Question

Show that the complete integral of z = px +qy - 2p - 3q represent all possible planes through the points (2, 3, 0)

Answer 1

Given that

$$z=px+qy-2p-3q\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

Its complete integral is

$$z=ax+by-2a-3b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

$a, b$ being arbitrary constants.

Differentiating $(2)$ partially with respect to $x$

$${\partial z \over \partial x}=a=>p=a$$

Differentiating $(2)$ partially with respect to $y$

$${\partial z \over \partial y}=b=>q=b$$

Substituting in $(2)$ we get

$$z=px+qy-2p-3q$$

The equation $(2)$ is a linear equation in $x,y,z.$ Hence it represents planes for various values of 𝑎 and 𝑏.  Substitute $x=2, y=3, z=0$

$$2a+3b-2a-3b=0=>0=0$$

The coordinates of the point (2, 3, 0) satisfy the equation $(2).$ Hence the complete integral $(2)$ of $(1)$ represents all possible planes passing through the point (2, 3, 0).