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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