Prove that the normals to the planes 4x + 11y + 2z + 3 = 0 and 3x - 2y + 5z = 8 are perpendicular to each other.

Given :


Equations of plane are :


4x + 11y + 2z + 3 = 0


3x – 2y + 5z = 8


To Prove : are perpendicular.


Formula :


1) Dot Product :


If are two vectors




then,



Note :


Direction ratios of the plane given by


ax + by + cz = d


are (a, b, c).


For plane


4x + 11y + 2z + 3 = 0


direction ratios of normal vector are (4, 11, 2)


therefore, equation of normal vector is



And for plane


3x – 2y + 5z = 8


direction ratios of the normal vector are (3, -2, 5)


therefore, the equation of normal vector is



Now,




= 12 – 22 + 10


= 0



Therefore, normals to the given planes are perpendicular.


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