Q. 5

# Find the equation of the plane passing through the point (1, 4, - 2) and parallel to the plane 2x – y + 3z + 7 = 0.

Answer :

Formula : Plane = r . (n) = d

Where r = any random point

n = normal vector of plane

d = distance of plane from origin

If two planes are parallel , then their normal vectors are same.

Therefore ,

Parallel Plane 2x – y + 3z + 7 = 0

Normal vector = (2i - j + 3k)

∴ Normal vector of required plane = (2i - j + 3k)

Equation of required plane r . (2i - j + 3k) = d

In cartesian form 2x - y + 3z = d

Plane passes through point (1,4, - 2) therefore it will satisfy it.

2(1) - (4) + 3( - 2) = d

d = 2 – 4 – 6 = - 8

Equation of required plane 2x - y + 3z = - 8

2x - y + 3z + 8 = 0

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