Question

The equation of a plane which cut off intercepts $a,b,c$ on $X,Y,Z$-axes respectively is

• A. $ax+by+cz=0$
• B. $ax+by+cz=1$
• C. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$
• D. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Answer 1

Answer: (D)

Let the equation of the plane be
$Ax+By+Cz+D=0(D\ne 0)$...(i)
Let the plane make intercepts $a,b,c$ on $X,Y$ and Z-axes, respectively (see figure).
Hence, the plane meets $X,Y$ and Z-axes at $(a,0,0)$, $(0,b,0),(0,0,c)$, respectively.

Therefore,
$Aa+D=0\text{ or }A=\frac{-D}{a}$
$Bb+D=0\text{ or }B=\frac{-D}{b}$
$Cc+D=0\text{ or }C=\frac{-D}{c}$
Substituting these values in the Eq. (i) of the plane and simplifying, we get
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
It is the required equation of the plane in the intercept form.