Question

Solve the following :

Find the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.

Answer 1

Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the cartesian equation of the plane is ax + by + cz = 0.      ...(1)
(1, 2, 3) d (3, 2, 1) lie on the plane.
∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ `a/|(2, 3),(2, 1)| = b/|(3, 1),(1, 3)| = c/|(1, 2),(3, 2)|`

∴ `a/(-4) = b/(8) = c/(-4)`

i.e. `a/(1) = b/(-2) = c/(1)`

∴ a, b, c are proprtional to 1, – 2, 1
∴ from (1), the required cartesian equation is x –2y + z = 0

Case 2 : Let he plane make non zero intercept p on each axis.

then its equation is `x/p + y/p + z/p` = 1

i.e. x + y + z = p                                                      ...(2)
Since this plane pass through (1, 2, 3) and (3, 2, 1)

∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p

∴ p = 6

∴ from (2), the required cartesian equation is
x + y + z = 6
Hence, the cartesian equations of required planes are
x + y + z = 6 and x – 2y + z = 0.