Problem solving using planes and scalar

The key fact. For any two vectors $\mathbf{a}$ and $\mathbf{b}$, $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \quad\Rightarrow\quad \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

where the dot product is computed component-by-component: $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = a_1 b_1 + a_2 b_2 + a_3 b_3.$

Two language tools you must keep straight.

• A line is described by its direction vector $\mathbf{d}$ (the way it points).
• A plane is described by its normal vector $\mathbf{n}$ (the way perpendicular to it). The plane equation $\mathbf{n}\cdot\mathbf{r} = d$ packages the normal $\mathbf{n}$ and the perpendicular distance information together.

Why this matters. Because a plane "speaks" through its normal — a vector that sticks straight out of it — every angle involving a plane is really an angle involving its normal. That single shift (plane $\to$ normal) is what makes the line-plane and plane-plane formulae look slightly different. Get it clear now and the rest of the subtopic falls into place.

Cambridge tip. Always take the modulus of the dot product in these formulae ($|\mathbf{d}\cdot\mathbf{n}|$). It guarantees an acute angle, which is what Cambridge wants unless the question explicitly asks for the obtuse one.

Angle between a line and a plane. The normal $\mathbf{n}$ is perpendicular to the plane, so the angle between the line $\mathbf{d}$ and the normal is $(90^\circ - \theta)$, where $\theta$ is the angle we actually want (line to plane). Since $\cos(90^\circ - \theta) = \sin\theta$: $\boxed{\;\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}\;}$

The $\sin$ (not $\cos$) is the whole trick — beginners who blindly write $\cos$ get the complement of the right answer.

Angle between two planes. The angle between two planes equals the angle between their normals: $\boxed{\;\cos\theta = \frac{|\mathbf{n_1}\cdot\mathbf{n_2}|}{|\mathbf{n_1}||\mathbf{n_2}|}\;}$

Here it really is $\cos$, because we compare normal directly to normal.

Cambridge tip. Write "$\sin\theta$" for line–plane and "$\cos\theta$" for plane–plane as your first line of working. If you mix them up you lose every accuracy mark even with perfect arithmetic.

Perpendicular distance from a point to a plane. If the plane is written as $\mathbf{n}\cdot\mathbf{r} = d$ and the point is $\mathbf{p}$: $\boxed{\;\text{distance} = \frac{|\mathbf{n}\cdot\mathbf{p} - d|}{|\mathbf{n}|}\;}$

In Cartesian form, the plane $ax + by + cz = d$ has normal $\mathbf{n} = (a, b, c)$, so for point $(x_0, y_0, z_0)$: $\text{distance} = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2+b^2+c^2}}.$

Foot of the perpendicular. The shortest path from $\mathbf{p}$ to the plane runs along the normal. So write the line through $\mathbf{p}$ in the direction $\mathbf{n}$: $\mathbf{r} = \mathbf{p} + \lambda\mathbf{n}.$ Substitute into $\mathbf{n}\cdot\mathbf{r} = d$, solve for $\lambda$, and put that $\lambda$ back in. The result $\mathbf{f} = \mathbf{p} + \lambda\mathbf{n}$ is the foot — the point on the plane closest to $\mathbf{p}$.

Reflection of a point in a plane. The foot $\mathbf{f}$ is the midpoint of $\mathbf{p}$ and its mirror image $\mathbf{p}'$. So you simply go twice as far: $\mathbf{p}' = \mathbf{p} + 2\lambda\mathbf{n} = 2\mathbf{f} - \mathbf{p}.$

Cambridge tip. The distance formula $\frac{|\mathbf{n}\cdot\mathbf{p}-d|}{|\mathbf{n}|}$ is fast and self-contained — use it whenever the question asks only for the distance. Switch to the "$\lambda$ method" the moment you also need the foot or the reflection.

Where a line meets a plane. A (non-parallel) line crosses a plane at exactly one point. The line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ gives every point on the line as the single parameter $t$ varies. To find where it pierces $\mathbf{n}\cdot\mathbf{r} = d$:

1. Substitute $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ into $\mathbf{n}\cdot\mathbf{r} = d$.
2. This gives one linear equation in $t$: $\mathbf{n}\cdot\mathbf{a} + t\,(\mathbf{n}\cdot\mathbf{d}) = d$. Solve for $t$.
3. Put $t$ back into the line to get the point.

If $\mathbf{n}\cdot\mathbf{d} = 0$ the line is parallel to the plane (it either lies in it or never meets it).

The line of intersection of two planes. Two non-parallel planes meet in a line. There are two clean methods:

• Direction first. The line lies in both planes, so it is perpendicular to both normals. Its direction is therefore $\mathbf{n_1}\times\mathbf{n_2}$ (the vector product — covered in the next subtopic). Then find any one common point by setting a convenient coordinate (say $z = 0$) and solving the two Cartesian equations simultaneously.
• Eliminate a variable. Solve the two Cartesian plane equations together, treating one coordinate as a free parameter; the parametric solution is the line.

Worked sketch (eliminate method). Planes $x + y + z = 6$ and $2x - y + z = 3$. Set $z = t$: $x + y = 6 - t, \qquad 2x - y = 3 - t.$ Adding: $3x = 9 - 2t \Rightarrow x = 3 - \tfrac{2}{3}t$. Then $y = (6 - t) - x = 3 - \tfrac{1}{3}t$. So the line is $\mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix} + t\begin{pmatrix} -\tfrac{2}{3} \\ -\tfrac{1}{3} \\ 1 \end{pmatrix},$ which you may scale to direction $(-2, -1, 3)$.

Cambridge tip. A line parallel to a plane is the classic trap. Before solving for $t$, glance at $\mathbf{n}\cdot\mathbf{d}$: if it is zero, there is no single intersection point and you must say so.