Vectors 5 Question 4
4. Let $\overrightarrow{\mathbf{p}}, \overrightarrow{\mathbf{q}}, \overrightarrow{\mathbf{r}}$ be three mutually perpendicular vectors of the same magnitude. If a vector $\overrightarrow{\mathbf{x}}$ satisfies the equation
$ \begin{aligned} & \overrightarrow{\mathbf{p}} \times[(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{q}}) \times \overrightarrow{\mathbf{p}}]+\overrightarrow{\mathbf{q}} \times[(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{r}}) \times \overrightarrow{\mathbf{q}}] \\ &+\overrightarrow{\mathbf{r}} \times[(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{p}}) \times \overrightarrow{\mathbf{r}}]=\overrightarrow{\mathbf{0}}, \text { then } \overrightarrow{\mathbf{x}} \text { is given by } \\ \end{aligned} $
(a) $\frac{1}{2}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}-2 \overrightarrow{\mathbf{r}})$
(b) $\frac{1}{2}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}})$
(c) $\frac{1}{3}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}})$
(d) $\frac{1}{3}(2 \overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{r}})$
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Answer:
Correct Answer: 4. (b)
Solution:
- Since, $\overrightarrow{\mathbf{p}}, \overrightarrow{\mathbf{q}}, \overrightarrow{\mathbf{r}}$ are mutually perpendicular vectors of same magnitude, so let us consider
$ \begin{array}{rl} |\overrightarrow{\mathbf{p}}|=|\overrightarrow{\mathbf{q}}|=|\overrightarrow{\mathbf{r}}|=\lambda \text { and } \\ \overrightarrow{\mathbf{p}} \cdot \overrightarrow{\mathbf{q}}=\overrightarrow{\mathbf{q}} \cdot \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{p}}=0 \end{array} $
Given, $\overrightarrow{\mathbf{p}} \times{(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{q}}) \times \overrightarrow{\mathbf{p}}}+\overrightarrow{\mathbf{q}} \times{(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{r}}) \times \overrightarrow{\mathbf{q}}}$
$ +\overrightarrow{\mathbf{r}} \times{(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{p}}) \times \overrightarrow{\mathbf{r}}}=\overrightarrow{\mathbf{0}} $
$\Rightarrow \quad(\overrightarrow{\mathbf{p}} \cdot \overrightarrow{\mathbf{p}})(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{q}})-{\overrightarrow{\mathbf{p}} \cdot(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{q}})} \overrightarrow{\mathbf{p}}+(\overrightarrow{\mathbf{q}} \cdot \overrightarrow{\mathbf{q}})(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{r}})$
$ -{\overrightarrow{\mathbf{q}} \cdot(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{r}})} \overrightarrow{\mathbf{q}}+(\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{r}})(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{p}})-{\overrightarrow{\mathbf{r}} \cdot(\overrightarrow{\mathbf{x}}-\overrightarrow{\mathbf{p}})} \overrightarrow{\mathbf{r}}=0 $
$\Rightarrow \overrightarrow{\mathbf{x}}{\overrightarrow{\mathbf{p}} \cdot \overrightarrow{\mathbf{p}})+(\overrightarrow{\mathbf{q}} \cdot \overrightarrow{\mathbf{q}})+(\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{r}})}-(\overrightarrow{\mathbf{p}} \cdot \overrightarrow{\mathbf{p}}) \overrightarrow{\mathbf{q}}$
$-(\overrightarrow{\mathbf{q}} \cdot \overrightarrow{\mathbf{q}}) r-(\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{r}}) \overrightarrow{\mathbf{p}}=(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{p}}) \overrightarrow{\mathbf{p}}+(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{q}}) \overrightarrow{\mathbf{q}}+(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{r}}) \overrightarrow{\mathbf{r}}$
$\Rightarrow \quad 3 \overrightarrow{\mathbf{x}}|\lambda|^{2}-(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}})|\lambda|^{2}=(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{p}}) \overrightarrow{\mathbf{p}}$
$ +(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{q}}) \overrightarrow{\mathbf{q}}+(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{r}}) \overrightarrow{\mathbf{r}} $
Taking dot of Eq. (ii) with $\overrightarrow{\mathbf{p}}$, we get
$ 3(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{p}})|\lambda|^{2}-|\lambda|^{4}=(\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{p}})|\lambda|^{2} \Rightarrow \overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{p}}=\frac{1}{2}|\lambda|^{2} $
Similarly, taking dot of Eq. (ii) with $\overrightarrow{\mathbf{q}}$ and $\overrightarrow{\mathbf{r}}$ respectively, we get
$ \overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{q}}=\frac{|\lambda|^{2}}{2}=\overrightarrow{\mathbf{x}} \cdot \overrightarrow{\mathbf{r}} $
$\therefore$ Eq. (ii) becomes
$ \begin{aligned} & 3 \overrightarrow{\mathbf{x}}|\lambda|^{2}-(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}})|\lambda|^{2}=\frac{|\lambda|^{2}}{2}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}}) \\ \Rightarrow 3 \overrightarrow{\mathbf{x}} & =\frac{1}{2}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}})+(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}}) \\ \Rightarrow \quad \overrightarrow{\mathbf{x}} & =\frac{1}{2}(\overrightarrow{\mathbf{p}}+\overrightarrow{\mathbf{q}}+\overrightarrow{\mathbf{r}}) \end{aligned} $
