Question

find the intercepts on the coordinate axes by the plane x + 2y - 2z = 5.

Answer 1

Step 1:

The given equation is x+2y−2z=5

Divide throughout by 5, we get

x/5 + 2y/5 – 2z/5 =1

Step 2:

Equation of a plane in intercept form is x/a +y/b +z/c =1

Where a, b, c are the intercepts cut off by the plane at x, y and z axes respectively.

a=5, b=2/5 and c=−2/5

Thus the intercepts cutoff by the plane are 2, 2/5 and −2/5

Answer 2

Answer:

     Given, the equation of plane is  

                           x + 2y - 2z = 5

Divide by 5, we get

                $\frac{X}{5} +\frac{2Y}{5} - \frac{2Z}{5} = 1$

                $\frac{X}{5} + \frac{Y}{\frac{5}{2} } + \frac{Z}{\frac{-5}{2} } = 1$  

which of the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

∴ a = 5, b = $\frac{5}{2}$ , c = $\frac{-5}{2}$

∴ Intercepts are $5, \frac{5}{2}, \frac{-5}{2}.$