Finding a plane containing $(-1,2,1)$, parallel to $\frac{x}{2}=-\frac{y}{3}=-z$, and making a $60^\circ$ angle with $x=y, z=0$

Question

I have to find the equation of the plane that passes through $$(-1,2,1),$$ is parallel to the line: $$\frac{x}{2}=-\frac{y}{3}=-z$$ and makes $60^\circ$ angle with $$x=y, z=0$$

I thought that I could take two points from the line and make a vector with them that would be parallel to the plane, but that means I'm just using one of the given conditions and also getting the wrong answer.

Isn't that last condition extra? If I have a point and a parallel line why do I need anything else? And why am I getting the wrong answer? I don't know what I'm doing wrong.

Could you please help me with this? Thanks a lot

Answer 1

The last condition is not extra. Parallel to the given line and passing through the given point, one may have infinitely many planes. These can be generated by rotating any one such plane about an axis passing through $(-1,2,1)$ and parallel to the line.

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Let the normal vector of the desired plane be $(a,b,c)$ with $a^2+b^2+c^2=1$. The plane is parallel to $l_1$, so its normal vector is perpendicular to $l_1$, i.e.$$(a,b,c)\cdot(2,-3,-1)=0\iff c=2a-3b$$$l_2$ makes $60^\circ$ with the plane and thus $30^\circ$ with the normal vector. So$$\frac{(a,b,c)\cdot(1,1,0)}{\sqrt2}=\cos(30^\circ)\iff a+b=\sqrt\frac32$$Can you solve these to get the normal vector? Once that is done, the equation of the plane is given by$$((x,y,z)-(-1,2,1))\cdot(a,b,c)=0$$

Answer 2

Hint.

Given two lines $L_1,L_2$ and a point $p_0\in \Pi$ such that

$$ \cases{ L_1\to p = \lambda_1\vec v_1\\ L_2\to p = \lambda_2\vec v_2\\ \Pi\to (p-p_0)\cdot\vec n=0 } $$

we need

$$ \cases{ \vec n_\cdot \vec v_1 = 0\\ \vec n\cdot\vec v_2 = |\vec n||\vec v_2|\cos\left(\frac{\pi}{3}\right)\\ |\vec n|= 1 } $$

or

$$ \cases{ \vec n_\cdot \vec v_1 = 0\\ \vec n\cdot\vec v_2 = |\vec v_2|\cos\left(\frac{\pi}{3}\right)\\ |\vec n|= 1 } $$

three equations and three unknowns $\vec n = (n_x,n_y,n_z)$. Here

$$ \cases{ \vec v_1 = (2,-3,-1)\\ \vec v_2 = (1,1,0) } $$