4.5 Intercept form of plane

2.1 Vectorial form of a line passing through a given point and parallel to a given vector
2.2 Vectorial form of a line passing through two given points
2.3 Cartesian form of a line passing through a given point and parallel to a given vector
2.4 Cartesian form of a line passing through two given points

3.1 Condition for perpendicularity
3.2 Condition for parallelism
3.3 Co-planarity of two lines
3.4 Shortest distance between two lines
3.5 Distance between two skew lines
3.6 Distance between parallel lines
3.7 Perpendicular distance of a point from a line
3.8 Image of a point in a straight line

4.1 Equation of plane in normal form
4.2 Equation of a plane perpendicular to a given vector and passing through a given point
4.3 Equation of plane passing through a given point and parallel to the two given vectors
4.4 Equation of plane passing through three non-collinear points
4.5 Intercept form of plane
4.6 Equation of plane passing through the intersection of two given planes

5.1 Angle between two planes
5.2 Angle between a line and a plane
5.3 Distance of a point from a plane
5.4 Distance between two parallel planes

The equation of a plane intercepting the lengths of $a$, $b$ and $c$ with $x$-axis, $y$-axis and $z$-axis respectively is $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

Proof: Let us assume the equation of plane in general form be $$Ax+By+Cz+D=0$$ where $D \ne 0$. The plane make intercepts of $a$, $b$ and $c$ on $X$-axis, $Y$-axis and $Z$-axis respectively.

Hence, the plane meets $X$-axis, $Y$-axis and $Z$-axis at $A(a,s0,0)$, $B(0,b,0)$ and $C(0,0,c)$ respectively as shown in figure.

Therefore, these co-ordinates must satisfy the equation of plane.

We get

$$\begin{equation} \begin{aligned} Aa + D = 0 \Rightarrow A = - \frac{D}{a} \\ Bb + D = 0 \Rightarrow B = - \frac{D}{b} \\ Cc + D = 0 \Rightarrow C = - \frac{D}{c} \\\end{aligned} \end{equation} $$

Put the above values in the equation of plane, we get $$\begin{equation} \begin{aligned} - \frac{D}{a}x - \frac{D}{b}y - \frac{D}{c}z + D = 0 \\ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \\\end{aligned} \end{equation} $$

Question 12. A variable plane moves in such a way that the sum of the reciprocals of its intercept on the three co-ordinate axes is constant. Show that the plane passes through the fixed point.

Solution: Let the equation of plane in intercept form be $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ Therefore, the intercepts made by the plane with axes are $$A(a,0,0);\quad B(0,b,0);\quad C(0,0,c)$$ It is given that the sum of reciprocals of its intercept on three co-ordinate axes is constant i.e., $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = {\text{constant(}}k{\text{)}}$$ We can write $$\frac{1}{{ka}} + \frac{1}{{kb}} + \frac{1}{{kc}} = 1$$ Compare it with $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ we get $$x = \frac{1}{k};\;\;y = \frac{1}{k};\;\;z = \frac{1}{k}$$ Therefore, we can say that the plane passes through the fixed point $$\left( {\frac{1}{k},\frac{1}{k},\frac{1}{k}} \right)$$