Question

The equations of the planes through the line of intersection of the planes$r\cdot (2\hat{i}+6\hat{j})+12=0$ and $r\cdot (3\hat{i}-\hat{j}+4\hat{k})=0$ Which are at unit distance from the origin are :

• A. $2x+y+2z+3=0,x-2y+2z-3=0$
• B. $2x+y+2z-3=0,x-2y+2z+3=0$
• C. $2x+y+2z+4=0,x-2y+2z-4=0$
• D. $2x+y+2z+1=0,x-2y+2z-3=0$

Answer 1

Answer: (A)

$2x+y+2z+3=0,x-2y+2z-3=0$

Given planes: $r\cdot \left(2\hat{i}+6\hat{j}\right)+12=0$ and $r\cdot \left(3\hat{i}-\hat{j}+4\hat{k}\right)=0$

Equation of plane passing through the intersection of two planes:

$\left(a_{1}x+b_{1}y+c_{1}z+d_1\right)+\lambda \left(a_{2}x+b_{2}y+c_{2}z+d_2\right)=0$

So, the equation of the plane:

$\left(2x+6y+12\right)+\lambda \left(3x-y+4z\right)=0$ ....($1$)

$x\left(2+3\lambda \right)+y\left(6-\lambda \right)+4z\lambda +12=0$

The perpendicular from origin:

$\begin{array}{c}P=\left|\frac{d}{\sqrt{a^2+b^2+c^2}}\right|\Rightarrow 1=\frac{12}{\sqrt{{\left(2+3\lambda \right)}^2+{\left(6-\lambda \right)}^2+{\left(4\lambda \right)}^2}}\\ \Rightarrow {\left(2+3\lambda \right)}^2+{\left(6-\lambda \right)}^2+{\left(4\lambda \right)}^2=144\\ \Rightarrow \lambda =\pm 2\end{array}$

Substituting the above in equation ($1$)

When $\lambda =+2$

$\begin{array}{c}\left(2x+6y+12\right)+2\left(3x-y+4z\right)=0\\ 8x+4y+8z+12=0\\ 2x+y+2z+3=0\end{array}$

When $\lambda =-2$

$\begin{array}{c}\left(2x+6y+12\right)-2\left(3x-y+4z\right)=0\\ -4x+8y-8z+12=0\\ x-2y+2z-3=0\end{array}$

So, the final answer is: $2x+y+2z+3=0$ and $x-2y+2z-3=0$