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Thread: Equation of plan

  1. #1
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    Equation of plan

    Finding the equation of a plane passing through M points (1,3,0) and N (4,0,0) making an angle of 30 with the plane x + y + z -1=0



    answer:
    $\displaystyle 5x+5y+(8+or-3\sqrt6)z-20=0$
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  2. #2
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    I must find the plan that way 30° with the plane x + y + z-1 = 0.


    $\displaystyle cos\theta=\frac{|n1.n2|}{|n1|.|n2|}$

    $\displaystyle \frac{\sqrt{3}}{2}=\frac{|(x,y,z).(1,1,1)|}{\sqrt{ x^2+y^2+z^2}.1}$

    $\displaystyle \sqrt{x^2+y^2+z^2} . \frac{\sqrt{3}}{2} = |x+y+z|$


    You are correct what I am doing?
    But as I continue to find (x, y, z)?
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  3. #3
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    Hello,

    Quote Originally Posted by Apprentice123 View Post
    I must find the plan that way 30° with the plane x + y + z-1 = 0.


    $\displaystyle cos\theta=\frac{{\color{red}|}n1.n2{\color{red}|}} {|n1|.|n2|}$ << SiMoon says : there is no absolute value in |n1.n2|

    $\displaystyle \frac{\sqrt{3}}{2}=\frac{{\color{red}|}(x,y,z).(1, 1,1){\color{red}|}}{\sqrt{x^2+y^2+z^2}.1}$

    $\displaystyle \sqrt{x^2+y^2+z^2} . \frac{\sqrt{3}}{2} = {\color{red}|}x+y+z{\color{red}|}$


    You are correct what I am doing?
    But as I continue to find (x, y, z)?
    It's not x,y,z.

    Let $\displaystyle ax+by+cz+d=0$

    $\displaystyle \bold{n_1}=(a,b,c)$

    --> $\displaystyle \bold{n_1} \cdot \bold{n_2}=a+b+c$

    Same here : $\displaystyle \bold{n_1}=\sqrt{a^2+b^2+c^2}$

    BUT, $\displaystyle \bold{n_2}=\sqrt{1+1+1}={\color{red}\sqrt{3}} \neq 1$, be careful !



    $\displaystyle \frac{\sqrt{3}}{2} \cdot \sqrt{a^2+b^2+c^2} \cdot {\color{red}\sqrt{3}}=a+b+c$

    $\displaystyle 3 \sqrt{a^2+b^2+c^2}=2(a+b+c)$ (1)

    -----------------

    Now, M(1,3,0) & N(4,0,0) are on the plane.

    --> $\displaystyle a+3b+d=0$ & $\displaystyle 4a+d=0$

    From the second one, $\displaystyle d=-4a$.
    Substituting in the first one :

    $\displaystyle -3a+3b=0 \implies a=b$.

    ------------------

    Letting $\displaystyle \boxed{d=-4}$ (we can, as soon as it keeps the proportionality), we can write :

    $\displaystyle \boxed{a=b=1}$

    ------------------

    Substituting in (1), we get :

    $\displaystyle 3 \sqrt{2+c^2}=2(2+c)$

    $\displaystyle 9(2+c^2)=4(2+c)^2$

    $\displaystyle 18+9c^2=16+16c+4c^2$

    $\displaystyle 5c^2-16c+2=0$

    $\displaystyle \Delta=16^2-40=256-40=216=(6 \sqrt{6})^2$

    So $\displaystyle \boxed{c}=\frac{16 \pm 6 \sqrt{6}}{10}\boxed{=\frac{8 \pm 3 \sqrt{6}}{5}}$


    ----------------------

    Equation of the plane is :

    $\displaystyle x+y+\frac{8 \pm 3 \sqrt{6}}{5}z-4=0$

    Multiplying by 5 :

    $\displaystyle 5x+5y+(8 \pm 3 \sqrt{6})-20=0$


    Last edited by Moo; Jun 25th 2008 at 07:54 AM.
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  4. #4
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    Thank you very much. I was calculated wrong
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