The foot of the perpendicular from a point on the circle x 2 + y 2 = 1 , z = 0 to the plane 2 x + 3 y + z = 6 lies on which one of the following curves?

Step-by-step solution

Let the point on the circle be cos θ , sin θ , 0 and foot of the perpendicular from a point on the circle x 2 + y 2 = 1 , z = 0 to the plane 2 x + 3 y + z = 6 be h , k , w , Now using foot of perpendicular formula we get, h - cos θ 2 = k - sin θ 3 = w - 0 1 = - 1 2 cos θ + 3 sin θ - 6 14   . . . . . . 1 ⇒ h = cos θ + - 2 2 cos θ + 3 sin θ - 6 14 ⇒ h = 10 cos θ - 6 sin θ + 12 14   . . . . . 2 Now again from equation 1 we get, k = sin θ