If $\vec{x} + \vec{c} \times \vec{y} = \vec{a} and \vec{y} + \vec{c} \times \vec{x} = \vec{b}$ where $\vec{c}$ is a non-zero vector ,then which of the following is not correct?

Moderate

A

$\vec{x} = \frac{\vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$
B

$\vec{x} = \frac{\vec{c} \times \vec{b} + \vec{b} + (\vec{c} \cdot \vec{a}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$
C

$\vec{y} = \frac{\vec{a} \times \vec{c} + \vec{b} + (\vec{c} \cdot \vec{b}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$
D

none of these

Solution

$\begin{array}{r} \vec{x} + \vec{c} \times \vec{y} = \vec{a} - - - - - (i) \\ \vec{y} + \vec{c} \times \vec{x} = \vec{b} - - - - (ii) \end{array}$
Taking cross with $\vec{c}$ , we have
$\begin{array}{l} \vec{c} \times \vec{y} + \vec{c} \times (\vec{c} \times \vec{x}) = \vec{c} \times \vec{b} \\ \Rightarrow (\vec{a} - \vec{x}) + (\vec{c} \cdot \vec{x}) \vec{c} - (\vec{c} \cdot \vec{c}) \vec{x} = \vec{c} \times \vec{b} \\ Also \vec{x} + \vec{c} \times \vec{y} = \vec{a} \\ \Rightarrow \vec{c} \cdot \vec{x} + \vec{c} \cdot (\vec{c} \times \vec{y}) = \vec{c} \cdot \vec{a} \\ or \vec{c} \cdot \vec{x} + 0 = \vec{c} \cdot \vec{a} \\ or \vec{c} \cdot \vec{x} = \vec{c} \cdot \vec{a} \\ \Rightarrow \vec{a} - \vec{x} + (\vec{c} \cdot \vec{a}) \vec{c} - (\vec{c} \cdot \vec{c}) \vec{x} = \vec{c} \times \vec{b} \\ or \vec{x} (1 + (\vec{c} \cdot \vec{c})) = \vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \cdot \vec{c} \\ or \vec{x} = \frac{\vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \vec{c}}{1 + \vec{c} \cdot \vec{c}} \end{array}$
Similarly on taking cross product of Eq. (i), we find
$\vec{y} = \frac{\vec{a} \times \vec{c} + \vec{b} + (\vec{c} \cdot \vec{b}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME