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Q.
The equation of a plane which cut off intercepts $a, b, c$ on $X, Y, Z$-axes respectively is
Three Dimensional Geometry
Solution:
Let the equation of the plane be
$A x+B y+C z+D=0(D \neq 0)$...(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,
$A a+D=0 \text { or } A=\frac{-D}{a}$
$B b+D=0 \text { or } B=\frac{-D}{b}$
$C c+D=0 \text { or } C=\frac{-D}{c}$
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.
