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Find the Equation of the Plane Through the Line of Intersection of `Vecr*(2hati-3hatj + 4hatk) = 1`And `Vecr*(Veci - Hatj) + 4 =0`And Perpendicular to the Plane `Vecr*(2hati - Hatj + Hatk) + 8 = 0`. Hence Find Whether the Plane Thus Obtained Contains the Line X − 1 = 2y − 4 = 3z − 12 - Mathematics

Find the equation of the plane through the line of intersection of `vecr*(2hati-3hatj + 4hatk) = 1`and `vecr*(veci - hatj) + 4 =0`and perpendicular to the plane `vecr*(2hati - hatj + hatk) + 8 = 0`. Hence find whether the plane thus obtained contains the line x − 1 = 2y − 4 = 3z − 12.

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Solution

The equation of the plane through the intersection of planes is

So, the vector normal to the plane and vector parallel to the line are perpendicular to each other.

Hence, the plane thus obtained contains the given line.

Concept: Vector and Cartesian Equation of a Plane
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