Question

If equation of a plane is given $4x+2y+12z=7$ then $x,y\&z$ intercepts 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 the equation of a plane in intercept form which is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

$a,b\text{and}c$ are the intercepts made by the plane on $x,y\text{and}z$ axes respectively.

In the question the given equation is not in the intercept form, so we’ll convert it into intercept form and then compare.

$4x+2y+12z=7$

On dividing the equation by 7

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

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

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.