Question

Find the point where the line of intersection of the planes $x-2y+z=1$ and $x+2y-2z=5$ intersects the plane $3x+2y+z+6=0$.

• A. $P(1,-2,-4)$
• B. $P(1,2,-4)$
• C. $P(1,-2,4)$
• D. None of these

Answer 1

The correct option is A $P(1,-2,-4)$

$P_1\equiv x-2y+z=1$

$P_2\equiv x+2y-2z=5$

Let $D{r}^{\prime }s$ of intersection of planes be $a,b,c$

$\Rightarrow a-2b+c=0$ and $a+2b-2c=0$

$\Rightarrow \frac{a}{4-2}=\frac{b}{1+2}=\frac{c}{2+2}\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{4}$

Calculating a point which lies on both plane $P_1$ and $P_2$,

$x-2y+z=1\Rightarrow 2y=x+z-1$ and $x+2y-2z=5\Rightarrow 2y=-x+2z+5$

$\Rightarrow x+z-1=-x+2z+5\Rightarrow 2x=z+6$

Now, $z=0\Rightarrow x=3\Rightarrow y=1$

$\therefore$ equation of line passing through $(3,1,0)$ and has $d{r}^{\prime }s$ $2,3,4$ is

$\therefore \frac{x-3}{2}=\frac{y-1}{3}=\frac{z}{4}=\lambda$(say)

$\therefore$ general point on it is $P(2\lambda +3,3\lambda +1,4\lambda )$

Now, solving with plane $2x+2y+z+6=0$

$\Rightarrow 2(2\lambda +3)+2(3\lambda +1)+4\lambda +6=0\Rightarrow 4\lambda +6+6\lambda +2+4\lambda +6=0\Rightarrow 14\lambda +14=0\Rightarrow \lambda =-1$

$\Rightarrow P(1,-2,-4)$ is the required point of intersection.