Question

If O be the origin and the coordinates of P be (1, 2, -3),then find the equation of the plane passing through P and perpendicular to OP.

Answer 1

The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, -3) respectively.

Therefore, the direction ratios of OP are (1-0) =1, (2-0)=2, and (-3-0)=-3

It is known that the equation of the plane passing through the point(x₁, y₁, z₁) is a(x-x₁) + b(y-y₁) + c(z-z₁) = 0
where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and -3, and the point P is (1, 2, -3).
Thus, the equation of the required plane is 1(x-1) + 2(y-2) - 3(z+3) = 0
⇒ x+2y-3z-14 = 0