Thread: Vector calculus - planes

A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.

Originally Posted by njr008

A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.

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Now substitute z = 4 - x + 2y. Expand. Simplify. Spadework for you to do.

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Let and find extrema in the usual way:

Solve simultaneously:

.... (1)

.... (2)

Spadework for you.

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Get the vector QP and show that QP is parallel to the normal to the plane (which is seen from the Cartesian equation of the plane to be i - 2j + k).