Straight Line in 3D — Angles and Distances

A railway track runs northeast from Jaipur. A highway cuts across the same region heading east. Stand at the crossing and look along both — there is an angle between them, and you can measure it with a protractor flat on the ground.

Now imagine the track is climbing uphill while the highway stays level. The two directions no longer share a flat plane. They tilt away from each other in three-dimensional space, and the angle between them is no longer something a flat protractor can handle. You need a tool that works in 3D.

That tool is the dot product. You already know that the dot product of two vectors encodes the cosine of the angle between them. The entire theory of angles between lines in space falls out of this one fact — combined with the equations of lines you built in the previous article.

Angle between two lines

Take two lines in space:

The direction of L_1 is given by \vec{b}_1 and the direction of L_2 by \vec{b}_2. The "angle between the two lines" means the angle between their directions — the vectors \vec{b}_1 and \vec{b}_2.

There is one subtlety. A line has no preferred "forward" direction: \vec{b} and -\vec{b} define the same line. The dot product \vec{b}_1 \cdot \vec{b}_2 might come out negative if the two direction vectors happen to point in roughly opposite senses, and that would give \cos\theta < 0, meaning an obtuse angle. But the angle between two undirected lines is always taken to be acute (or right). So you take the absolute value of the dot product.

Why the absolute value? Without it, you get the angle between the two directed vectors, which could be obtuse. The absolute value forces the result to be in [0, \pi/2], giving the acute angle between the undirected lines. If a problem specifically asks for the angle between two directed vectors (say, velocity vectors), drop the absolute value. But for lines, keep it.

Deriving it. The dot product formula says \vec{b}_1 \cdot \vec{b}_2 = |\vec{b}_1||\vec{b}_2|\cos\alpha, where \alpha is the angle between the vectors. Then \cos\alpha = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}. The angle between the lines is \theta = \alpha if \alpha is acute, and \theta = \pi - \alpha if \alpha is obtuse. In both cases, \cos\theta = |\cos\alpha|, which gives the formula.

Parallel and perpendicular lines

Two special cases deserve their own names, because they come up constantly.

Parallel lines

Two lines are parallel if they have the same direction (or opposite directions — a line does not care about sign). This means their direction vectors are proportional:

Equivalently, \vec{b}_1 = k\vec{b}_2 for some non-zero scalar k, or \vec{b}_1 \times \vec{b}_2 = \vec{0} (their cross product is zero).

When lines are parallel, \cos\theta = 1, so \theta = 0.

Perpendicular lines

Two lines are perpendicular if the angle between them is \pi/2. Since \cos(\pi/2) = 0, the condition is:

No absolute value is needed here — if the dot product is zero, it is zero regardless of sign.

Here is a quick summary:

Perpendicular distance from a point to a line

This is the most useful distance formula in 3D line geometry, and it shows up in nearly every exam.

The setup. You have a line L given in vector form \vec{r} = \vec{a} + \lambda\vec{b}, and a point P not on the line. You want the shortest distance from P to the line.

The geometric idea. Drop a perpendicular from P to the line. The foot of this perpendicular — call it M — is the point on the line closest to P. The distance PM is the perpendicular distance.

The cross-product formula

Let A be the known point on the line (with position vector \vec{a}), and let \vec{b} be the direction vector. The vector from A to P is \vec{AP} = \vec{p} - \vec{a}, where \vec{p} is the position vector of P.

Consider the parallelogram formed by \vec{AP} and \vec{b}. Its area is |\vec{AP} \times \vec{b}|. But this same parallelogram has base |\vec{b}| and height d (the perpendicular distance). So

Solving for d:

This is the formula. It is clean, it is fast, and it works every time.

Reading the formula. The cross product \vec{AP} \times \vec{b} gives you a vector whose magnitude is the area of the parallelogram spanned by \vec{AP} and \vec{b}. Dividing by the base length |\vec{b}| gives the height — which is exactly the perpendicular distance.

The Cartesian version

If the line is \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} and the point is P = (\alpha, \beta, \gamma), then \vec{AP} = (\alpha - x_1)\hat{i} + (\beta - y_1)\hat{j} + (\gamma - z_1)\hat{k} and \vec{b} = a\hat{i} + b\hat{j} + c\hat{k}. Compute the cross product, take its magnitude, and divide by \sqrt{a^2 + b^2 + c^2}. There is no need to memorise a separate Cartesian formula — the vector formula does the same job once you plug in components.

Foot of perpendicular

Often you need not just the distance but the foot of the perpendicular — the actual point M on the line closest to P.

The parametric method

Write a general point on the line as M = (x_1 + a\lambda,\; y_1 + b\lambda,\; z_1 + c\lambda) for some unknown parameter \lambda.

The vector \vec{PM} must be perpendicular to the line's direction \vec{b}. So \vec{PM} \cdot \vec{b} = 0.

