Question

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

Answer 1

Hint: To find the equation of a plane, we need the direction ratios of a vector that is perpendicular to this plane and we need a point which is lying on this plane. We are given two points O and P. Find the direction ratio of the vector OP. Since OP is perpendicular to the plane, we can get the direction ratio of a vector perpendicular to the plane. Also, we are given that point P is lying on the plane. Using the normal form of the plane, we can find the equation of the plane.

Complete step-by-step answer:

Before proceeding with the question, we must know the formula that will be required to solve this question.

In 3D geometry, if we are given the direction ratios of a vector that is perpendicular to a plane as (l, m, n) and also we are given the a point (a, b, c) that is lying on the plane, the equation of the plane is given by the formula,

lx + my + nz = la + mb + nc . . . . . . . . . . . . . (1)

In the question, we are given that OP is perpendicular to the plane. The coordinate of O is (0, 0, 0) and the coordinate of P is (1, 2, -3). So, the direction ratio of OP i.e. vector perpendicular to the plane is (1-0, 2-0, -3-0) i.e. (1, 2, -3). Also, it is given that point P (1, 2, -3) lies on the plane. Using formula (1), the equation of plane is,

1x + 2y + (-3)z = (1)(1) + (2)(2) + (-3)(-3)

$\Rightarrow $ x + 2y – 3z = 1 + 4 + 9

$\Rightarrow $ x + 2y – 3z = 14

Hence, the equation of the plane is x + 2y – 3z = 14.


Note: If we were given two vectors and a point and also, we were given that these two vectors lies along the plane, then, in that case, to find the vector perpendicular to the plane, we would have found the cross product of these two vectors from which we can get a vector perpendicular to the two vectors or plane. Then using normal form, we could have found the equation of the plane.