A plane in R 3 is a subspace generated by two nonlinear vectors. Let P be the plane in R 3 that goes through the points (0, 0, 0), (1, 2, 0)

Question

A plane in R 3 is a subspace generated by two nonlinear vectors. Let P be the plane in R 3 that goes through the points (0, 0, 0), (1, 2, 0), and (0, 0, 1). Does the point (3, 6, 4) lie on this plane

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Caroline 2 months 2021-10-15T14:48:18+00:00 1 Answer 0 views 0

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    2021-10-15T14:49:28+00:00

    Answer:

    Yes, it lies on the plane

    Step-by-step explanation:

    Note that the generator (1,2,0) will always make vectors whose third coordinate is always 0. In order to obtain a 4 in the third coordinate, we need to use the generator (0,0,1). Specifically, we need (0,0,1) to be multiplied by 4, if we do that we obtain 4*(0,0,1) = (0,0,4).

    We still need to obtain (3,6,4)-(0,0,4) = (3,6,0) from the generator (1,2,0), which is possible because (3,6,0) = 3*(1,2,0). Therefore

    3*(1,2,0) + 4 *(0,0,1) = (3,6,4)

    Thus, the point (3,6,4) is generated by the two vectors and, as a result, this points lies in the plane.

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