Q113 / 170 · Manipal (MET) 2015 · Mathematics
The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $X$-axis, is
- Correct answerA$y-3 z+6=0$
- B$3 y-z+6=0$
- C$y+3 z+6=0$
- D$3 y-2 z+6 \quad 0$
Solution
The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ is
$(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$
or $(2 \lambda+1) x+(3 \lambda+1) y+(1-\lambda) z+4 \lambda-1=0 \ldots$ ...(i)
It is parallel to $X$-axis.
i.e.
$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$
$\begin{array}{ll}\therefore & 1(2 \lambda+1)+0 \times(3 \lambda+1)+0(1-\lambda)=0 \\ \Rightarrow & \lambda=-\frac{1}{2}\end{array}$
On substituting $\lambda=-\frac{1}{2}$ in Eq. (i), we get
$\begin{array}{r}\left(2 x-\frac{1}{2}+1\right) x+\left(3 \times \frac{-1}{2}+1\right) y+\left(1+\frac{1}{2}\right) \\ z+4 \times\left(-\frac{1}{2}\right)-1=0\end{array}$
$\begin{aligned} \Rightarrow & \frac{-1}{2} y+\frac{3}{2} z-3 \\ = & =0 \\ \Rightarrow & y-3 z+6\end{aligned}=0$
which is the equation of the required plane.
Hence, the correct answer is option: 'A'.