Find a Cartesian equation for the plane P in R3 that passes through the points A(1,0,0), B(0,3,0) and C(0,0,4).

I there any way to do this question without using the cross product?

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- Sep 20th 2010, 11:34 AMSyNtHeSiSEquation of plane
Find a Cartesian equation for the plane P in R3 that passes through the points A(1,0,0), B(0,3,0) and C(0,0,4).

I there any way to do this question without using the cross product? - Sep 20th 2010, 01:48 PMemakarov
See this Wikipedia article, especially Method 2. It uses matrix algebra to solve a system of linear equations, but this system can be solved using any method, such as the elimination of variables, which I believe is taught in school. You could put d = -1 (or any other non-zero number).

- Sep 20th 2010, 08:06 PMSoroban
Hello, SyNtHeSiS!

Quote:

$\displaystyle \text{Find a Cartesian equation for the plane }P\text{ in }R^3\text{ that passes}$

$\displaystyle \text{through the points }A(1,0,0),\:B(0,3,0)\text{ and }C(0,0,4).$

$\displaystyle \text{Is there any way to do this question without using the cross product?}$

The general equation of a plane is: .$\displaystyle ax + by + cz \:=\:d$

Substitute the three given points:

$\displaystyle \begin{array}{ccccccccc}

A(1,0,0)\!: & a(1) + b(0) + c(0) \;=\; d & \Longrightarrow& a \;=\; d \\ \\[-3mm]

B(0,3,0)\!: & a(0) + b(3) + c(0) \;=\; d & \Longrightarrow & b \;=\; \frac{d}{3} \\ \\[-3mm]

C(0,0,4)\!: & a(0) + b(0) + c(4) \;=\; d & \Longrightarrow & c \;=\; \frac{d}{4}

\end{array}$

So we have: .$\displaystyle dx + \dfrac{d}{3}y + \dfrac{d}{4}z \;=\;d$

Multiply through by $\displaystyle \frac{12}{d}\!:\;\;12x + 4y + 3z \;=\;12$

- Sep 21st 2010, 12:57 AMMathoMan
Form vectors $\displaystyle \vec{a}=\vec{AB}=(-1,3,0)$ and $\displaystyle \vec{b}=\vec{AC}=(-1,0,4)$. Now you have one point (A with its radius vector $\displaystyle \vec{r}_A=(1,0,0)$) and two vectors that lie in the same plane. Using them you can form a parametric equation of that plane, describing a radius vector $\displaystyle \vec{r}_T=x\vec{i}+y\vec{j}+z\vec{k}$ of an arbitrary point $\displaystyle T(x,y,z)$ in that plane.

$\displaystyle \vec{r}_T=\vec{r}_A+\lambda \vec{AB}+\mu \vec{AC}.$

Componentwise it looks like this.

$\displaystyle x=1-\lambda - \mu$

$\displaystyle y=3\lambda$

$\displaystyle z=4\mu$

And from there it is easy to form the equation of the plane, just substituting $\displaystyle \lambda$ and $\displaystyle \mu$ in the first equation with the expression from second and third eq.

$\displaystyle x=1-\frac{y}{3}-\frac{z}{4}$

$\displaystyle 12x+4y+3z-12=0.$

There you have it mate.