1. Let a = 3i + j - k and b = i - j + 2k.
A. Determine if a and b are orthogonal, parallel or neither.
B. Compute the scalar a vector projection of b onto a, i.e. comp_a (b) and proj_a(b).
2. Find the area of the parallelogram ABCD with the points A(0,1), B(3,0), C(5,-2), D(2,-1).
B. Since a and b are orthogonal, there is no component of
b in the direction of a, thus
comp_a(b) = 0 and proj_a(b) = 0(vector).
2. Let a = AB = (3,-1,0) and b = AD = (2,-2,0), then the
area is |a x b|.
a x b = (0, 0, 3(-2)-(-1)2 = -4), thus the area is |-4| = 4.
Note that even if you used the wrong vectors, like a = AB and b = AC (for the parallelogram ABDC) you would get the same area!
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