A plane P contains the line


A plane P contains the line

$x+2 y+3 z+1=0=x-y-z-6$

and is perpendicular to the plane $-2 x+y+z+8=0$. Then which of the following points lies on P?

  1. $(-1,1,2)$

  2. $(0,1,1)$

  3. $(1,0,1)$

  4. $(2,-1,1)$

Correct Option: , 2


Equation of plane P can be assumed as

$P: x+2 y+3 z+1+\lambda(x-y-z-6)=0$

$\Rightarrow \mathrm{P}:(1+\lambda) \mathrm{x}+(2-\lambda) \mathrm{y}+(3-\lambda) \mathrm{z}+1-6 \lambda=0$

$\Rightarrow \overrightarrow{\mathrm{n}}_{1}=(1+\lambda) \hat{\mathrm{i}}+(2-\lambda) \hat{\mathrm{j}}+(3-\lambda) \hat{\mathrm{k}}$

$\therefore \quad \vec{n}_{1} \cdot \vec{n}_{2}=0$

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

$\Rightarrow 2+2 \lambda-2+\lambda-3+\lambda=0 \Rightarrow \lambda=\frac{3}{4}$\

$\Rightarrow P: \frac{7 x}{4}+\frac{5}{4} y+\frac{9 z}{4}-\frac{14}{4}=0$

$\Rightarrow 7 x+5 y+9 z=14$

$(0,1,1)$ lies on $\mathrm{P}$

Leave a comment