3D Geometry Ques 79
79. Let $P$ be the plane, which contains the line of intersection of the planes, $x+y+z-6=0$ and $2 x+3 y+z+5=0$ and it is perpendicular to the $X Y$-plane. Then, the distance of the point $(0,0,256)$ from $P$ is equal to
(2019 Main, 9 April II)
(a) $63 \sqrt{5}$
(b) $205 \sqrt{5}$
(c) $\frac{11}{\sqrt{5}}$
(d) $\frac{17}{\sqrt{5}}$
Show Answer
Answer:
(c)
Solution:
- Equation of plane, which contains the line of intersection of the planes
$ \begin{gathered} x+y+z-6=0 \text { and } 2 x+3 y+z+5=0, \text { is } \\ (x+y+z-6)+\lambda(2 x+3 y+z+5)=0 \end{gathered} $
$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+\lambda) z+(5 \lambda-6)=0$
$\because$ The plane (i) is perpendicular to $X Y$-plane (as DR’s of normal to $X Y$-plane is $(0,0,1))$.
$\therefore 0(1+2 \lambda)+0(1+3 \lambda)+1(1+\lambda)=0$
$\Rightarrow \quad \lambda=-1$
On substituting $\lambda=-1$ in Eq. (i),
we get
$ -x-2 y-11=0 $
$ \Rightarrow \quad x+2 y+11=0 $
which is the required equation of the plane.
Now, the distance of the point $(0,0,256)$ from plane $P$ is
$ \frac{0+0+11}{\sqrt{1+4}}=\frac{11}{\sqrt{5}} $
$\left[\because\right.$ distance of $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane
$ a x+b y+c z-d=0, \text { is }\left|\frac{a x_{1}+b y_{1}+c z_{1}-d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| $
BETA
