Q. 264.8( 4 Votes )

Find the equation

Answer :

Let P1 be the first plane and equation of P1 is


or x + 3y – 6 = 0


And P2 be the second plane and its equation is


or 3x – y – 4z = 0


As the equation of the plane passing through intersection of 2 planes P1 and P2 is given by


P1 + λP2 = 0


x + 3y – 6 + λ(3x – y – 4z) = 0


x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0


Given the distance of this plane from (0,0,0) is 1.


Length of perpendicular from a point (x1,y1,z1) to a plane


ax + by + cz + d = 0 is given by –


p =


length of perpendicular from (0, 0, 0) to plane


x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0 is given by-


p =


As p = 1 (given)


Squaring both sides -



1 + 6λ + 9λ2 + 9 + λ2 – 6λ + 16λ2 = 36


26λ2 + 10 = 36


26λ2 = 26


λ2 = 1


λ = ±1


equation of planes are given by –


x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0


putting λ = 1


we have: 4x + 2y – 4z = 6


and putting λ = - 1


we have: -2x + 4y + 4z – 6 = 0


OR


Let P1 be the first plane and equation of P1 is


or x – y + 2z - 5 = 0


Normal vector to this plane is


And P2 be the second plane and its equation is


or 3x + y + z = 6


Normal vector to this plane is


As line is parallel to both the planes so it will be perpendicular to both normal vector and parallel to the cross product of


So we will first find



Expanding about the first row, we get –



=


As we have one point and got a line parallel to the line for which equation is asked. So we can give the equation easily.


Equation of the line is given by –


where is P.V of any random point on line. is the vector parallel to the line whose equation is given and is the P.V o any point on line.


[given]


equation of line is –



or


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