Question

Find the coordinate of the point $P$ where the line through $A(3,-4,-5)$ and $B(2,-3,1)$ crosses the plane passing through three points $L(2,2,1),M(3,0,1)$ and $N(4,-1,0)$. Also, find the ratio in which $P$ divides the line segment $AB$.

Answer 1

The equation of line $AB$ is given by $\overrightarrow{r}=(3\hat{i}-4\hat{j}-5\hat{k})+\lambda (\hat{i}-\hat{j}-6\hat{k})$

The normal to the plane will be $\overrightarrow{LM}\times \overrightarrow{MN}=(\hat{i}-2\hat{j})\times (-\hat{i}+\hat{j}+\hat{k})=0+\hat{k}-\hat{j}-2\hat{k}-2\hat{i}=-2\hat{i}-\hat{j}-\hat{k}$

The equation of the plane is given by $2x+y+z+d=0$

Since $(2,2,1)$ lies on the plane, we get $4+2+1+d=0$

$\Rightarrow d=-7$

The plane becomes $2x+y+z-7=0$

Let the point $P$ on the line be $(3+t,-4-t,-5-6t)$

Since this lies on the plane, $6+2t-4-t-5-6t-7=0$

$\Rightarrow -5t-10=0$ or $t=-2$

$P=(1,-2,7)$

$1=\frac{2m+3n}{m+n}$

$\Rightarrow m+n=2m+3n$ or $m=-2n$

$m:n=-2:1$