Proving an equation - Plane

Apr 2010
7
0
I'm not too sure what I should've put as the title...
My teacher presented the class with a question that any of us has yet to solve.


Show that the plane that passes through the axis at \(\displaystyle x=a\) , \(\displaystyle y=b\) , \(\displaystyle z=c\) and is described by the equation,

\(\displaystyle (x/a)+(y/b)+(z/c)=1\)

I have absolutely no idea how to approach this question and would appreciate any suggestions.
 

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
I'm not too sure what I should've put as the title...
My teacher presented the class with a question that any of us has yet to solve.


Show that the plane that passes through the axis at \(\displaystyle x=a\) , \(\displaystyle y=b\) , \(\displaystyle z=c\) and is described by the equation,

\(\displaystyle (x/a)+(y/b)+(z/c)=1\)

I have absolutely no idea how to approach this question and would appreciate any suggestions.
1. The plane passes through the points A(a,0,0), B(0,b,0) and C(0,0,c).

2. The parametric equation of the plane is:

\(\displaystyle \vec r = (0,0,a)+s \cdot (-a,b,0) + t \cdot (-a,0,c)\)

3. The normal vector of this plane is:

\(\displaystyle (-a,b,0) \times (-a,0,c) = (bc, ac, ab)\)

4. Therefore the equation of the plane is:

\(\displaystyle (bc, ac, ab) \cdot ((x,y,z)-(a,0,0))=0\)

Expand:

\(\displaystyle bcx + acy +abz = abc\)

Divide by (abc) and you'll get the given result.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, deovolante!

Show that the plane that passes through the axes at \(\displaystyle x=a\) , \(\displaystyle y=b\) , \(\displaystyle z=c\)

is described by the equation: .\(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\:1\:\)

The equation of a plane has the form: .\(\displaystyle Ax + By + Cz +D \:=\:0\) .[1]


Substitute the three given intercepts:

. . \(\displaystyle \begin{array}{ccccccccc}
(a,0,0)\!: & A(a) + B(0) + C(0) + D &=& 0 \\
(0,b,0)\!: & A(0) + B(b) + C(0) + D &=& 0 \\
(0,0,c)\!: & A(0) + B(0) + C(c) + D &=& 0
\end{array}\)


And we have: .\(\displaystyle \begin{array}{ccccccc}
Aa + D \:=\:0 & \Rightarrow & A \:=\: -\dfrac{D}{a} \\ \\[-3mm]
Bb + D \:=\:0 & \Rightarrow & B \:=\: -\dfrac{D}{b} \\ \\[-3mm]
Cc + D \:=\:0 & \Rightarrow & C \:=\: -\dfrac{D}{c}
\end{array}\)



Substitute into [1]: . \(\displaystyle -\dfrac{D}{a}x - \dfrac{D}{b}y - \dfrac{D}{c}z + D \:=\:0\)


Divide by \(\displaystyle -D\!:\;\;\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} - 1 \:=\:0\)


Therefore: . \(\displaystyle \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} \;=\;1\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You are not asked to derive the equation of the plane, just to show that this is the equation of that plane.

Since \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= 1\) is linear in each variable, it represents a plane.

If x= a, y= 0, z= 0, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{a}{a}+ \frac{0}{b}+ \frac{0}{c}= 1\) so the plane contains (a, 0, 0).

If x= 0, y= b, z= 0, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{0}{a}+ \frac{b}{b}+ \frac{0}{c}= 1\) so the plane contains (0, b, 0).

If x= 0, y= 0, z= c, then \(\displaystyle \frac{x}{a}+ \frac{y}{b}+ \frac{z}{c}= \frac{0}{a}+ \frac{0}{b}+ \frac{c}{c}= 1\) so the plane contains (0, 0, c).

Since a plane is determined by three points, this is the required plane.