Find the point on the square of the curve X = t, y = t, and the third power of Z = t, so that the tangent of the point is parallel to the plane x + 2Y + Z = 4

Find the point on the square of the curve X = t, y = t, and the third power of Z = t, so that the tangent of the point is parallel to the plane x + 2Y + Z = 4

The normal vector of the plane is (1,2,1). Let the point be (T, T ^ 2, T ^ 3), then the tangent vector is (1,2t, 3T ^ 2). If it is perpendicular to the normal vector of the plane, then 1 + 2 * 2T + 3T ^ 2 = 0, T1 = - 1, T2 = - (1 / 3). So the point is (- 1,1, - 1) and (- 1 / 3,1 / 9, - 1 / 27)

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