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Suppose that you want to find the angle between vectors A = 3i-5j+6k, B =
2i+j-3k, you could try the following operation (angular measure set to degrees)
in ALG mode:
1 - Enter vectors [3,-5,6], press `, [2,1,-3], press `.
2 - DOT(ANS(1),ANS(2)) calculates the dot product
3 - ABS(ANS(3))*ABS((ANS(2)) calculates product of magnitudes
4 - ANS(2)/ANS(1) calculates cos(θ)
5 - ACOS(ANS(1)), followed by ,NUM(ANS(1)), calculates θ
The steps are shown in the following screens (ALG mode, of course):
!!!
Thus, the result is θ = 122.891
o
. In RPN mode use the following:
[3,-5,6] ` [2,1,-3] ` DOT
[3,-5,6] ` BS [2,1,-3] ` BS *
/ COS NUM
Moment of a force
The moment exerted by a force F about a point O is defined as the cross-
product M = r×F, where r, also known as the arm of the force, is the position
vector based at O and pointing towards the point of application of the force.
Suppose that a force F = (2i+5j-6k) N has an arm r = (3i-5j+4k)m. To
determine the moment exerted by the force with that arm, we use function
CROSS as shown next: