Question

Find the equation of the plane whose x-intercept, y-intercept and z-intercept are 2, 3, 1. respectively.

Answer 1

Hint: Equation of a plane with a, b and c as the intercepts on the x, y and z- axes respectively is given as \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Intercepts are the corresponding coordinates of the intersections of a geometrical figure with a coordinate axis. In the xy-plane, the x-intercept of a line or a curve is the x-coordinates of its intersection with the x-axis.
In this question intercept of x, y, z axes are given, so substitute the value of intercept in equation \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\] and simplify the equation to get the equation of the required plane.

Complete step-by-step answer:
X-intercept parallel to the yz-plane \[x = 2\]
Y-intercept parallel to the xz-plane \[y = 3\]
Z-intercept parallel to the xy-plane \[z = 1\]
Equation of a plane with a, b and c is given as: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1 - - - - (i)\]
Now substitute the value of intercepts as \[x = 2\], \[y = 3\] and \[z = 1\] in the equation (i), we get:
\[\dfrac{x}{2} + \dfrac{y}{3} + \dfrac{z}{1} = 1 - - - - (ii)\]
Evaluating equation (ii) by taking LCM of the denominator, we get:
\[
  \dfrac{x}{2} + \dfrac{y}{3} + \dfrac{z}{1} = 1 \\
  \dfrac{{3x + 2y + 6z}}{6} = 1 - - - - (iii) \\
 \]
Cross-multiplying the terms in the equation (iii), we get:
\[
  \dfrac{{3x + 2y + 6z}}{6} = 1 \\
  3x + 2y + 6z = 6 \\
  3x + 2y + 6z - 6 = 0 \\
 \]
Hence, the equation of the plane whose x-intercept, y-intercept and z-intercept are 2, 3 and 1 respectively is \[3x + 2y + 6z - 6 = 0\]

Note: A plane is a two dimensional flat surface infinitely far extended on which straight line joining any two points would lie. A plane is the two dimensional analogue of a point, a line and the three dimensional space. Students should not misunderstand the intercepts here, with the direction ratios. Direction ratios are denoted by m,n and l for the x, y and z axes.