^{2} = 9\left( b^{2} + \frac{a^{2}}{81} + \frac{2(\overrightarrow{a} \cdot \overrightarrow{b})}{9} \right) = 44 }$$
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