Vectors Ques 108
- Let $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ be non-coplanar unit vectors, equally inclined to one another at an angle $\theta$. If $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=p \overrightarrow{\mathbf{a}}+q \overrightarrow{\mathbf{b}}+r \overrightarrow{\mathbf{c}}$, then find scalars $p, q$ and $r$ in terms of $\theta$.
(1997, 5M)
Show Answer
Solution:
Formula:
- Since, $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are non-coplanar vectors.
$\begin{array}{ll}\Rightarrow & {[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \neq 0} \ \text { Also, } & \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=p \overrightarrow{\mathbf{a}}+q \overrightarrow{\mathbf{b}}+r \overrightarrow{\mathbf{c}}\end{array}$
Taking dot product with $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ respectively both sides, we get
$$ \begin{aligned} & p+q \cos \theta+r \cos \theta=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \\ & p \cos \theta+q+r \cos \theta=0 \end{aligned} $$
and $\quad p \cos \theta+q \cos \theta+r=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
On adding above equations,
$$ p+q+r=\frac{2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]}{2 \cos \theta+1} $$
On multiplying Eq. (iv) by $\cos \theta$ and subtracting Eq. (i), we get
$p(\cos \theta-1)=\frac{2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \cos \theta}{2 \cos \theta+1}-[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$$ \begin{array}{ll} \Rightarrow & p=\frac{[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]}{(1-\cos \theta)(2 \cos \theta+1)} \\ \text { Similarly, } & q=\frac{-2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \cos \theta}{(1+2 \cos \theta)(1-\cos \theta)} \end{array} $$
and
$$ r=\frac{[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]}{(1+2 \cos \theta)(1-\cos \theta)} $$
Now, $\quad[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]^{2}=\left|\begin{array}{lll}\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} \ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} \ \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}}\end{array}\right|=\left|\begin{array}{ccc}1 & \cos \theta & \cos \theta \ \cos \theta & 1 & \cos \theta \ \cos \theta & \cos \theta & 1\end{array}\right|$
Applying $R _1 \rightarrow R _1+R _2+R _3$
$$ =(1+2 \cos \theta)\left|\begin{array}{ccc} 1 & 1 & 1 \\ \cos \theta & 1 & \cos \theta \\ \cos \theta & \cos \theta & 1 \end{array}\right| $$
Applying $C _2 \rightarrow C _2-C _1, C _3 \rightarrow C _3-C _1$
$$ =(1+2 \cos \theta) \cdot\left|\begin{array}{ccc} 1 & 0 & 0 \\ \cos \theta & 1-\cos \theta & 0 \\ \cos \theta & 0 & 1-\cos \theta \end{array}\right| $$
$$ \begin{gathered} =(1+2 \cos \theta) \cdot(1-\cos \theta)^{2} \\ \Rightarrow \quad[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=(\sqrt{1+2 \cos \theta}) \cdot(1-\cos \theta) \\ \therefore p=\frac{1}{\sqrt{1+2 \cos \theta}}, q=\frac{-2 \cos \theta}{\sqrt{1+2 \cos \theta}} \text { and } r=\frac{1}{\sqrt{1+2 \cos \theta}} \end{gathered} $$
BETA
