3D Geometry 3 Question 11
11. The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4 z=5$, which is perpendicular to the plane $x-y+z=0$ is
(2019 Main, 8 April II)
(a) $\mathbf{r} \cdot(\hat{\mathbf{i}}-\hat{\mathbf{k}})-2=0$
(b) $\mathbf{r} \times(\hat{\mathbf{i}}+\hat{\mathbf{k}})+2=0$
(c) $\mathbf{r} \times(\hat{\mathbf{i}}-\hat{\mathbf{k}})+2=0$
(d) $\mathbf{r} \cdot(\hat{\mathbf{i}}-\hat{\mathbf{k}})+2=0$
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Answer:
(d)
Solution:
- Since, equation of planes passes through the line of intersection of the planes
$ x+y+z=1 $
and $2 x+3 y+4 z=5$, is
$ (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 $
$\Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-(1+5 \lambda)=0$
$\because \quad$ The plane (i) is perpendicular to the plane
$ x-y+z=0 \text {. } $
$\therefore \quad(1+2 \lambda)-(1+3 \lambda)+(1+4 \lambda)=0$
$\left[\because\right.$ if plane $a _1 x+b _1 y+c _1 z+d _1=0$ is perpendicular to plane $a _2 x+b _2 y+c _2 z+d _2=0$, then $\left.a _1 a _2+b _1 b _2+c _1 c _2=0\right]$
$ \Rightarrow \quad 3 \lambda+1=0 $
$ \Rightarrow \quad \lambda=-\frac{1}{3} $
So, the equation of required plane, is
$ \begin{aligned} & 1-\frac{2}{3} x+1-\frac{3}{3} y+1-\frac{4}{3} z-1-\frac{5}{3}=0 \\ & \Rightarrow \frac{1}{3} x-\frac{1}{3} z+\frac{2}{3}=0 \Rightarrow x-z+2=0 \end{aligned} $
Now, vector form, is $\mathbf{r} \cdot(\hat{\mathbf{i}}-\hat{\mathbf{k}})+2=0$
