SATHEE: Vectors 5 Question 4

Vectors 5 Question 4

4. Let $\vec{p}, \vec{q}, \vec{r}$ be three mutually perpendicular vectors of the same magnitude. If a vector $\vec{x}$ satisfies the equation

$\begin{aligned} \vec{p} \times [(\vec{x} - \vec{q}) \times \vec{p}] + \vec{q} \times [(\vec{x} - \vec{r}) \times \vec{q}] \\ + \vec{r} \times [(\vec{x} - \vec{p}) \times \vec{r}] = \vec{0}, then \vec{x} is given by \end{aligned}$

(a) $\frac{1}{2} (\vec{p} + \vec{q} - 2 \vec{r})$

(b) $\frac{1}{2} (\vec{p} + \vec{q} + \vec{r})$

(d) $\frac{1}{3} (2 \vec{p} + \vec{q} - \vec{r})$

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Answer:

Correct Answer: 4. (b)

Solution:

Since, $\vec{p}, \vec{q}, \vec{r}$ are mutually perpendicular vectors of same magnitude, so let us consider

$\begin{aligned} | \vec{p} | = | \vec{q} | = | \vec{r} | = λ and \\ \vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{r} = \vec{r} \cdot \vec{p} = 0 \end{aligned}$

Given, $\vec{p} \times (\vec{x} - \vec{q}) \times \vec{p} + \vec{q} \times (\vec{x} - \vec{r}) \times \vec{q}$

$+ \vec{r} \times (\vec{x} - \vec{p}) \times \vec{r} = \vec{0}$

$\Rightarrow (\vec{p} \cdot \vec{p}) (\vec{x} - \vec{q}) - \vec{p} \cdot (\vec{x} - \vec{q}) \vec{p} + (\vec{q} \cdot \vec{q}) (\vec{x} - \vec{r})$

$- \vec{q} \cdot (\vec{x} - \vec{r}) \vec{q} + (\vec{r} \cdot \vec{r}) (\vec{x} - \vec{p}) - \vec{r} \cdot (\vec{x} - \vec{p}) \vec{r} = 0$

$\Rightarrow \vec{x} \vec{p} \cdot \vec{p}) + (\vec{q} \cdot \vec{q}) + (\vec{r} \cdot \vec{r}) - (\vec{p} \cdot \vec{p}) \vec{q}$

$- (\vec{q} \cdot \vec{q}) r - (\vec{r} \cdot \vec{r}) \vec{p} = (\vec{x} \cdot \vec{p}) \vec{p} + (\vec{x} \cdot \vec{q}) \vec{q} + (\vec{x} \cdot \vec{r}) \vec{r}$

$\Rightarrow 3 \vec{x} | λ |^{2} - (\vec{p} + \vec{q} + \vec{r}) | λ |^{2} = (\vec{x} \cdot \vec{p}) \vec{p}$

$+ (\vec{x} \cdot \vec{q}) \vec{q} + (\vec{x} \cdot \vec{r}) \vec{r}$

Taking dot of Eq. (ii) with $\vec{p}$ , we get

$3 (\vec{x} \cdot \vec{p}) | λ |^{2} - | λ |^{4} = (\vec{x} \cdot \vec{p}) | λ |^{2} \Rightarrow \vec{x} \cdot \vec{p} = \frac{1}{2} | λ |^{2}$

Similarly, taking dot of Eq. (ii) with $\vec{q}$ and $\vec{r}$ respectively, we get

$\vec{x} \cdot \vec{q} = \frac{| λ |^{2}}{2} = \vec{x} \cdot \vec{r}$

$∴$ Eq. (ii) becomes

$\begin{aligned} 3 \vec{x} | λ |^{2} - (\vec{p} + \vec{q} + \vec{r}) | λ |^{2} = \frac{| λ |^{2}}{2} (\vec{p} + \vec{q} + \vec{r}) \\ \Rightarrow 3 \vec{x} & = \frac{1}{2} (\vec{p} + \vec{q} + \vec{r}) + (\vec{p} + \vec{q} + \vec{r}) \\ \Rightarrow \vec{x} & = \frac{1}{2} (\vec{p} + \vec{q} + \vec{r}) \end{aligned}$