Question

If $a,b$ and $c$ be non-coplanar unit vectors equally inclined to one another at an acute angle $\theta$. If $a\times b+b\times c=pa+qb+rc$, then

• A. $p=r$
• B. $p=\frac{1}{\sqrt{1+2\cos \theta }},q=\frac{2\cos \theta }{\sqrt{1+2\cos \theta }}$
• C. $r=\frac{1}{\sqrt{1+2\cos \theta }}$
• D. $p=\frac{2\cos \theta }{\sqrt{1+2\cos \theta }}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $p=r$
B $p=\frac{1}{\sqrt{1+2\cos \theta }},q=\frac{2\cos \theta }{\sqrt{1+2\cos \theta }}$
C $r=\frac{1}{\sqrt{1+2\cos \theta }}$

We are given that
$a\times b+b\times c=pa+qb+rc....\left(1\right)$
Taking dot product with a we have
$\left[abc\right]=pa\cdot a+qa\cdot b+ra\cdot c$
$=p+q\cos \theta +r\cos \theta .....\left(2\right)$
since $a\cdot a={\left|a\right|}^2=1,a\cdot b=\left|a\right|\left|b\right|\cos \theta$ and
$a\cdot c=\cos \theta .$
Similarly, taking dot product with $b$ and $c$ we get
$0=p\cos \theta +q+r\cos \theta \left(3\right)$
and $\left[abc\right]=p\cos \theta +q\cos \theta +r\left(4\right)$
Therefore, from $\left(2\right),\left(4\right)$ we get $p=r$
Adding $\left(2\right),\left(3\right)$ and $\left(4\right)$ and we get
$2\left[abc\right]=(2\cos \theta +1)(p+q+r)$
$\Rightarrow \frac{2\left[abc\right]}{2\cos \theta +1}=p+q+r\left(5\right)$
Mnltiplying $\left(5\right)$ by $\cos \theta$and subtracting from $\left(2\right)$ we get
$\left[abc\right]-\frac{2\cos \theta \left[abc\right]}{2\cos \theta +1}=p(1-\cos \theta )$
$\Rightarrow -\frac{\left[abc\right]}{(2\cos \theta +1)(1-\cos \theta )}=p$
Similarly, $q=\frac{-2\left[abc\right]\cos \theta }{(2\cos \theta +1)(1-\cos \theta )}$
Now,
$[abc{]}^2=\left|\begin{array}{ccc}a\cdot a& a\cdot b& a\cdot c\\ b\cdot a& b\cdot b& b\cdot c\\ c\cdot a& c\cdot b& c\cdot c\end{array}\right|$
$=\left|\begin{array}{ccc}1& \cos \theta & \cos \theta \\ \cos \theta & 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right|$ $=(1+2\cos \theta )\left|\begin{array}{ccc}1& 1& 1\\ \cos \theta & 1& \cos \theta \\ \cos \theta & \cos \theta & 1\end{array}\right|$
$=(1+2\cos \theta )\left|\begin{array}{ccc}1& 0& 0\\ \cos \theta & 1-\cos \theta & 0\\ \cos \theta & 0& 1-\cos \theta \end{array}\right|$
$=(1+2\cos \theta )(1-\cos \theta {)}^2$
$\Rightarrow \left[abc\right]=\left(\sqrt{1+2\cos \theta }\right)(1-\cos \theta )$ ( since $\theta$ is acute, $0<\cos \theta <1)$
Hence $p=r=\frac{1}{\sqrt{1+2\cos \theta }},q=-\frac{2\cos \theta }{\sqrt{1+2\cos \theta }}$