Question

Plane XOZ divides the join of $\left( 1,-1,5 \right)$ and $\left( 2,3,4 \right)$ in the ratio $\lambda :1$ , then $\lambda $ is:
A. -3
B. $\dfrac{1}{4}$
C. 3
D. $\dfrac{1}{3}$

Answer 1

Hint: Section Formula: The co-ordinates of a point $P\left( x,y,z \right)$ which divides the line joining two points $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ in the ratio $AP:PB=m:n$, are given as:
For internal division: $P(x,y)=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\dfrac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)$ .
For external division: $P(x,y)=\left( \dfrac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\dfrac{m{{y}_{2}}-n{{y}_{1}}}{m-n}\dfrac{m{{z}_{2}}-n{{z}_{1}}}{m-n} \right)$ .
The equation of the plane XOZ is $y=0$ .

Complete step-by-step answer:
Let's say that the line joining the points $A\left( 1,-1,5 \right)$ and $B\left( 2,3,4 \right)$ meets the plane XOZ at a point $P\left( x,y,z \right)$ , which divides the line in the ratio $AP:PB=\lambda :1$ .
Since the point P is on the plane XOZ, its y co-ordinate of must be 0. i.e. The point P is $\left( x,0,z \right)$ .
Also, since the y co-ordinate of A is negative (-1) and the y co-ordinate of B is positive (3), the point P with y co-ordinate 0 must be in-between A and B and thus, must divide the line AB internally in the given ratio.
Now, using the section formula, we must have:
 $P(x,0,z)=\left( \dfrac{\lambda (2)+1(1)}{\lambda +1},\dfrac{\lambda (3)+1(-1)}{\lambda +1},\dfrac{\lambda (4)+1(5)}{\lambda +1} \right)$
Equating the y coordinates, we get the following equation:
 $0=\dfrac{3\lambda -1}{\lambda +1}$
On multiplying both sides of the equation by $\lambda +1$ , we get:
 $0=3\lambda -1$
⇒ $3\lambda =1$
⇒ $\lambda =\dfrac{1}{3}$
Hence, the correct answer option is D. $\dfrac{1}{3}$ .

Note: The XOY, XOZ and YOZ planes are also called XY, XZ and YZ planes respectively.
The general equation of a plane is $Ax+By+Cz=D$ .
The general equation of the XY plane is $0x+0y+z=0$ OR$z=0$ , and so on for YZ and XZ planes.
The distance between a point $P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and the $Ax+By+Cz+D=0$ is given by: $\text{Distance}=\dfrac{|A{{x}_{1}}+B{{x}_{2}}+C{{z}_{2}}+D|}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$ .