HP 50g User Manual page 294

Thus, M = (10i+26j+25k) m N. We know that the magnitude of M is such that
|M| = |r||F|sin( ), where is the angle between r and F. We can find this
-1
angle as, = sin (|M| /|r||F|) by the following operations:
1 – ABS(ANS(1))/(ABS(ANS(2))*ABS(ANS(3)) calculates sin( )
2 – ASIN(ANS(1)), followed by NUM(ANS(1)) calculates
These operations are shown, in ALG mode, in the following screens:
o
Thus the angle between vectors r and F is = 41.038 . RPN mode, we can
use: [3,-5,4] ` [2,5,-6] ` CROSS BS [3,-5,4] `
BS [2,5,-6] ` BS * / SIN NUM
Equation of a plane in space
Given a point in space P (x ,y ,z ) and a vector N = N i+N j+N k normal to
0 0 0 0 x y z
a plane containing point P , the problem is to find the equation of the plane.
0
We can form a vector starting at point P and ending at point P(x,y,z), a
0
generic point in the plane. Thus, this vector r = P P = (x-x )i+ (y-y )j + (z-z )k,
0 0 0 0
is perpendicular to the normal vector N, since r is contained entirely in the
plane. We learned that for two normal vectors N and r, N r =0. Thus, we can
use this result to determine the equation of the plane.
To illustrate the use of this approach, consider the point P (2,3,-1) and the
0
normal vector N = 4i+6j+2k, we can enter vector N and point P as two
0
vectors, as shown below. We also enter the vector [x,y,z] last:
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