Engineering Statics

B-9


  Given the vectors P = -2i + 3j + 4k and Q = -i + 4j - 6k calculate the angle between vectors P and Q using the vector product.

The vector or cross product is defined the the product of 
the absolute values of the magnitudes of the two vectors times 
the sine of the angle between them times a unit vector in the 
direction of the cross product.  The absolute value of the 
vector product does not need the unit vector.
The cross product of P and Q was found in problem b-06.  To find the 
magnitude, the square root of the sum of the squares of the components 
of the cross product (34<sup>2</sup>)+(-16)<sup>2</sup>)+(-5)<sup>2</sup>) 
are taken.  This gives 37.9

The magnitude of Q is given by the square root of the sum of the 
squares of the components of Q.
The sine of <font face=symbol>q</font> can then be found by substitution.
The angle may then be found using the inverse or arcfunction 
on the calculator.  Notice that since the sine is positive, the 
angle is less than 90 degrees.