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Thread: Equation of plane

  1. #1
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    Equation 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?
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  2. #2
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    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).
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  3. #3
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    Hello, SyNtHeSiS!

    $\displaystyle \text{Find a Cartesian equation for the plane }P\text{ in }R^3\text{ that passes}$
    $\displaystyle \text{through the points }A(1,0,0),\:B(0,3,0)\text{ and }C(0,0,4).$

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

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

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    Quote Originally Posted by SyNtHeSiS View Post
    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?
    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.
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