We know that the general equation of a plane is given by,
Ax + By + Cz + D=0, where D ≠ 0.....(1)
Here, A, B, C are the co - ordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any general point through which the plane passes.
Now let us say, this plane is making intercepts at points P, Q, and R on the x, y, and z - axes respectively at (a, 0, 0), (0, b, 0) and (0, 0, c).
So, the plane cuts the x - axis, y - axis and z - axis at three points P(a, 0, 0), Q(0, b, 0) and R(0, 0, c) respectively.
Since the plane also passes through each of these three points, we can substitute them into equation (1) i.e. general equation of the plane and we have,
(i) Aa + D = 0
⇒ A = \(-\cfrac{D}a\)
(ii) Bb + D = 0
⇒ B = \(-\cfrac{D}b\)
(iii) Cc + D=0
⇒ C = \(-\cfrac{D}c\)
Substituting these values of A, B, and C in equation (1) of the plane, we shall get the equation of a plane in intercept form, which is given by,
if the plane makes intercepts at (a, 0, 0), (0, b, 0) and (0, 0, c) with the x - , y - and z - axes respectively.
