Equation Of A Plane In Space - HP F2226A - 48GII Graphic Calculator User Manual

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Thus the angle between vectors r and F is θ = 41.038
use: [3,-5,4] ` [2,5,-6] ` CROSS
ABS [2,5,-6] ` ABS * / ASIN

Equation of a plane in space

Given a point in space P
plane containing point P
We can form a vector starting at point P
generic point in the plane. Thus, this vector r = P
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
normal vector N = 4i+6j+2k, we can enter vector N and point P
vectors, as shown below. We also enter the vector [x,y,z] last:
Next, we calculate vector P
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:
) and a vector N = N
(x
,y
,z
0
0
0
0
, the problem is to find the equation of the plane.
0
and ending at point P(x,y,z), a
0
P = r as ANS(1) – ANS(2), i.e.,
0
o
. RPN mode, we can
ABS [3,-5,4] `
NUM
i+N
j+N
k normal to a
x
y
z
P = (x-x
)i+ (y-y
)j + (z-z
0
0
0
(2,3,-1) and the
0
as two
0
Page 9-18
)k,
0

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