A given parallelogram has vertices at A(2,-1,4), B(1,0,-1), C(1,2,3), D(2,1,8).

Question

A given parallelogram has vertices at A(2,-1,4), B(1,0,-1), C(1,2,3), D(2,1,8). A.Find the cosine of the interior angle at B, B.the vector projection of BA onto BC, C.the area of the parallelogram and D.an equation for the plane of the parallelogram. answers should be equal to a.3/sqrt(15) b.2j+4k c.6sqrt6 d. 7x+y -2=8

Explanation / Answer

Cosine ==> dot product of A(2,-1,4), B(1,0,-1), C(1,2,3) vectors intertior angle ==> cos-1(root(3/5)) BA = i-j+5k BC = 2j + 4k Hence the projection will be BC itself (BC doesnt have i) c.Area = 2abcosx = 36/root(6) = 6root(6) d.plane of ABC = 7x+y-z -8 =0

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