3.18 Intersections and angles of lines and planesIB Maths: Analysis and Approaches HL: Revision notes

Write the line in terms of its parameter, e.g. $(1 + 2t,\ t,\ 2 - t)$, substitute into the Cartesian equation of the plane and solve for $t$. Then put $t$ back into the line.

There are three possibilities, decided by $\mathbf{d}\cdot\mathbf{n}$ (direction of line, normal of plane):

• $\mathbf{d}\cdot\mathbf{n} \neq 0$: one point of intersection.
• $\mathbf{d}\cdot\mathbf{n} = 0$ and a point of the line is not on the plane: the line is parallel to the plane and never meets it (the equation for $t$ becomes something like $5 = 9$).
• $\mathbf{d}\cdot\mathbf{n} = 0$ and a point of the line is on the plane: the line lies in the plane (the equation becomes $9 = 9$).

The angle $\theta$ between a line and a plane is measured between the line and its projection on the plane. It is the complement of the angle between the line and the normal, so $\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.$ Example: $\mathbf{d} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$, $\mathbf{n} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}$ gives $\sin\theta = \frac{3}{6}$, so $\theta = 30^\circ$.

If you prefer the scalar product formula for $\cos$, find the angle $\phi$ between $\mathbf{d}$ and $\mathbf{n}$ and then use $\theta = 90^\circ - \phi$ (taking the acute value of $\phi$).

Two planes are either parallel (normals parallel; distinct or identical) or they meet in a line.

• Direction of the line: it lies in both planes, so it is perpendicular to both normals: $\mathbf{d} = \mathbf{n}_1\times\mathbf{n}_2$.
• A point on the line: fix one variable (e.g. $y = 1$) and solve the two equations for the other two. Alternatively, solve the system with one variable as the parameter, or use a GDC.

The angle between two planes is the acute angle between their normals: $\cos\theta = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}.$ For $x + 2y + 2z = 5$ and $2x - y + 2z = 1$: $\cos\theta = \frac{4}{9}$, $\theta = 63.6^\circ$, and $M$: $\mathbf{r} = (-1, 1, 2) + s(6, 2, -5)$.

Solve the three equations simultaneously (elimination, row reduction or a GDC) and interpret the result:

• Unique solution: the planes meet at a single point.
• Infinitely many solutions with one parameter: they meet in a line (a sheaf of planes). This happens when one equation is a combination of the other two, e.g. $3(x + y + z = 6) - (2x - y + z = 3)$ gives $x + 4y + 2z = 15$.
• Infinitely many with two parameters: all three are the same plane.
• No solution: the system is inconsistent. Either at least two planes are parallel, or no two are parallel and they form a triangular prism (each pair meets in a line, and the three lines are parallel).

The shortest route from a point S to a plane is along the normal. Take the line $\mathbf{r} = \overrightarrow{OS} + s\mathbf{n}$, find where it meets the plane (the foot of the perpendicular F) and calculate $|\overrightarrow{SF}|$.

Example: S$(3, -1, 8)$ and $x - 2y + 2z = 3$: $(3 + s) - 2(-1 - 2s) + 2(8 + 2s) = 3$ gives $s = -2$, so F is $(1, 3, 4)$ and the distance is $|s||\mathbf{n}| = 2\times3 = 6$.