Meet a line and a plane

Classify a line and a plane

Test the line direction against the plane normal first. A nonzero dot product gives one intersection. A zero dot product needs a second test: substitute a point from the line to distinguish containment from a disjoint parallel line.

Let the line be $\mathbf r=\mathbf a+\lambda\mathbf m$ and the plane $\mathbf r\cdot\mathbf n=d$, with $\mathbf m\ne\mathbf0$ and $\mathbf n\ne\mathbf0$.

$\mathbf m\cdot\mathbf n\ne0$

Point test: No additional test needed

Relationship: One crossing

Common points: Exactly one

$\mathbf m\cdot\mathbf n=0$

Point test:

$$\mathbf a\cdot\mathbf n\ne d$$

Relationship: Parallel and disjoint

Common points: None

$\mathbf m\cdot\mathbf n=0$

Point test:

$$\mathbf a\cdot\mathbf n=d$$

Relationship: Line lies in plane

Common points: Infinitely many

Solve the one-crossing case

For the one-crossing case, follow the source’s three steps. First write the intersection position as $\overrightarrow{OP}=\mathbf a+\lambda\mathbf m$. Then substitute it into the plane:

$$\begin{aligned}(\mathbf a+\lambda\mathbf m)\cdot\mathbf n&=d,\\\mathbf a\cdot\mathbf n+\lambda(\mathbf m\cdot\mathbf n)&=d,\\\lambda&=\frac{d-\mathbf a\cdot\mathbf n}{\mathbf m\cdot\mathbf n}.\end{aligned}$$

Finally substitute this parameter back into the line to obtain P. The source summary’s last instruction says “find $\Pi$”; the required result is the intersection point P, not the plane.

Interpret a zero direction test

When $\mathbf m\cdot\mathbf n=0$, do not divide by it. Substitution instead gives either a contradiction (no intersection) or an identity (every line point is in the plane). The source uses “parallel” for both an in-plane direction and a disjoint parallel line; the point test distinguishes those cases. A nonparallel line cannot be disjoint from an infinite plane in three-dimensional Euclidean space.

Derive the angle between a line and a plane

The source defines $\alpha$ using the acute angle between the line and its projection in the plane. For nonzero direction $\mathbf m$ and normal $\mathbf n$, first find the directed-vector angle $\theta\in[0,\pi]$:

$$\cos\theta=\frac{\mathbf m\cdot\mathbf n}{|\mathbf m||\mathbf n|}.$$

The source refers back to its earlier vector-angle section 12.2. It gives two cases, each with a complement method and a trigonometric derivation.

Acute $\theta$. Here $\alpha=\pi/2-\theta$. Using $\cos(\pi/2-\alpha)=\sin\alpha$ gives

$$\sin\alpha=\frac{\mathbf m\cdot\mathbf n}{|\mathbf m||\mathbf n|}.$$

Obtuse $\theta$. Here $\alpha=\theta-\pi/2$. Expand the cosine of a sum:

$$\begin{aligned}\cos(\pi/2+\alpha)&=\cos(\pi/2)\cos\alpha-\sin(\pi/2)\sin\alpha\\&=-\sin\alpha.\end{aligned}$$

Thus $-\sin\alpha=(\mathbf m\cdot\mathbf n)/(|\mathbf m||\mathbf n|)$, and the negative dot product must be negated before taking the nonnegative line–plane angle. Both cases combine as

$$\alpha=\sin^{-1}\!\left(\frac{|\mathbf m\cdot\mathbf n|}{|\mathbf m||\mathbf n|}\right).$$

Reversing either vector does not change the geometric line–plane angle. The source’s word “acute” describes its general examples; the formula also includes the boundary cases $\alpha=0$ for a direction parallel to the plane and $\alpha=\pi/2$ for a perpendicular line. In the latter case the projected direction is zero, so the projection-angle description is understood by this perpendicular convention. Neither zero vector defines the required direction or normal.

Explain the three possible intersection equations

For $\Pi:x+2y+2z=6$, classify the lines $l_1:\mathbf r=(0,0,0)+t(2,-1,0)$, $l_2:\mathbf r=(6,0,0)+t(2,-1,0)$ and $l_3:\mathbf r=(0,0,0)+t(1,0,0)$. Predict the outcome using direction and point tests, then interpret the equation obtained by substitution.

After your attempt: compare the reasoning

With normal $\mathbf n=(1,2,2)$, the first two directions have dot product zero with the normal.

The origin is off the plane, so $l_1$ is parallel and disjoint: substitution gives $0=6$.

The base point $(6,0,0)$ is in the plane, so all of $l_2$ lies in it: substitution gives $6=6$.

For $l_3$, the direction dot product is 1, so there is one crossing. Substitution gives $t=6$, at $(6,0,0)$.

An inconsistent equation, an identity and a unique parameter correspond to different geometric situations.

Compare a line angle with its normal angle

A line has direction $\mathbf m=(1,0,\sqrt3)$ and a plane has normal $\mathbf n=(0,0,1)$. Find their angle. Repeat for line directions $(1,0,0)$ and $(0,0,1)$. Explain why blindly reporting the angle to the normal would exchange the two boundary cases.

After your attempt: compare the reasoning

For $(1,0,\sqrt3)$ the normal angle is $30^\circ$, so the line–plane angle is $60^\circ$.

The dot-product cosine gives the angle to the normal; take its complement for the plane.

For $(1,0,0)$ the normal angle is $90^\circ$ but the line–plane angle is $0^\circ$; for $(0,0,1)$ the normal angle is $0^\circ$ but the line–plane angle is $90^\circ$.

Use $\sin\alpha=|\mathbf m\cdot\mathbf n|/(|\mathbf m||\mathbf n|)$ for the line–plane angle.

A direction test alone says nothing about whether a parallel line is contained in the plane.