Question

A point on the plane that passes through the points $(1,-1,6),(0,0,7)$ and perpendicular to the plane $x-2y+z=6$ is

• A. $(1,-1,2)$
• B. $(1,1,2)$
• C. $(-1,1,2)$
• D. $(1,1,-2)$

Answer 1

Answer: (B)

$(1,1,2)$

Let the direction ratios of the required plane be $(a,b,c)$ and it passes through the point $(0,0,7)$.

Equation of such a plane can be written as $ax+by+c(z-7)=0$

Now, this passes through the point $(1,-1,6)$

=> $a(-1)+b(-1)+c(6-7)=0$

=> $a=b+c$------(1)

As this plane is perpendicular to the plane $x-2y+z-6=0$, dot product of direction ratios of both the planes will be 0.

Direction ratios of this plane are $(1,-2,1)$

=> $1.a-2.b+1.c=0$

=> $a=2b-c$

=>$b+c=2b-c$

=> $b=2c$-----(2)

Putting this in (1), we get

=> $a=3c$

Equation can be written as

$3cx+2cy+c(z-7)=0$

=>$3x+2y+z-7=0$

passes through the point $(1,1,2)$