Let vectors A = (2,-1,1), B = (3,0,5) and C = (1,4,-2), where (x,y,z) are the co

Question

Let vectors A = (2,-1,1), B = (3,0,5) and C = (1,4,-2), where (x,y,z) are the components of the vectors along i, j, and k respectively. Calculate the following:

a) Express your answer as an ordered triplet absolute value (A), abs(B), abs (C) with commas to separate the magnitudes.

b) Vectors A*B

c) Determine the angle between B and C

d) A*(B x C)

Explanation / Answer

abs(A) = sqrt(2^2 + (-1)^2 + 1^2) = 2.44948974; abs(B) = sqrt(9 + 25) = 5.83095189 abs(C) = sqrt(1 + 16 + 4) = 4.58257569 b) A.B = 2*3 + -1*0 + 1*5 = 6 + 5 = 11; c) angle between B and C cos(theta)= B.C/abs(B)*abs(C) ==> angle = cos^-1( 3*1 + 0*4 + 5*-2 / (5.83095189 * 4.58257569) ); ==> theta = 105.186893 degrees d) B X C = i (0*-2 - 4*5) - j (5*1 - 3*-2) + k (3*4 - 1*0) = -20 i -11 j + 12 k; A*(B X C) = 2*-20 + -1*-11 + 1*12 = -18;

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