Compute \vec{PM} = M - P:

The perpendicularity condition is:

This is a single linear equation in \lambda. Solve it:

Substitute this \lambda back into the parametric equations of the line to get M. Then d = PM.

Reading this formula. The numerator a(\alpha - x_1) + b(\beta - y_1) + c(\gamma - z_1) is the dot product \vec{AP} \cdot \vec{b}. The denominator a^2 + b^2 + c^2 is |\vec{b}|^2. So \lambda = \frac{\vec{AP} \cdot \vec{b}}{|\vec{b}|^2}, which is precisely the scalar projection of \vec{AP} onto \vec{b} divided by |\vec{b}| — the signed length along \vec{b} where the foot lands.

In vector form, the foot of the perpendicular is:

This is the projection of P onto the line — the same projection formula from the dot product article, applied to lines.

Two worked examples

Common confusions

A few things students reliably get tripped up by.

• "The angle between two lines in 3D can be obtuse." No — the angle between two undirected lines is always taken as the acute angle (or exactly 90°). The formula with the absolute value on the dot product enforces this. If a problem asks for the angle between two directed vectors, that is a different question and the answer can be obtuse.

• "If two lines have the same direction ratios, they are the same line." Not necessarily. They are parallel — same direction — but they might not pass through the same points. Two parallel lines are the same line only if one of them passes through a point that also lies on the other.

• "The perpendicular from P to a line always meets the line." In 3D, yes — there always exists a unique foot of perpendicular from an external point to a line. (This is different from asking whether two lines meet — two lines in 3D can be skew.) The foot is found by the parametric method above, and it is unique because the equation for \lambda is linear and always has exactly one solution.

• "I can use the 2D distance formula \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} for a line in 3D." That formula is specific to lines in 2D given by ax + by + c = 0. In 3D, a single linear equation defines a plane, not a line. Use the cross-product formula instead.

• Forgetting the absolute value in the angle formula. Without it, you might report an obtuse angle as the answer. The absolute value is part of the formula for angles between lines. If you are computing the angle between vectors, you may want the signed version — but for lines, always take |\vec{b}_1 \cdot \vec{b}_2|.

Going deeper

Why the cross product gives the distance

The cross-product distance formula d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|} has a clean geometric proof that is worth seeing in full.

The magnitude of the cross product |\vec{AP} \times \vec{b}| equals the area of the parallelogram formed by \vec{AP} and \vec{b}. This parallelogram has base |\vec{b}| and height equal to the perpendicular distance from P to the line through A in the direction of \vec{b}. So

and therefore d = \frac{\text{Area}}{|\vec{b}|} = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}.

The same idea — interpreting a cross product magnitude as a parallelogram area and then extracting a height — will reappear when you study the shortest distance between two skew lines.

The projection interpretation

The parametric method reveals something about what "foot of perpendicular" really means. When you compute \lambda = \frac{\vec{AP} \cdot \vec{b}}{|\vec{b}|^2} and then set \vec{m} = \vec{a} + \lambda\vec{b}, you are projecting the point P onto the line. The formula for \lambda is the same projection formula from the dot product article.

This means the foot of the perpendicular is the orthogonal projection of P onto L. The perpendicular distance is the "rejection" — the component of \vec{AP} that is perpendicular to \vec{b}. In fact:

This is Pythagoras's theorem in disguise: the hypotenuse is |\vec{AP}|, the base (projection) is \frac{|\vec{AP} \cdot \vec{b}|}{|\vec{b}|}, and the height (rejection) is d.

Direction cosines version

If the direction of the line is given by direction cosines (l, m, n) instead of direction ratios, the formulas simplify slightly because l^2 + m^2 + n^2 = 1. The angle formula becomes \cos\theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|, and the perpendicular distance formula becomes d = |\vec{AP} \times \hat{b}| — no division needed, because |\hat{b}| = 1.

Connection to planes

The angle between a line and a plane is a closely related problem. If the line has direction \vec{b} and the plane has normal \vec{n}, the angle \phi between the line and the plane satisfies \sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}. Notice the sine instead of cosine — because the angle between a line and a plane is defined as the complement of the angle between the line and the normal. You will see this in the article on Plane — Distances and Angles.

Where this leads next

You now know how to measure angles and distances involving a single line and a point. The next layer of complexity involves two lines at once — lines that may not meet, may not be parallel, and may twist past each other in three-dimensional space.

• Straight Line in 3D — Advanced — shortest distance between two skew lines, the condition for two lines to intersect, the image of a point in a line, and the distance between parallel lines.
• Plane — Basic Equations — the vector and Cartesian equations of a plane, and how three points determine one.
• Dot Product — a deeper look at the operation that powers angle measurement in any dimension.
• Cross Product — the operation behind the distance formula, with its full geometric interpretation.
• Distance Formulas — a unified view of all the distance formulas in 2D and 3D geometry.