Planes
in Space
Let ro = OP be a vector from the origin to point P. So OP
= i + j + k Let OQ be a vector from the origin to point Q. So OQ
= i ˗
j + 3
k Since both vectors OP and OQ
are vectors in the plane, the cross product yields the normal vector, n. n
= OP x OQ (vector cross product) i j k n =
det
1 1 1 where det denotes the 3x3 determinant 1 - 1
3 n = i (3 + 1) - j (3
– 1) + k (-1 -1) = 4
i - 2 j
-2 k Also let r be a vector to an arbitrary point (x, y, z)
in the plane. So r = x
i + y j + z
k So r -
ro =
i
(x - 1) - j (y – 1) + k (z -1)
Note: r -
ro is a vector in the desired
plane. and therefore, n and
r -
ro are perpendicular to each other so the dot
product is zero. n ∙ ( r -
ro
) = 0 =
4(x - 1) - 2(y – 1) - 2(z -1)
or 2(x - 1)
- (y – 1) - (z -1)
= 0 2x -
y - z
= 0 (result for equation of the plane) |
Copyright © 2017 Richard C. Coddington
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