Find the coordinates of a point on Y-axis which is at a distance of 3√3 from the point P(1, -2, 1).

Question

Question

Answer 1

Concept:

The distance between the two points P(x₁, x₂, x₃) and Q(y1, y2, y3) is given by

\(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

Let the point be A(0, y, 0). Then,

AP = 3√3

\(⇒ \sqrt{(0-1)^2+(y+2)^2+(0-1)^2}=3\sqrt3\)

On squaring both sides, we get

1 + (y + 2)² + 1 = 27

⇒ (y + 2)² = 25

⇒ y + 2 = ±5

⇒ y = 3, -7

Hence, the coordinates of the required point are (0, 3, 0) and (0, -7, 0).