Question

Equation of a plane passing through the point (2,1,-1) and (1,1,-2) and perpendicular to the plane x + 2y + 3z =4, is

• A. x + 2y + z - 3 = 0
• B. x + y - z - 4 = 0
• C. 2x = y + z - 4 = 0
• D. 2x + yt - z - 5 = 0

Answer 1

The correct option is B x + y - z - 4 = 0

The equation of the plane passing through the point (2,1,-1) $P_1$is

a(x-2) + b(y-1) + c(z+1)=0--(1)

$P_1$ is also pass through (1,1,-2)

Therefore, a(1-2) + b(1-1) + c(-2+1) =0

=>a+c=0

=>a=-c--(2)

$P_1$ is perpendicular to x+2y+3z=4.So,

=>$(a\hat{i}+b\hat{j}+c\hat{k})\cdot (1\hat{i}+2\hat{j}+3\hat{k})$=0

=>a$\cdot$1 + b$\cdot$2 + 3$\cdot$c =0

=>a+2b+3c=0--(3)

put the value of 2 in 3 ,we get:

=>-c+2b+3c=0

=>b=-c

Therefore,a = b = -c=k

=>a=k,b=k,c=-k--(4)

put the value of 4 in 1,

=>k(x-2)+k(y-1)+(-k)(z+1)=0

=>(x-2)+(y-1)-(z+1)=0

=>x+y-z-4=0

Hence,x+y-z-4=0 is the required plane.