Vector Product 02 012 (part 1 of 3) 10.0 points Consider vectors A and B with co

Question

Vector Product 02 012 (part 1 of 3) 10.0 points Consider vectors A and B with coordinate components shown in the illustration below. 12 Note: Each coordinate has a length of 5 units. To indicate the coordinates of each vector, a line is projected to the horizontal plane then two lines are projected to the hor- izontal coordinates. As well, a line is directly projected to the vertical coordinate. Find the z component of A × B 013 (part 2 of 3) 10.0 points What is the magnitude of the vector product of these two vectors? 014 (part 3 of 3) 10.0 points What is the angle between A and B? Answer in units of.

Explanation / Answer

Answering Q1:

vector A:

x = 3*5 = 15
y = -2*5 = -10
z = 3*5 = 15

vector B:
x = -4*5 = -20
y = 4*5 = 20
z = 5*5 = 25

Thus, vector A = 15i - 10j + 15k
vector B = -20i + 20j + 25k

Now, |A| = (15^2 + 10^2 + 15^2)^0.5 = 23.45
|B| = (20^2 + 20^2 + 25^2)^0.5 = 37.75

Dot product, A.B = (15i - 10j + 15k).(-20i + 20j + 25k) = -300 - 200 + 375 = 125

A.B = |A|.|B|*cos(theta)
cos(theta) = 125/(23.45*37.75) = 0.14
theta = 81.88 ~ 82 degrees

Magnitude of cross product = |A|.|B|.sin(theta) = 23.45 * 37.75 * sin82 = 876.37

The answers are 876.37 and 82 deg

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