3D Geometry 3 Question 67

####67. If the distance between the plane $A x-2 y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$, then $|d|$ is equal to….

(2010)

Show Answer

Answer:

$|D|=6$

Solution:

  1. Equation of the plane containing the lines $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$ \text { is } a(x-2)+b(y-3)+c(z-4)=0 $

where, $3 a+4 b+5 c=0$

$ 2 a+3 b+4 c=0 $

and $a(1-2)+b(2-3)+c(2-3)=0$

i.e.

$ a+b+c=0 $

From Eqs. (ii) and (iii), $\frac{a}{1}=\frac{b}{-2}=\frac{c}{1}$, which satisfy

Eq. (iv).

Plane through lines is $x-2 y+z=0$.

Given plane is $A x-2 y+z=d$ is $\sqrt{6}$.

$\therefore$ Planes must be parallel, so $A=1$ and then

$ \frac{|d|}{\sqrt{6}}=\sqrt{6} \Rightarrow|d|=6 $



NCERT Chapter Video Solution

Dual Pane