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Question

Mathematics Question on Equation of a Line in Space

The equation of the plane through the points (2,2,1)(2, 2,1) and (9,3,6)(9, 3, 6) and perpendicular to the plane 2x+6y+6z1=02x + 6y + 6 z - 1 = 0 is

A

3x+4y+5z+9=03x + 4y + 5z + 9 = 0

B

3x+4y5z+9=03x + 4y - 5z + 9 = 0

C

3x4y+5z+9=03x - 4y + 5z + 9 = 0

D

3x +4y 5z 9 = 03x\ + 4y\ -5z\ -9\ =\ 0

Answer

3x +4y 5z 9 = 03x\ + 4y\ -5z\ -9\ =\ 0

Explanation

Solution

The equation of plane through the point (2,2,1) is
a(x-2) + b(y-2) + c(z-1) = 0 ………(1)
Since this line passes through (9,3,6)
a(9-2) + b(3-2) + c(6-1) = 0
7a+b+5c = 0 ………(2)
Since plane (1) is perpendicular to the plane 2x+6y+6z = 9
a(2) + b(6) + c(6) = 0
2a + 6b +6c = 0
a + 3b + 3c = 0 ………(3)
from eq (2) and eq (3)
a315\frac {a}{3-15} = b521\frac {b}{5-21} = c211\frac {c}{21-1}
a12\frac {a}{-12} = b16\frac {b}{-16} = c20\frac {c}{20}
a3\frac a3= b4\frac {b}{4} = c5\frac {c}{-5}
Let a3\frac a3 = b4\frac b4 = c5\frac {c}{-5} = k
Then a = 3k, b= 4k and C= -5k
From eq (1)
3k(x-2) + 4k(y-2) + (-5k)(z-1) = 0
3x - 6 +4y - 8 - 5z + 5 = 0
3x + 4y - 5z - 9 = 0
This is the required equation of plane.

So, the correct answer is (D): 3x + 4y - 5z - 9 = 0