Question

Reduce the equation of the plane 4x – 3y + 2z = 12 to the intercept form, and hence find the intercepts made by the plane with the coordinate axes.

Answer 1

Given : 

Equation of plane : 4x – 3y + 2z = 12 

To Find : 

1) Equation of plane in intercept form 

2) Intercepts made by the plane with the co-ordinate axes. 

Formula :

If \(\frac{x}a\) + \(\frac{x}b\) + \(\frac{x}c\) = 1

is the equation of a plane in intercept form then intercept made by it with co-ordinate axes are 

X-intercept = a 

Y-intercept = b 

Z-intercept = c 

Given the equation of plane: 

4x – 3y + 2z = 12 

Dividing the above equation throughout by 12

∴ \(\frac{4x}{12}\) + \(\frac{-3y}{12}\) + \(\frac{2z}{12}\) = 1

∴ \(\frac{x}{3}\) + \(\frac{y}{-4}\) + \(\frac{z}{6}\) = 1

This is the equation of a plane in intercept form. 

Comparing the above equation with

\(\frac{x}a\) + \(\frac{x}b\) + \(\frac{x}c\) = 1

We get, 

a = 3 

b = -4 

c = 6 

Therefore, intercepts made by plane with co-ordinate axes are 

X-intercept = 3 

Y-intercept = - 4 

Z-intercept = 6