Question

Find the equation of the plane through the points $ \left( {0,4, - 3} \right) $ and $ \left( {6, - 4,3} \right) $ other than the plane through the origin and which cuts off intercept on the axes whose sum is 0.

Answer 1

Hint: The equation of the plane in intercept form is given as $ \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1 $ .The two points given will satisfy the equation of the plane and also the sum of the intercepts given will enable us to solve the value of $ a,b $ and $ c $ .

Complete step-by-step answer:
Given information
The plane passes through the points $ \left( {0,4, - 3} \right) $ and $ \left( {6, - 4,3} \right) $ .
The plane makes an intercept on the three axes such that, the sum of the intercepts is equal to
The intercept form of the equation of the plane is given by,
\[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1 \cdots \left( 1 \right)\]
Where $ a,b $ and $ c $ are the $ x,y $ and $ z $ intercepts of the plane.
The equation of the plane given in equation (1) will satisfy the two given points.
Put $ \left( {0,4, - 3} \right) $ in equation (1), we get
 $ \dfrac{0}{a} + \dfrac{4}{b} + \dfrac{{ - 3}}{c} = 1 \cdots \left( 2 \right) $
On simplifying equation (2) it becomes as
 $ - \dfrac{4}{b} + \dfrac{3}{c} = - 1 \cdots \left( 3 \right) $
Put $ \left( {6, - 4,3} \right) $ in equation (3), we get
  $ \dfrac{6}{a} - \dfrac{4}{b} + \dfrac{3}{c} = 1 \cdots \left( 4 \right) $
Now substitute the value of
 $ \dfrac{6}{a} - 1 = 1 \cdots \left( 5 \right) $
Solving equation (5) for
 $
  2a = 6 \\
  a = 3 \cdots \left( 6 \right) \\
  $
Sum of the intercepts made by the plane on three axes is equal to $ 0 $ . It can be written as
 $ \Rightarrow a + b + c = 0 \cdots \left( 7 \right) $
Substituting $ a = 3 $ in equation (7), we get the equation in the form of $ b $ and $ c $ only as
 $
\Rightarrow 3 + b + c = 0 \\
\Rightarrow b = - c - 3 \cdots \left( 8 \right) \\
  $
Substitute the value of $ b = - c - 3 $ in equation (3), we get
 $ - \dfrac{4}{{ - c - 3}} + \dfrac{3}{c} = - 1 \cdots \left( 9 \right) $
Solving in terms of, we get
 $
  \dfrac{4}{{c + 3}} + \dfrac{3}{c} = - 1 \\
  \dfrac{{4c + 3\left( {c + 3} \right)}}{{c\left( {c + 3} \right)}} = - 1 \\
  4c + 3c + 9 = - {c^2} - 3c \\
  {c^2} + 10c + 9 = 0 \cdots \left( {10} \right) \\
  $
Equation (10) is a quadratic equation in terms of .
 $
\Rightarrow {c^2} + 9c + c + 9 = 0 \\
  c\left( {9 + c} \right) + 1\left( {c + 9} \right) = 0 \\
  \left( {c + 9} \right)\left( {c + 1} \right) = 0 \\
  $
Therefore, two values of $ c $ are $ - 9 $ and $ - 1 $ .
Value of $ b $ can be obtained from equation (8) by substituting the two values of $ c = - 9, - 1 $ as,
 $
  b = - \left( { - 9} \right) - 3 \\
  b = 6 \\
  $
Also ,
 $
  b = - \left( { - 1} \right) - 3 \\
  b = - 2 \\
  $
We have two values of $ b $ and $ c $ . Therefore, two equations of planes are possible.
Substitute the value of $ a = 3,b = 6 $ and $ c = - 9 $ in equation (1), we get the first equation of plane
 $ \Rightarrow \dfrac{x}{3} + \dfrac{y}{6} - \dfrac{z}{9} = 1 \cdots \left( {11} \right) $
Substitute the value of $ a = 3,b = - 2 $ and $ c = - 1 $ in equation (1), we get the second equation of the plane.
 $ \Rightarrow \dfrac{x}{3} - \dfrac{y}{2} - \dfrac{z}{1} = 1 \cdots \left( {12} \right) $
Thus, the two equation of plane are $ \dfrac{x}{3} + \dfrac{y}{6} - \dfrac{z}{9} = 1 $ and $ \dfrac{x}{3} - \dfrac{y}{2} - \dfrac{z}{1} = 1 $ .

Note: The important concept is to remember the equation of the plane in the intercept form of the plane as the condition is given in terms of the intercept made by the plane.
The equation of the plane in intercept form is
 $ \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1 $ where $ a,b $ and $ c $ are $ x,y $ and $ z $ are intercepts of the plane.