Vectors: lines and perpendicular feet

Build a line from a displacement

Let P have position vector p and let $\overrightarrow{PQ}=\mathbf m$.

Point	Position on the line	Position vector
Q	one displacement from P	$\mathbf p+\mathbf m$
E	beyond Q, $PE=3PQ$	$\mathbf p+3\mathbf m$
F	midpoint of PQ	$\mathbf p+\frac12\mathbf m$
G	beyond P away from Q, $PG:PQ=2:1$	$\mathbf p-2\mathbf m$

The point order is G,P,F,Q,E. Thus any point R on the line has $\overrightarrow{OR}=\mathbf p+\lambda\mathbf m$ for real λ. For distinct P,Q, m is nonzero. Negative λ moves opposite P→Q; $0<\lambda<1$ lies inside PQ; $\lambda>1$ lies beyond Q.

From a polygon mesh to a ray

The notes motivate vector geometry with film graphics such as Jurassic Park and Lord of the Rings. A modelled surface is built from connected flat polygons, often triangles, whose vertices are stored as coordinates. A wire mesh records how those pieces connect.

For shading, the notes describe following a ray backwards from the viewpoint to a surface facet and onward towards a light source. In their simplified picture, a facet reached by a light path is shaded brightly and one without such a path more darkly.

This is a motivation for line–plane geometry, not a complete lighting model. The two schematic illustrations below show the polygon mesh and the light–surface–eye path.

Describing the ray as a line and a facet as part of a plane turns the intersection into vector equations; the lines and planes chapters develop those descriptions. The printed source credit is retained as a further-reading pointer.

Derive the vector equation of a line

Let A be a fixed point with position vector $\mathbf a$ and let $\mathbf m\ne\mathbf0$ be parallel to the line. For a general point R with position vector $\mathbf r$, vector addition gives $\mathbf r=\mathbf a+\overrightarrow{AR}$. Since $\overrightarrow{AR}$ is parallel to $\mathbf m$, write $\overrightarrow{AR}=\lambda\mathbf m$ for a real scalar. Therefore $\mathbf r=\mathbf a+\lambda\mathbf m$, $\lambda\in\mathbb R$.

The line is infinitely long. Each real parameter gives a unique point, and each point on this fixed representation has one parameter because the direction is nonzero. The symbol r varies over the position vectors of every point on the line; it is not one fixed vector.

The same geometric line can have different equations. Choose another point B on it with position vector b and any nonzero parallel direction n; then $\mathbf r=\mathbf b+\mu\mathbf n$ also describes it. The base point, direction and parameter need not look identical. A parameter measures a displacement multiple, not a distance unless the direction is a unit vector.

Convert through the component equations

Write $\mathbf r=(x,y,z)$, $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf m=(m_1,m_2,m_3)$. Equating components in $\mathbf r=\mathbf a+\lambda\mathbf m$ gives

$$x=a_1+\lambda m_1,\qquad y=a_2+\lambda m_2,\qquad z=a_3+\lambda m_3.$$

These are parametric equations: x,y,z are functions of the same parameter. When all three direction components are nonzero, solve each for it:

$$\lambda=\frac{x-a_1}{m_1}=\frac{y-a_2}{m_2}=\frac{z-a_3}{m_3}.$$

This is the symmetric Cartesian form. If a direction component is zero, retain its coordinate as a separate constant equation instead of dividing by zero. For example, if $m_3=0$ and $m_1,m_2\ne0$, use $(x-a_1)/m_1=(y-a_2)/m_2$ together with $z=a_3$. If only one direction component is nonzero, the other two coordinates are constant.

Vector, parametric and Cartesian forms are equivalent representations. Choose the form that makes the next calculation convenient, and convert by equating components rather than guessing denominators.

Check whether a point lies on a line

For $l:\mathbf r=\mathbf a+\lambda\mathbf m$, assume the point P is on the line and set $\mathbf p=\mathbf a+\lambda\mathbf m$. Compare all x,y,z components. A single λ must satisfy every equation; solve for it and check all remaining coordinates. When a direction component is zero, that equation checks a constant coordinate rather than determining λ. Example 22 shows both the vector and Cartesian tests.

Find the foot of a perpendicular

Let P be off the line and F its perpendicular foot.

Method 1: solve on the line. Since F lies on l, write $\overrightarrow{OF}=\mathbf a+\lambda\mathbf m$. Find $\overrightarrow{PF}=\overrightarrow{OF}-\overrightarrow{OP}$. Perpendicularity gives $\overrightarrow{PF}\cdot\mathbf m=0$. Solve for λ, then substitute into OF. This direction need not be a unit vector.

Method 2: use a projection vector. Choose A on l and let $\widehat{\mathbf m}=\mathbf m/|\mathbf m|$. Then $\overrightarrow{AF}=(\overrightarrow{AP}\cdot\widehat{\mathbf m})\widehat{\mathbf m}$. As $\overrightarrow{AF}=\overrightarrow{OF}-\overrightarrow{OA}$,

$$\overrightarrow{OF}=\overrightarrow{OA}+(\overrightarrow{AP}\cdot\widehat{\mathbf m})\widehat{\mathbf m}.$$

Equivalently, $\mathbf f=\mathbf a+[(\mathbf p-\mathbf a)\cdot\mathbf m/(\mathbf m\cdot\mathbf m)]\mathbf m$, with m nonzero. A negative projection coefficient places F opposite the chosen direction from A. Example 23 follows both routes, finds the distance PF and reflects P using $\mathbf p'=2\mathbf f-\mathbf p$.

