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


My attempt:
$\vec{a}\times\vec{b}+\vec{b}\times\vec{c}=p\vec{a}+q\vec{b}+r\vec{c}$

$(\vec{a}\times\vec{b}).\vec{c}=p(\vec{a}.\vec{c})+q(\vec{b}.\vec{c})+r$
Since the angle between $\vec{a},\vec{b}$ and $\vec{b},\vec{c}$ and $\vec{c},\vec{a}$ is $\theta$

$(\vec{a}\times\vec{b}).\vec{c}=p\cos\theta+q\cos\theta+r.............(1)$
$\hspace{1.5cm}0=p\cos\theta+q+r\cos\theta.............(2)$
$(\vec{b}\times\vec{c}).\vec{a}=p+q\cos\theta+r\cos\theta.............(3)$
We know that $(\vec{a}\times\vec{b}).\vec{c}=(\vec{b}\times\vec{c}).\vec{a}$
From equation $(2),q=-p\cos\theta-r\cos\theta$
Put this in equation $(1)$ and $(3)$
$(\vec{a}\times\vec{b}).\vec{c}=p\cos\theta+(-p\cos\theta-r\cos\theta)\cos\theta+r$
$(\vec{a}\times\vec{b}).\vec{c}=p(\cos\theta-\cos^2\theta)+r(1-\cos^2\theta)$
$(\vec{a}\times\vec{b}).\vec{c}=p(\cos\theta-\cos^2\theta)+r\sin^2\theta...........(4)$
$(\vec{b}\times\vec{c}).\vec{a}=p+(-p\cos\theta-r\cos\theta)\cos\theta+r\cos\theta$
$(\vec{b}\times\vec{c}).\vec{a}=p(\sin^2\theta)+r(\cos\theta-\cos^2\theta)...........(5)$
Dividing equation $(4)$ by equation $(5)$,and solving we get $p=r$

But after that i am stuck.And could not solve further.In the answers,$p=\dfrac{-1}{\sqrt{1+2\cos\theta}},q=\dfrac{2\cos\theta}{\sqrt{1+2\cos\theta}},r=\dfrac{-1}{\sqrt{1+2\cos\theta}}$ is given.

But i do not know how it came.Please help me.Thanks.

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  • $\begingroup$ dot-multiply your first equation by $\vec{b}\times\vec{c}$. Since it's orthogonal to both $\vec b$ and $\vec c$, two terms disappear. The coefficient of $p$ is not zero since the vectors are not coplanar (it's a determinant). The cross product and the dot product have an easy trigonometric interpretation, there remains the determinant. $\endgroup$ – Jean-Claude Arbaut Nov 12 '15 at 16:10
  • $\begingroup$ You can just edit the title as you like! :) $\endgroup$ – H. R. Nov 12 '15 at 16:13
  • $\begingroup$ @Jean-ClaudeArbaut,i got the answer by your hint.Thank you Sir. $\endgroup$ – Vinod Kumar Punia Nov 12 '15 at 16:27
  • $\begingroup$ You are welcome :) $\endgroup$ – Jean-Claude Arbaut Nov 12 '15 at 16:31
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Hint

You have

$$\vec a \times \vec b + \vec b \times \vec c = p\vec a + q\vec b + r\vec c$$

now dot product with ${\vec a \times \vec b}$, ${\vec b \times \vec c}$, and ${\vec c \times \vec a}$ respectively to obtain

$$\eqalign{ & \left( {\vec a \times \vec b} \right).\left( {\vec a \times \vec b} \right) + \left( {\vec a \times \vec b} \right).\left( {\vec b \times \vec c} \right) = r\left( {\vec a \times \vec b} \right).\vec c \cr & \left( {\vec b \times \vec c} \right).\left( {\vec a \times \vec b} \right) + \left( {\vec b \times \vec c} \right).\left( {\vec b \times \vec c} \right) = p\left( {\vec b \times \vec c} \right).\vec a \cr & \left( {\vec c \times \vec a} \right).\left( {\vec a \times \vec b} \right) + \left( {\vec c \times \vec a} \right).\left( {\vec b \times \vec c} \right) = q\left( {\vec c \times \vec a} \right).\vec b \cr} $$

Now, you should just simplify . :)

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