Q. 74.0( 2 Votes )

Find the equation of the plane through (2, 3, – 4) and (1, – 1, 3) and parallel to the x – axis.

Answer :

We know that the equation of plane passing through (x1,y1,z1) is given by

a(x – x1) + b(y – y1) + c(z – z1) = 0 ……(1)


So, equation of plane passing through (2,3, – 4) is


a(x – 2) + b(y – 3) + c(z + 4) = 0 ……(2)


It also passes through (1, – 1, – 3)


So, equation (2) must satisfy the point (1, – 1, – 3)


a(1 – 2) + b(– 1 – 3) + c(– 3 + 4) = 0


– a – 4b + c = 0


a + 4b – 7c = 0 ……(3)


We know that line is parallel to plane


a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 = 0 ……(4)


Here, equation(2) is parallel to x axis,


……(5)


Using (2) and (5) in equation (4)


a×1 + b×0 + c×0 = 0


a = 0


Putting the value of a in equation (3)


a – 4b + 7c = 0


0 – 4b + 7c = 0


– 4b = – 7c


b =


Now, putting the value of a and b in equation (2)


a(x – 2) + b(y – 3) + c(z + 4)


0(x – 2) + (y – 3) + c(z + 4) = 0



7cy – 21c + 4cz + 16c = 0


Dividing by c we have,


7y – 21 + 4z + 16 = 0


7y + 4z – 5 = 0


Equation of required plane is 7y + 4z – 5 = 0


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