Calc 3

Consider the points below.
P(2, 0, 2), Q(−2, 1, 4), R(7, 2, 6)
(a) Find a nonzero vector orthogonal to the plane through the points
P, Q,
and R.

(b) Find the area of the triangle PQR.

* I answered part a and got the correct answer with the vector <0,26,-13> I found this by solving the cross product. I can not get part B right I solved the answer to be 14.5 but this is apparently wrong. Please help.

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  1. Q-P = (-4,1,2)
    R-P = (5,2,4)

    Q-P x R-P = (0,26,-13)

    |Q-P x R-P| is the area of the parallelogram, so the area of PQR is ha;f that

    |Q-P x R-P| = 13 sqrt(5), so PQR hs area 6.5 sqrt(5) = 14.5

    Hmmm. Maybe the given answer is wrong.

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