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Find an equation of the plane. the plane through the origin and the points (3, −1, 8) and (6, 4, 1)

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Final answer:

To find the equation of a plane through the origin and two given points, we can use the cross product of the vectors formed by the two given points and the origin. The equation of the plane is -41x - 41y + 18z = 0.

Step-by-step explanation:

To find the equation of a plane through the origin and two given points, we can use the cross product of the vectors formed by the two given points and the origin.

Let's denote the two given points as A(3, -1, 8) and B(6, 4, 1).

The vector AB can be found by subtracting the coordinates of A from the coordinates of B: AB = (6 - 3, 4 - (-1), 1 - 8) = (3, 5, -7).

The vector AO can be found by subtracting the coordinates of A from the origin (0, 0, 0): AO = (0 - 3, 0 - (-1), 0 - 8) = (-3, 1, -8).

Now, we can find the cross product of AB and AO:

AB x AO = (5(-8) - (-7)(1), (-7)(-3) - 5(-8), 3(1) - 5(-3)) = (-41, -41, 18).

Therefore, the equation of the plane is -41x - 41y + 18z = 0.

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User Kstandell
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