# Math Help - Equation of plan

1. ## 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

$5x+5y+(8+or-3\sqrt6)z-20=0$

2. I must find the plan that way 30° with the plane x + y + z-1 = 0.

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

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

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

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

3. Hello,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{a=b=1}$

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Substituting in (1), we get :

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

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

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

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

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

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

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Equation of the plane is :

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

Multiplying by 5 :

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

4. Thank you very much. I was calculated wrong