3 years ago

Find the linear equation of the plane through the origin and the points (5,2,2) and (6,−1,1). equation:

1 answer:

3 0

The cross of the position vectors is
.. [5, 2, 2] × [6, -1, 1] = [4, 7, -17] . . . . . the normal vector of the desired plane

Since the origin is a point in the plane, its equation can be written as
.. 4x +7y -17z = 0

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