Discussion Forum

Let a,b and c be three  non-coplanar unit vectors such that  the  angle betwwen every  pair of them is   $\pi$ /3, if a x b +b x c  =pa+qb+rc, where p,q, r are scalars , then  the value of   $\frac{p^{2}+2q^{2}+r^{2}}{q^{2}}$ is

Answer:

Explanation:

Plan

(i)a.b= |a|.|b| cosθ

(ii)   [a b c]²=  $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$

a.a=|a|² =1,similarly b.b=c.c=1

a.b=|a||b|cos $\left(\frac{\pi}{3}\right)=\left(\frac{1}{2}\right)$similarly

b.c=c.a=$\frac{1}{2}$

  [a b c]²= $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$

  = $\begin{bmatrix}1 & \frac{1}{2} &\frac{1}{2}\\\frac{1}{2} & 1 &\frac{1}{2} \\ \frac{1}{2} &\frac{1}{2}&1\end{bmatrix}=\frac{3}{4}-\frac{1}{2}=\frac{1}{2}$

         $\therefore$             $[ a.b.c]=\frac{1}{\sqrt{2}}$         .........(i)

As , given a x b+b x c=pa+qb+rc

  Take dot  product  with a

a.(axb)+a.(bxc)  = p a²+q b²+r c.a

 $\Rightarrow$                                  $0+\frac{1}{\sqrt{2}}=p+\frac{q}{2}+\frac{r}{2}$

$\therefore$                   [a a b]=0

Now, take dot product  with b and c

           $0=\frac{p}{2}+q+\frac{r}{2}$         ......(iii)

 and            $\frac{1}{\sqrt{2}}=\frac{p}{2}+\frac{q}{2}+r$        .......(iv)

 On subtracting Eq(ii) from Eq. (iv), we get

  $\frac{p}{2}-\frac{r}{2}=0\Rightarrow p=r$

$\Rightarrow$ p+r=0   [By Eq.(iii)]

$\therefore$   $\frac{p^{2}+2q^{2}+r^{2}}{q^{2}}$

                                                   $=\frac{p^{2}+2p^{2}+p^{2}}{p^{2}}=4$