Discussion Forum

Let $a =-\widehat{i}-\widehat{k},b=-\widehat{i}+\widehat{j}$ and $c= \widehat{i}+2 \widehat{j}+3 \widehat{k}$  be the three vectors. If r is a  vector such that  r x b=c x b and r.a=0 , then the value of r.b is 

Answer:

Explanation:

 r x b= c x b

 $\Rightarrow$  (r-c) x b=0  $\Rightarrow$ $r-c+\lambda b$

  or  $r=c+\lambda b$..........(i)

 Given, r.a=0 , taking dot product with a for eq.(i)

 Now, $r.a=a.c+\lambda a.b$

$\therefore$  $\lambda=\frac{\overrightarrow{a}.\overrightarrow{c}}{a.b}$ $[\because \overrightarrow{r}.\overrightarrow{a}=0]$......(ii)

  From Eq.(i) and (ii) , we get

 $r=c-\frac{a.c}{a.b}b$

 Taking dot with b, we get

 $r.b=c.b-\frac{a.c}{a.b} (b.b)$

 where, $\begin{bmatrix}a=-\widehat{i}-\widehat{k}  \\b=-\widehat{i}+\widehat{j}\\c=\widehat{i}+2\widehat{j}+3\widehat{k} \end{bmatrix}$

 =$(-1+2)- \frac{(-1-3)}{(1)}(1+1)=1+8=9$