I have to find all the planes that contain the straight line
$x-1 = y + 1 = z$
and that have and angle of $60deg$ with the plane
$ 2x + 4y -6z +9 = 0$
Correct Answer:
a) $2x - 3y + z - 5 = 0$
b) $3x - y -2z -4 = 0$
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For solve the problem
1) I found the bundle of planes containing the straight line
$ x(A) + y(B-A) +z (-B) -2(A) + B = 0$
with director parameters
$ \vec n = ( A, B - A, -B )$
2) I found the given plane director parameters
$\vec v = (2, 4, -6)$ that i transformed to $\vec v =(1, 2, -3)$
3) I used the formula to calculate the angle between planes
$cos(t) = {\vec v * \vec n\over||\vec n|| ||\vec v||} = cos(60deg)$
Solving this equation i arrived to
$2A^2 - 6B^2 + AB = 0$
I used B as constant with value $B=1$ and solved the equation using A as variable, doing so i found the vectors of the planes
$\vec q = (2, -1, -1), \vec w = (-3/2, 5/2, -1)$
that are wrong.
What am i missing?
After solving the problem myself, I discovered that the elementary method I was talking about is not that elementary and your's is the best one with a slight error.
Let the equation of plane passing through the point $(1,-1,0)$ be $A(x-1)+B(y+1)+Cz=0$. Now, the direction vector of the line is perpendicular to this plane. So, $A+B+C=0$. $$\therefore Ax+By+z(-A-B)+B-A=0\tag{1}$$ Now, after applying the condition of $60^\circ$, we get the quadratic as $3A^2+11AB+B^2=0$ $$\therefore A=-3B\;\text{or}\,A=-\frac{2}{3}B$$ Substitute this values in eq.$1$ and you get your answer.