Q. 363.5( 13 Votes )

Find the equation

Answer :

Given: A plane passing through the points A(2,1, 2) and B(4, -2, 1) and perpendicular to plane


To Find: Equation of the plane. Also find coordinates of the point where the line passing through the points (3,4,1) and (5,1,6) crosses the plane thus obtained.


Let, P(x,y,z) be a point on the plane.


A and B both lie on the same plane.


Normal to the plane is =


And this plane is perpendicular to the plane


Normal of this plane is,


So, we can say,


--- (i)


Here, and



Putting the values in (i) we get,



-(y – 1) + 3(z – 2) -2{-3(x – 2) – 2(y – 1)} = 0


-y + 1 + 3z – 6 + 6(x – 2) + 4(y – 1) = 0


-y + 1 + 3z – 6 + 6x – 12 + 4y – 4 = 0


6x + 3y + 3z = 21


2x + y + z = 7


So, the equation of the plane is 2x + y + z = 7 [Answer(i)]


Equation of line passing through the points (3,4,1) and (5,1,6) is,



So, x = 2k + 3


y = -3k + 4


z = 5k + 1


Putting the values of x, y, z in the previous plane equation we get,


2(2k + 3) – 3k + 4 + 5k + 1 = 7


4k + 6 – 3k + 4 + 5k + 1 = 7


6k = -4


k =


So, the coordinates of the point crossing the plane are,


x = 2 + 3 =


y = -3 + 4 = 6


z = 5 + 1 =


So, the point is = (, 6, ) [Answer(ii)]


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