HP 49g+ - Manual (Page 289)

Page 9-18
Thus the angle between vectors r and F is θ = 41.038 o . RPN mode, we can
use: [3,-5,4] ` [2,5,-6] ` CROSS ABS [3,-5,4] `
ABS [2,5,-6] ` ABS * / ASIN NUM
Equation of a plane in space
Given a point in space P
0
(x
0
,y
0
,z
0
) and a vector N = N
x
i+N
y
j+N
z
k normal to a
plane containing point P
0
, the problem is to find the equation of the plane.
We can form a vector starting at point P
0
and ending at point P(x,y,z), a
generic point in the plane. Thus, this vector r = P
0
P = (x-x
0
) i+ (y-y
0
) j + (z-z
0
) k,
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
0
(2,3,-1) and the
normal vector N = 4i+6j+2k, we can enter vector N and point P
0
as two
vectors, as shown below. We also enter the vector [x,y,z] last:
Next, we calculate vector P
0
P = r as ANS(1) – ANS(2), i.e.,
Finally, we take the dot product of ANS(1) and ANS(4) and make it equal to
zero to complete the operation
N•r =0: