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

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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