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I am asked the following question:

There are two planes $\pi_1$ and $\pi_2$ such that each of them contain the line $t \{ (-1,4,0) + \lambda (1,2,3)$ and form the same angle with the lines $r \{ (0,0,0) + \beta (1,1,1)$ and $s \{ (-1,2,3) + \alpha (1,1,-1)$. Find the angle formed between those planes

My solution:

Let's say the normal vector of $\pi$ is $\vec{n} = (a,b,c)$. Since the angle between the plane(s) and the lines are equal,

$$ \begin{align*} \measuredangle \left( \pi, r \right) &= \measuredangle \left( \pi, s \right)\\ \frac{\left( a,b,c \right) \cdot (1,1,1)}{\sqrt{a^2+b^2+c^2} \ \sqrt{3}} &= \frac{\left( a,b,c \right) \cdot (1,1,-1)}{\sqrt{a^2+b^2+c^2} \ \sqrt{3}}\\ a+b+c &= a+b-c\\ c &= 0 \end{align*} $$

Also, since $t$ is contained in the plane, the dot product between its vector and the normal vector of the plane is zero:

$$ \begin{align*} (a,b,c) \cdot (1,2,3) &= 0\\ a+2b+3c&=0\\ a &= -2b \end{align*} $$

Assuming that the normal vector has length one,

$$a^2 + b^2 + c^2 = 1$$

So we have three equations with three variables and the solution for those is

$$a = \mp \frac{2}{\sqrt{5}} \quad a = \pm \frac{1}{\sqrt{5}} \quad c = 0$$

so the normal vectors of the two planes found are

$$\left( \frac{2}{\sqrt{5}} , - \frac{1}{\sqrt{5}} , 0 \right) \quad \left( - \frac{2}{\sqrt{5}} , \frac{1}{\sqrt{5}} , 0 \right)$$

and the angle between them is

$$\theta = \arccos \left(\frac{\left \vert - \frac{2}{\sqrt{5}} \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} \frac{1}{\sqrt{5}} + 0 \right \vert}{1} \right) = \arccos(1) = 0^\circ $$

My answer does not agree with the textbook's solution.

Textbook's solution:

$$\theta = \arccos \left( \frac{9}{\sqrt{95}} \right)$$

Did I make a mistake somewhere?

Thank you.

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The 2 planes you found this way are the same, reversing the normal vector does not change it.

Note that

$$ |(a,b,c) \cdot (1,1,1)| = |(a,b,c) \cdot (1,1,-1)| \\ |a+b+c| = |a+b-c| $$ yields

$$a+b+c = a+b-c \quad \text{ and } \quad a+b+c = -a-b+c$$

To find the second plane you would use the second alternative.

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