Choose a route to perpendicular distance

Write h for the perpendicular distance and use a unit direction $\widehat{\mathbf m}$.

1. Cross product: $h=|\overrightarrow{AP}\times\widehat{\mathbf m}|=|\overrightarrow{AP}\times\mathbf m|/|\mathbf m|$, with A any point on the line. The source refers back to its cross-product area construction in Section 8.4.3 of Vectors: Basic.
2. Known foot: find F first, as in Example 23, then $h=|\overrightarrow{PF}|$. This is convenient when the foot is already available.
3. Projection and Pythagoras: find the projection length $d=|\overrightarrow{AP}\cdot\widehat{\mathbf m}|$, then $h=\sqrt{|\overrightarrow{AP}|^2-d^2}=\sqrt{|\overrightarrow{AP}|^2-|\overrightarrow{AP}\cdot\widehat{\mathbf m}|^2}$. This is convenient when the projection length is already available.

Source qualification: on printed page 59, the Student copy puts hats on the direction in the cross-product and projection formulas; the Teacher copy omits them. Those length formulas require a unit direction. We retain the Student normalisation and the explicit denominator versions above; the worked Example 24 uses $(1,2,1)/\sqrt6$ in both copies. A dot product with an arbitrary direction alone is a scaled projection, not a projection length.

Classify the relationship between two lines

For two distinct lines the possibilities are distinct parallel, intersecting at one point, or skew (neither parallel nor intersecting).

Intersecting and distinct parallel lines are coplanar: they lie in a common plane.

Skew lines do not lie in one plane.

The printed list also mentions coincident lines despite introducing distinct lines; coincidence means two equations describe the same line and is an additional case when distinctness has not been established.

Use $l_1:\mathbf r=\mathbf a_1+\lambda\mathbf m_1$ and $l_2:\mathbf r=\mathbf a_2+\mu\mathbf m_2$, with both directions nonzero.

1. Test whether $\mathbf m_1=k\mathbf m_2$ for a nonzero real k. If so, directions are parallel. If needed, test whether a base point from one line lies on the other to distinguish coincident from distinct parallel lines.
2. If directions are not parallel, equate $\mathbf a_1+\lambda\mathbf m_1=\mathbf a_2+\mu\mathbf m_2$. In 3 D this is three linear equations in two unknowns. Use the source G.C. PlySmlt 2 simultaneous-equation method, or solve two independent component equations and check the third. An arbitrary pair might be dependent, so choose a pair that determines the two parameters.
3. If one parameter pair satisfies all three equations, the lines intersect. For the unique solution λ=λ₀,μ=μ₀, $\overrightarrow{OP}=\mathbf a_1+\lambda_0\mathbf m_1=\mathbf a_2+\mu_0\mathbf m_2$. Substitute into either line to find P.
4. If the full system is inconsistent and directions are nonparallel, the lines are skew. Do not infer this just because two parameter values differ: λ and μ belong to different representations.

The cuboid question below tests all eight printed pairs, including face diagonals and space diagonals. A crossing in its flat projection is not sufficient evidence of a spatial intersection. Example 25 retains both the analytic checks and the calculator systems and outputs.

Find the smaller angle between lines

Let the nonzero direction vectors be m₁,m₂. First find their vector angle θ from $\mathbf m_1\cdot\mathbf m_2=|\mathbf m_1||\mathbf m_2|\cos\theta$, so $\theta=\cos^{-1}[(\mathbf m_1\cdot\mathbf m_2)/(|\mathbf m_1||\mathbf m_2|)]$. The source refers back to its vector-angle formula in Section 7.4.1 of Vectors: Basic.

If θ is acute, the smaller line angle α is θ. If θ is obtuse, use $\alpha=\pi-\theta$ (or $180^\circ-\theta$). Equivalently,

$$\alpha=\cos^{-1}\left(\frac{|\mathbf m_1\cdot\mathbf m_2|}{|\mathbf m_1||\mathbf m_2|}\right).$$

The lines need not intersect: the formula compares directions and applies to skew lines too. The notes call α the acute angle; in the boundary cases, parallel directions give 0° and perpendicular directions give 90°, so in general it is the smaller nonnegative angle. Reversing one direction leaves the line unchanged while replacing θ by its supplement.

The notes invite you to explore angles in 3D. Use the chapter’s existing 3D scene to distinguish spatial lines from their flat projections.

Example 26 uses direction vectors only. Example 27(ii) asks an internal angle at A, so use AO and AB without forcing the result to be acute. The Teacher note on parallelogram labelling is integrated into the first step: follow O→A→B→C successively around the boundary (anticlockwise in its schematic); reversing the viewpoint changes the apparent orientation but not the vector identities.