Answer:
6 5/7 ≈ 6.714
Step-by-step explanation:
You want the distance from point A(5, -4, 8) to the plane defined by B(2, 3, 1), C(4, 1, -2), and D(6, 3, 7).
Direction vector
The vector in the direction perpendicular to the plane can be found as ...
U = (CB×CD) . . . . . . vector in the direction ⊥ to BCD
u = U/||U|| . . . . . . unit vector in that direction
The values of these are ...
U = (-2, 2, 3) × (2, 2, 9) = (12, 24, -8) = 4(3, 6, -2)
u = (3, 6, -2)/√(3² +6² +2²)
u = (3/7, 6/7, -2/7) . . . . unit vector perpendicular to the plane
Length
Then the distance from A is the dot product of a vector between A and any point on the plane with the unit vector we just found.
AB•u = (-3, 7, -7)•(3/7, 6/7, -2/7) = -9/7 +6 +2 = 6 5/7
The distance from A to the plane BCD is 6 5/7 units.
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Additional comment
The distance is the absolute value of the dot product. We chose to use vector AB so the dot product would be positive.
