Question

If equation of a plane is given as 4x + 2y + 12z = 7, then the x, y and z intercepts by the given plane will be .

• A. $\frac{7}{2},\frac{7}{4},\frac{7}{12}$
• B. $\frac{7}{4},\frac{7}{2},\frac{7}{12}$
• C. $\frac{4}{7},\frac{2}{7},\frac{12}{7}$
• D. $\frac{2}{7},\frac{4}{7},\frac{12}{7}$

Answer 1

The correct option is B $\frac{7}{4},\frac{7}{2},\frac{7}{12}$

We know that in the intercept form which is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, a,b & c are the intercepts made by the plane on x, y & z axes respectively.

The given plane is 4x + 2y + 12z = 7
On dividing the equation by 7

$\frac{4x}{7}+\frac{2y}{7}+\frac{12z}{7}=1$

$\Rightarrow \frac{x}{\left(\frac{7}{4}\right)}+\frac{y}{\left(\frac{7}{2}\right)}+\frac{z}{\left(\frac{7}{12}\right)}=1$

On comparing it with the standard intercept form we’ll get -

$a=\frac{7}{4},b=\frac{7}{2},c=\frac{7}{12}$

Thus the intercept on x , y & z axes will be $\frac{7}{4},\frac{7}{2}\&\frac{7}{12}$ respectively.