Question

Find the intercepts made on the coordinate axes by the plane 2x + y – 2z = 3 and also find the direction cosines of the normal to the plane.

Answer 1

The given equation of the plane is 2x + y – 2z = 3

Dividing by 3 on both the sides, we get

We know that, if a, b, c are the intercepts by the plane on the coordinate axes, new equation of the plane is

Comparing the equation (i) and (ii), we get

Again the given equation of the plane is

2x + y - 2z = 3

Writing this in the vector form, we get

So vector normal to the plane is given by

Direction vector of  \(\vec n\) = 2, 1, -2

Direction vector of \(\vec n\) 

So,

Intercepts by the plane on the coordinate axes are = \(\cfrac23,3,-\cfrac32\)

Direction cosines of normal to the plane are = \(\cfrac23,\cfrac13,-\cfrac23\)