Here's what I did: For case 1, 2 and 3, I'm ok now, but for case 4, I still have trouble finding the right way to solve it. My solution to case 4 is theoretically possible, but as for me, I've no idea how to do it.

Case 1: Angle between the given plane and the given line is greater than 30 degrees: No solution

Case 2: Angle between the given plane and the given line is 0 degree: 2 planes satisfy

A line with a direction vector perpendicular to the given line, and is 30 degree to the plane.

Case 3: Angle between the given plane and the given line is 30 degree: 1 plane satisfies

Reverse the process from case 2. Use the projection of the given line on the plane, find the vector 90 to it that lies on the plane.

Case 4: Angle between the given plane and the given line is between 0 and 30 degrees exclusive: 2 planes satisfy

Step 1: find the intersection between the plane the and the given line

Step 2: use the equation of the plane and the point determined from step 1, find the lines that pass through the point and makes a 30-degree angle to the plane. The equation obtained should describe two conics with their heads pointing perpendicular to the plane.

Step 3: Get the equation for the infinite vectors that is perpendicular to the lines from step 2 and lies on the given plane.

Step4: Find the solutions to the equation where the cross product of the vectors from step 3 and 4 is perpendicular to the given line.



$$ \Pi_1\to (p-p_1)\cdot \vec n_1 = 0\\ L_1\to p = p_2+\lambda \vec v $$

with $p = (x,y,z), \ ||\vec n_1|| = ||\vec n_2 || = 1$ the sought plane is $\Pi_2\to (p-p_2)\cdot \vec n_2 = 0$

Note that $p_2 \in \Pi_2$. Now we have the conditions

$$ L_1\in \Pi_2\to (p_2-p_2+\lambda\vec v)\cdot \vec n_2 = 0\ \ \forall \ \ \lambda $$


$$ \vec v\cdot \vec n_2 = 0 $$

the last condition is

$$ \cos\phi = \frac{\sqrt 3}{2} = \vec n_1\cdot \vec n_2 $$

or grouping

$$ \vec v\cdot \vec n_2 = 0\\ \vec n_1\cdot \vec n_2 = \frac{\sqrt 3}{2}\\ ||\vec n_2 || = 1 $$

giving three equations and three unknowns which are the $\vec n_2$ components.


For your case 4, notice that when you take one of the vectors from step 2, and the corresponding vector from step 3, and take their cross product, you will always get a vector at a $60$-degree angle to the given plane.

(In the more general case, where the given line is at angle $\theta$ to the given plane, where $\theta$ is not necessarily equal to $30$ degrees, the angle between the cross-product vector and the given plane will be a right angle minus $\theta.$ But that's more than you need for this question.)

The $60$-degree-angle vectors form a cone. The vectors perpendicular to the given line form a plane. You want a vector in the intersection of that cone and that plane.

  • $\begingroup$ Thank you for answering, but Sesareo has a solution that is easier for me to understand. $\endgroup$ – Danny Yang Oct 20 '18 at 18:24

Express the line as the intersection of two planes, i.e., as a system of two linear equations $$a_1x+b_1y+c_1z+d_1=0\\a_2x+b_2y+c_2z+d_2=0.$$ There are several ways to do this if you’re starting from some other representation, which I won’t go into here. (Let me know if you’d like more details about how to do that.) The normals of these two planes are, of course, $\mathbf n_1=(a_1,b_1,c_1)^T$ and $\mathbf n_2=(a_2,b_2,c_3)^T$. Every plane that contains this line has an equation that’s a linear combination of the two equations above (and vice-versa), so we can parameterize their possible normals as $\mathbf n(\lambda,\mu) = \lambda\mathbf n_1+\mu\mathbf n_2.$ (We can in fact use a single parameter that represents an angle, but I find it a bit more convenient to start with two.) If the normal of the given plane is $\mathbf n_0$, the condition that the angles between the planes be $30°$ can be expressed as $$\mathbf n_0\cdot\mathbf n(\lambda,\mu) = \pm\|\mathbf n_0\|\,\|\mathbf n(\lambda,\mu)\|\cos(30°) = \pm\frac{\sqrt3}2\|\mathbf n_0\|\|\mathbf n(\lambda,\mu)\|.$$ The $\pm$ appears because the two normals might be pointing in “opposite” directions: normals with an angle of $180°-30°=150°$ between them also produce a dihedral angle of $30°$. Squaring both sides and rearranging gives you a quadratic equation that’s homogeneous in $\lambda$ and $\mu$, which you can then solve by making the substition $t=\lambda/\mu$ or $t=\mu/\lambda$. Alternatively, you could make this a system of quadratic equations by requiring that $\lambda^2+\mu^2=1$. The corresponding plane equation for each solution is the linear combination $$\lambda(a_1x+b_1y+c_1z+d_1)+\mu(a_2x+b_2y+c_2z+d_2)=0.$$ Note that even if you get two solutions for $\lambda$, they might both correspond to the same plane.

  • $\begingroup$ Thank you for answering, but Cesare's solution is easier for me to understand $\endgroup$ – Danny Yang Oct 20 '18 at 18:23

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