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Find the equations of the plane passing through the origin and parallel to the plane 2x – 3y + 7z + 13 = 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 – 3y + 7z + 13 = 0 

Normal vector = (2i - 3j + 7k) 

∴ Normal vector of required plane = (2i - 3j + 7k) 

Equation of required plane r . (2i - 3j + 7k) = d 

In cartesian form 2x - 3y + 7z = d 

Plane passes through point (0,0,0) therefore it will satisfy it. 

2(0) - (0) + 3(0) = d 

d = 0 

Equation of required plane 2x - 3y + 7z = 0



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