Question

Find the co ordinates 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

Equation of line$\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$

$\Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$

Equation of point $P=(3-k,k-4,6k-5)$

Equation of vectors lying on the plane

$=(3-2)\hat{i}+(0-2)\hat{j}+(1-1)\hat{k}$

$=\hat{i}-2\hat{j}$

and:

$=(4-3)\hat{i}+(-1-0)\hat{j}+(0-1)\hat{k}$

$=\hat{i}-\hat{j}-\hat{k}$

Equation of normal$=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 0\\ 1& -1& -1\end{array}\right]$

$=2\hat{i}-(-1)\hat{j}+(+1)\hat{k}$

$=2\hat{i}+\hat{j}+\hat{k}$

$\therefore$Equation of plane will be of form:

$2x+y+z+d=0$

Inserting $N(4,-1,0)$

$2\left(4\right)-1+0+d=0$

$d=-7$

Equation : $2x+y+z-7=0$

Now P lies in the plane

Thus, $2(3-k)+(k-4)+(6k-5)-7=0$

$6-2k+k-4+6k-5-7=0$

$5k=10$

$k=2$

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

$PA=\sqrt{{(3-1)}^2+{(-4+2)}^2+{(-5-7)}^2}$

$=\sqrt{2^2+2^2+{12}^2}$

$=\sqrt{4+4+144}$

$=\sqrt{152}=2\sqrt{38}$

$PB=\sqrt{{(2-1)}^2+{(-3+2)}^2+{(1-7)}^2}$

$=\sqrt{1^2+1^2+6^2}$

$=\sqrt{1+1+36}$

$=\sqrt{38}$

$\therefore PA:PB=2:1$

P divides the line in the ratio $2:1$