Equation of a plane

[ic3st0rm]

The exact question is :
Write the equation of the plane with x-intercept 3, y-intercept -1, and z-intercept 6.

I'm not sure how to start this problem >.<

 

[tkhunny]

The exact question is :
Write the equation of the plane with x-intercept 3, y-intercept -1, and z-intercept 6.

I'm not sure how to start this problem >.<

(x-3) = (y+1) = (z-6)

Or, if you learned the "Intercept Form"

\(\displaystyle \L\;\frac{x}{3} + \frac{y}{-1} + \frac{z}{6} = 1\)

 

[ic3st0rm]

ohhh.... haha, oops. I overlooked the part where they put all the steps together into a single equation, thank you for clarifying it for me

 

[pka]

Use the points (3,0,0), (0,-1,0) & (0,0,6).
The the vectors <-3,-1,0> & <-3,0,6> to get the normal <2,-6,1>.
Now write the equation of the plane.

 

[soroban]

Hello, ic3st0rm!

If you're waiting for a "Three point formula" for planes, good luck!

If you know anything about planes, you can work this out . . .

Write the equation of the plane with x-intercept 3, y-intercept -1, and z-intercept 6.

You're expected to know the general equation of a plane: \(\displaystyle \:Ax\,+\,By\,+\,Cz\,+\,D\:=\:0\)

You are given three points: \(\displaystyle \,(3,0,0),\;(0,-1,0),\;(0,0,6)\)

Plug them into the general equation.

\(\displaystyle \begin{array}{ccc}(3,0,0):\; &A(3)\,+\,B(0)\,+\,C(0)\,+\,D\:=\:0 & \;\Rightarrow\; & 3A\,+\,D\:=\:0 &\;\Rightarrow\; & A\,=\,-\frac{D}{3}\\
(0,-1,0):\; & A(0)\,+\,B(-1)\,+\,C(0)\,+\,D\:=\;0 & \;\Rightarrow\; &-B\,+\,D\:=\:0 & \;\Rightarrow\; & B\,=\,D\\
(0,0,6):\; & A(0)\,+\,B(0)\,+\,C(6)\,+\,D\:=\:0 & \;\Rightarrow\; & 6C\,+\,D\:=\:0 & \;\Rightarrow\; & C\,=\,-\frac{D}{6}
\end{array}\)


The equation is: \(\displaystyle \L\:-\frac{D}{3}x\,+\,Dy\,-\,\frac{D}{6}z\,+\,D\:=\:0\)

Divide by \(\displaystyle D:\;\;\L-\frac{1}{3}x\,+\,y\,-\,\frac{1}{6}z\,+\,1\:=\:0\)

Multiply by -\(\displaystyle 6:\;\;\L\fbox{2x\,-\,6y\,+\,z\,-\,6\:=\:0}\)


BTW, tkhunny's answer is correct.