Find the equations of the plane passing through the point ( - 1, 0, 7) and parallel to the plane 3x – 5y + 4z = 11.
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 3x – 5y + 4z = 11
Normal vector = (3i - 5j + 4k)
∴ Normal vector of required plane = (3i - 5j + 4k)
Equation of required plane r . (3i - 5j + 4k) = d
In cartesian form 3x - 5y + 4z = d
Plane passes through point ( - 1,0,7) therefore it will satisfy it.
3( - 1) - 5(0) + 4(7) = d
d = - 3 + 28 = 25
Equation of required plane 3x - 5y + 4z = 25
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