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I am given the following question:

Find the equations of the two planes that contain the line $r \{ x=z+1=y+2$ and form a $60^\circ$ angle with the plane $ \pi \{ x+2y-3z+2=0$

My solution:

Lets call the normal vector of the planes wanted $\vec{n}=(a,b,c)$

The vector of the line is $\vec{v} =(1,1,1)$ and since it is perpendicular to the planes we want,

$$ (1,1,1) \cdot (a,b,c) = 0 \therefore a+b+c = 0 $$

Also, since the angles between the plane $\pi$ and our new planes is $60^\circ$ we have

$$ \frac{1}{2}=\frac{\vert (a,b,c) \cdot (1,2,-3) \vert}{\sqrt{a^2+b^2+c^2} \sqrt{14}} $$

I need another equation to finish the exercise, but I'm not sure what is that third equation.

Textbook's answer

$$ \pi_1 \{ 2x-3y+z-5=0\\ \pi_2 \{ 3x -y-2z-4=0 $$

Thank you.

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    $\begingroup$ The reason you are missing an equation is because there are many normal vectors to a plane, all proportional to each other. So you need to fix one paramater, for instance, look for the normalized vector, where $\sqrt{a^2+b^2+c^2}=1$ $\endgroup$
    – justt
    Sep 16, 2016 at 12:27

2 Answers 2

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Since as has been oft remarked the lenght of $(a,b,c)$ is variable let us chose the length to be $\sqrt{14}$ in order to cancel with the other $\sqrt{14}$ So we have the equations:

$$a+b+c=0$$ $$a+2b-3c=\frac{1}{2}\sqrt{14}\sqrt{a^2+b^2+c^2}=\frac{1}{2}14=7$$ $$a^2+b^2+c^2=14$$

If we take the first two $$a+b+c=0$$ $$a+2b-3c=7$$ and solve them we get

$$a=-5c-7$$

$$b=4c+7$$

If we now substitute into the third equation and simplify we get $$c^2+3c+2=0$$ so $c=-1,-2$.

This gives us the two vectors $$(-2,3,-1) \text{and} (3,-1,-2).$$

And the two equations $$2x-3y+z=d$$ $$3x-y-2z=d$$

The value of $d$ can then be calculated in each case since the plane passes through $(0,-2,-1)$.

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  • $\begingroup$ Thank you Rene, I appreciate that. Best Regards, have a great day. $\endgroup$
    – bru1987
    Sep 16, 2016 at 14:32
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Remember that there is no "one" normal vector to a plane. There are infinitely many, but all have the same direction. A third equation may come from choosing the norm of your vector (example, by setting it to $1/\sqrt{14}$ : $\sqrt{a^2+b^2+c^2} = 1/\sqrt{14}$

Now you have 2 equations, easily solvable given that this third equation simplifies your second one :)

But, I believe the question should state that the 60° angle can be either positive or negative (otherwise, there would be only one plane).

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