3D Geometry 3 Question 27
####27. The plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $Y$-axis also passes through the point (2019 Main, 9 Jan I)
(a) $(3,3,-1)$
(b) $(-3,1,1)$
(c) $(3,2,1)$
(d) $(-3,0,-1)$
Show Answer
Answer:
(c)
Solution:
- The plane through the intersection of the planes $x+y+z-1=0$ and $2 x+3 y-z+4=0$ is given by $(x+y+z-1)+\lambda(2 x+3 y-z+4)=0$, where $\lambda \in R$
$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(4 \lambda-1)=0$, where
$\lambda \in R$
Since, this plane is parallel to $Y$-axis, therefore its normal is perpendicular to $Y$-axis.
$ \begin{array}{lc} \Rightarrow & {(1+2 \lambda) \hat{\mathbf{i}}+(1+3 \lambda) \hat{\mathbf{j}}+(1-\lambda) \hat{\mathbf{k}}} \cdot \hat{\mathbf{j}}=0 \\ \Rightarrow & 1+3 \lambda=0 \Rightarrow \quad \lambda=-\frac{1}{3} \end{array} $
Now, required equation of plane is
$ \begin{aligned} 1-\frac{2}{3} x+1-\frac{3}{3} y+1+\frac{1}{3} z+-\frac{4}{3}-1 & =0 \\ \text { [substituting } \lambda & =\frac{-1}{3} \text { in Eq. (i)] } \end{aligned} $
$\Rightarrow x+4 z-7=0$
Here, only $(3,2,1)$ satisfy the above equation.
