Question

If the position vector of the point of intersection of the line $\overrightarrow{r}=(i+2j+3k)+\lambda (2i+j+2k)$ and the plane $\overrightarrow{r}\cdot (2i-6j+3k)+5=0$ is $ai+bj+ck$ , then ${(50a+60b+75c)}^2$ is equal to

Answer 1

given eq of line

$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}+2\hat{k})$

comparing with eq of line we get

$positionvectorA(1,2,3)a=2,b=1,c=2$

eq become

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$

passing the given eq to point$a\hat{i}+b\hat{j}+c\hat{k}$ we get

$a=2\lambda +1,b=\lambda +2,c=2\lambda +3$

passing given points to eq of planes

$2a-6b+3c+5=0$

putting values of a,b,c

$2(2\lambda +1)-6(\lambda +2)+3(2\lambda +3)+5=0$

$4\lambda +2-6\lambda -12+6\lambda +9+5=0$

$4\lambda =-4$

$\lambda =-1$

$a=2\lambda +1=-2+1=-1$

$b=\lambda +2=-1+2=1$

$c=2\lambda +3=-2+3=1$

$(50a+60b+75c{)}^2$

$(-50+60+75{)}^2$

${85}^2=7225$