Question

Line passing through the point $2\overline{a}+3\overline{b}-\overline{c},3\overline{a}+4\overline{b}-2\overline{c}$ intersects the line passing through the point $\overline{a}-2\overline{b}+3\overline{c},\overline{a}-6\overline{b}+6\overline{c}$ at P. Position vector of P=

• A. $2\overline{a}+\overline{b}$
• B. $\overline{a}+2\overline{b}$
• C. $3\overline{a}+4\overline{b}$
• D. $\overline{a}-2\overline{b}$

Answer 1

The correct option is B $\overline{a}+2\overline{b}$

Let line passing through $(2,3,-1)$ and $(3,4,-2)$ be

$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z+1}{-1}=\alpha$ (Let)

So, $x=\alpha +2,y=\alpha +3,z=-\alpha -1$

and line passing through $(1,-2,3)$ and $(1,-6,6)$ be

$\frac{x-1}{0}=\frac{y+2}{-4}=\frac{z-3}{3}=\beta$ (Let)

On comparing we get

$\frac{x-1}{0}=\beta x=1x=2+\alpha =1\alpha =-1y=\alpha +3=2z=-\alpha -1=0$

So, $P\equiv (+1,2,0)$

Then position vector of $P=\overline{a}+2\overline{b}$