Discussion Forum

Let  $a= \widehat{i}+\widehat{j}+\widehat{k},b=\widehat{i}-\widehat{j}+\widehat{k}$ ans $c=\widehat{i}-\widehat{j}-\widehat{k}$ be three vectors. A vector v in the plane of a and b whose projection  of c is $\frac{1}{\sqrt{3}}$ is given by 

Answer:

Explanation:

 let  v$a+ \lambda b$

 $v= (1+ \lambda)\widehat{i}+(1-\lambda) \widehat{j}+(1+\lambda)\widehat{k}$

 Projection of v on c= $\frac{1}{\sqrt{3}}$

$\Rightarrow$   $\frac{v.c}{|c|}=\frac{1}{\sqrt{3}}$

$\Rightarrow$ $\frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}} =\frac{1}{\sqrt{3}}$

$\Rightarrow$  $1+ \lambda-1+\lambda-1-\lambda=1$

$\Rightarrow$   $\lambda-1=1 \Rightarrow \lambda=2$

$\therefore$  $v=3 \widehat{i}-\widehat{j}+3 \widehat{k}$