Question

Let $\overrightarrow{ a }, \overrightarrow{ b }$ and $\overrightarrow{ c }$ be non-coplanar unit vectors, equally inclined to one another at an angle $\theta$. If $\overrightarrow{ a } \times \overrightarrow{ b }+\overrightarrow{ b } \times \overrightarrow{ c }=p \overrightarrow{ a }+q \overrightarrow{ b }+r \overrightarrow{ c }$, then find scalars $p$ $q$ and $r$ in terms of $\theta$.

Answer 1

Since, $\overrightarrow{ a }, \overrightarrow{ b }, \overrightarrow{ c }$ are non-coplanar vectors. $\Rightarrow$
$[\overrightarrow{ a } \,\,\overrightarrow{ b } \,\,\overrightarrow{ c }] \neq 0$
Also, $ \overrightarrow{ a } \times \overrightarrow{ b }+\overrightarrow{ b } \times \overrightarrow{ c }=p \overrightarrow{ a }+q \overrightarrow{ b }+r \overrightarrow{ c }$
Taking dot product with $\overrightarrow{ a }, \overrightarrow{ b }$ and $\overrightarrow{ c }$ respectively both sides, we get
$p+q \cos \theta+r \cos \theta =[\overrightarrow{ a }\,\, \overrightarrow{ b } \,\,\overrightarrow{ c }].....$(i)
$p \cos \theta+q+r \cos \theta =0 .....$(ii)
and $ p \cos \theta+q \cos \theta+r =[\overrightarrow{ a }\,\, \overrightarrow{ b } \,\,\overrightarrow{ c }].....$(iii)
On adding above equations,
$p+q+r=\frac{2[\overrightarrow{ a } \,\,\overrightarrow{ b }\,\, \overrightarrow{ c }]}{2 \cos \theta+1} ....$(iv)
On multiplying Eq. (iv) by $\cos \theta$ and subtracting Eq. (i), we get
$p(\cos \theta-1) =\frac{2[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] \cos \theta}{2 \cos \theta+1}-[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] $
$\Rightarrow p=\frac{[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]}{(1-\cos \theta)(2 \cos \theta+1)} $
Similarly, $ q =\frac{-2[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] \cos \theta}{(1+2 \cos \theta)(1-\cos \theta)} $
and $ r=\frac{[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]}{(1+2 \cos \theta)(1-\cos \theta)}$
Now,
$[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]^{2}=\begin{vmatrix}\overrightarrow{ a } \cdot \overrightarrow{ a } & \overrightarrow{ a } \cdot \overrightarrow{ b } & \overrightarrow{ a } \cdot \overrightarrow{ c } \\ \overrightarrow{ b } \cdot \overrightarrow{ a } & \overrightarrow{ b } \cdot \overrightarrow{ b } & \overrightarrow{ b } \cdot \overrightarrow{ c } \\ \overrightarrow{ c } \cdot \overrightarrow{ a } & \overrightarrow{ c } \cdot \overrightarrow{ b } & \overrightarrow{ c } \cdot \overrightarrow{ c }\end{vmatrix}=\begin{vmatrix}1 & \cos \theta & \cos \theta \\ \cos \theta & 1 & \cos \theta \\ \cos \theta & \cos \theta & 1\end{vmatrix}$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$
$=(1+2 \cos \theta)\begin{vmatrix} 1 & 1 & 1 \\\cos \theta & 1 & \cos \theta \\\cos \theta & \cos \theta & 1\end{vmatrix}$
Applying $C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1}$
$=(1+2 \cos \theta) \cdot \begin{vmatrix} 1 & 0 & 0 \\\cos \theta & 1-\cos \theta & 0 \\\cos \theta & 0 & 1-\cos \theta\end{vmatrix}$
$=(1+2 \cos \theta) \cdot(1-\cos \theta)^{2} $
$ \Rightarrow [\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }] =(\sqrt{1+2 \cos \theta}) \cdot(1-\cos \theta) $
$\therefore p =\frac{1}{\sqrt{1+2 \cos \theta}} $
$ q =\frac{-2 \cos \theta}{\sqrt{1+2 \cos \theta}} $
and $ r =\frac{1}{\sqrt{1+2 \cos \theta}} $