or1 points 1 Previous Answers SCalcET8 125.026. 4/6 Submissions Used Find an equ

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or1 points 1 Previous Answers SCalcET8 125.026. 4/6 Submissions Used Find an equation of the plane. The plane through the point (7, 0, 4) and perpendicular to the line x-2, y = 5-t, z = 6 + 6t 3x Need Help? Deena Lunnel Submit Answer Save Progress to a Tutor 10. 1/1 points S Previous Answers SCalcET8 12.5 031· : ss submissions used Find an equation of the plane. The plane through the points (0, 8, 8), (8,0, 8), and (8, 8, 0) x+y+z= 16, 11. O 1/1 points 1 Previous Acswers SCaldETS 12.5.033. 1/6 Submissions Used Find an equation of the plane The plane through the points (2, 1, 2), (3,-8, 6), and (-2, -3, 1) 3 6

Explanation / Answer

10. We can get two vectors in the plane by subtracting pairs of points in the plane:

(0, 8, 8) - (8, 0, 8) = (-8, 8, 0)   and (8, 8, 0) - (8, 0, 8) = (0, 8, -8)

The cross product of these two vectors will be in the unique direction orthogonal to both, and hence in the direction of the normal vector to the plane.

(-8, 8, 0) x (0, 8, -8) = (-64, -64, -64)

The equation of a plane is ax+by+cz = d ==> -64x - 64y - 64z = d.

Plug in the point (8, 0, 8) to get -64*8 - 0 - 64*8 = d ==> d = -1024

then plane -64x - 64y - 64z = -1024 ==> x + y + z = 16.

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