# Find the vector equation of the plane through the points (2, 1, – 1) and (– 1, 3, 4) and perpendicular to the planex – 2y + 4z = 10.[CBSE 2013]

Vector equation of a plane is given as Where is any point on the plane and is a vector perpendicular to the plane.

Now, the given plane x – 2y + 4z = 10 is perpendicular to required plane. So, the normal vector of x – 2y + 4z = 10 will be parallel to the required plane. Hence, is parallel to the required plane.

Points say A(2,1, – 1) and B(– 1,3,4) are on the plane hence the vector is also parallel to the required plane so, is parallel to the required plane.

Hence as both and are parallel to the plane so the direction of is the cross product of the two vectors. So, the equation of required plane is,     Hence, the vector equation of required plane is .

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Find the coordinate of the point P where the line through and crosses the plane passing through three points and Also, find the ratio in which P divides the line segment AB.

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