Problem 1/4 Page 13, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Vectors V1 and V2 shown in the figure.

Find: V1 + V2, V1 - V2, V1 X V2, V2 X V1 and V1 . V2

V1 is along a line with a slope of 3/4.  Recalling that 32 + 42 = 52, we can see that a triangle with sides 4, 3, 5 is similar to a triangle with sides V1X, V1Y, and V1.  As the magnitude of V1 is known to be 15:

V1X/V1 = 4/5     V1X = 4/5 V1 = 12
V1Y/V1 = 3/5     V1Y = 3/5 V1 = 9

V1 = 12 i + 9 j

The vector V2 is along a line that makes an angle of 60 degrees with the negative X axis.  We can use the sine and cosine of 60 degrees to determine the X and Y components of V2.  The magnitude of V2 is given as 12.

V2X = -V2 cos(60) = -12 * 1/2 = - 6
V2Y = V2 sin(60) = 12 * 3^1/2 = 6 3^1/2

V2 = -6 i + 6 3^1/2  j

V1 + V2 = ( 12 - 6) i + ( 9 + 6 3^1/2) j = 6 i + 19.39 j

V1 - V2 = ( 12 + 6) i + ( 9 - 6 3^1/2) j = 18 i - 13.92 j

V1 X V2 = 12 * 6 3^1/2 k - 9*(-6) k = 178.7 k

V2 X V1 = -6 * 9 k - 6 3^1/2 * 12 k = -178.7 k

V1 . V2 = 12 * -6 + 9 * 6 3^1/2 = 21.5