Find the equation of the plane through the line of intersection of r (2^ i 3^ j + 4^ k) = 1 and r (^ i ^ j) + 4 = 0 and perpendicular to the plane r (2^ i ^ j + ^ k) + 8 = 0. Hence find whether the plane thus obtained contains the line x 1 = 2y 4 = 3z 12.
Answers
Equation of plane is 5x + 2y - 4z + 47 = 0 & this plane doesn't contain the line x /1 = 2y /4 = 3z/ 12.
•Equation of planes
P1 : r (2^ i 3^ j + 4^ k) = 1
2x + 3y + 4z = 1
P2: r (^ i ^ j) + 4 = 0
x + y + 4 = 0
•Equation of plane passing through line of intersection of two planes is:
P1 + h(P2) = 0
( 2x + 3y + 4z - 1 ) + h ( x + y + 4 ) = 0
x(2+h) + y(3+h) + 4z + 4h -1 = 0
______(1)
•This plane is perpendicular to
r (2^ i ^ j + ^ k) + 8 = 0
•dot product of direction ratios of the given planes will be zero
2(2+h) + (3+h) + 1(4) = 0
4 + 2h + 3 + h + 4 = 0
3h = -11
h = -11/3
•So, equation of plane is
x(-5/3) + y(-2/3) + 4z + (-47/3) = 0
5x + 2y - 4z + 47 = 0
____________(2)
•If a plane contains the line then the Dot product of direction ratios of the plane and line is 0 & plane passes through the point
•Now Drs of line are 1,2,4
Drs of plane are 5,2,-4
Dot product of Drs =
(1)+(5) + (2)(2) + (4)(-4)
= -7
•Hence , plane doesn't contain the line