The equation of the plane which passes through ( 2 , − 3 , 1 ) and is normal to the line joining the points ( 3 , 4 , − 1 ) and ( 2 , − 1 , 5 ) is given by

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The required plane passes through ( 2 , − 3 , 1 ) . It is normal to the line having d.r.s. ( 1 , 5 , − 6 ) . ∴ x + 5 y − 6 z = k ⇒ 2 + 5 ( − 3 ) − 6 ( 1 ) = 1 i.e. k = − 19 Hence equation of plane is x + 5 y − 6 z + 19 = 0

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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