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