HP (Hewlett-Packard) 50g ユーザーズマニュアル

Page 9-17

Thus, M = (10i+26j+25k) m

⋅N.  We know that the magnitude of M is such that 

|M| = |r||F|sin(

θ), where θ is the angle between r and F.  We can find this 

angle as, 

θ = sin

-1

(|M| /|r||F|) by the following operations:

1 – ABS(ANS(1))/(ABS(ANS(2))*ABS(ANS(3)) calculates sin(

θ)

2 – ASIN(ANS(1)), followed by 

NUM(ANS(1)) calculates θ

These operations are shown, in ALG mode, in the following screens:

     

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    BS [3,-5,4] `

BS [2,5,-6] ` BS *  / SIN   NUM

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: