Three Dimensional Geometry · Lesson 5 of 5

Chapter Summary and Practice

“Connect directions, line equations, angles, and distances through a complete chapter review and mixed problem solving.”

Learning Objectives

• Connect direction ratios, direction cosines, and coordinate displacements. • Choose a line representation suited to the given information. • Use dot and cross products for angle, perpendicularity, and parallelism questions. • Classify a pair of lines and choose the appropriate shortest-distance formula. • Solve mixed problems with checks for orientation, zero components, and nonzero directions.

One picture behind the chapter

The chapter is about separating location from direction, then comparing the directions and locations of lines. Coordinates locate points. A nonzero vector supplies a direction. Line equations combine these two ingredients, while vector products reveal angles and perpendicular gaps. Seeing these links makes the formulas easier to reconstruct and harder to confuse.

Begin with a displacement between two points. Its raw components are direction ratios; dividing by its length gives direction cosines. Attach a real multiple of this displacement to an anchor to obtain a line. Compare two line directions by a dot product for an angle or a cross product for a common normal. Finally, include the difference of the anchor vectors when measuring separation.

What you know or want	Useful representation or test	What it means
A direction from two points	$d = Q - P$	Coordinate differences give a nonzero direction when $P \neq = Q$ .
Direction cosines	$d /∣ d ∣$	The unit components obey $l^{2} + m^{2} + n^{2} = 1$ .
Point and direction	$r = a + t d$	The anchor locates the line; the direction supplies its travel.
Two points	$r = a + t (b - a)$	$t = 0$ and $t = 1$ recover the given points.
A point check	Use one common parameter in all coordinates.	Matching only one or two coordinates is insufficient.
Smaller line angle	$cos θ = ∣ u \cdot v ∣/ (∣ u ∣∣ v ∣)$	Anchor vectors are not used.
Perpendicular directions	$u \cdot v = 0$	The line angle is a right angle.
Parallel directions	$u \times v = 0$	The nonzero direction vectors are proportional.
Nonparallel distance	$d = ∣Δ a \cdot (u \times v) ∣/∣ u \times v ∣$	Project the joining vector onto the common normal.
Parallel distance	$d = ∣ u \times Δ a ∣/∣ u ∣$	Cross-product area divided by base gives perpendicular height.
Zero distance	Combine the distance result with the direction test.	Nonparallel lines intersect; parallel equations describe one coincident line.

Directions, signs, and zero components

A direction-cosine triple has unit length, while a direction-ratio triple can have any nonzero length. Rescaling all components together preserves the undirected direction. Reversing all signs changes orientation but not the line itself. These facts explain why many different equations may describe one line.

The coordinate-axis directions provide useful boundary cases. A line parallel to the $x$ -axis has direction $(1, 0, 0)$ , so its $y$ - and $z$ -coordinates stay fixed. Zero components should become fixed-coordinate conditions, not fractions with zero denominators. The normalization formula also requires a genuinely nonzero vector: the zero triple supplies neither an angle nor a line direction.

Example — An axis-parallel line through the origin

Problem
Find the equation of the line through the origin parallel to the $x$ -axis.

1.Choose anchor $(0, 0, 0)$ and direction $(1, 0, 0)$ . The vector equation is $r = t \hat{i}$ , with $t \in R$ .
2.The coordinates are $x = t$ , $y = 0$ , and $z = 0$ . Thus a Cartesian description is simply $y = 0$ , $z = 0$ , with $x$ unrestricted.
3.This example shows why symmetric fractions are not always the best form. The two fixed coordinates describe the line completely without dividing by zero.

Example — Connect normalization and line construction

Problem
Let $A = (1, 2, 0)$ and $B = (3, 1, 2)$ . Find a unit direction from $A$ to $B$ , a line equation, and the point reached after moving six units in that direction from $A$ .

1.The displacement is $A B = (2, - 1, 2)$ , whose length is $3$ . The unit direction is $e = (2/3, - 1/3, 2/3)$ .
2.A line equation is $r = (1, 2, 0) + t (2, - 1, 2)$ . In this form a parameter change of one travels three units.
3.To travel six units, add $6 e = (4, - 2, 4)$ to the anchor, giving $(5, 0, 4)$ . This corresponds to $t = 2$ in the unnormalized line equation.
4.The same line can be written $r = (1, 2, 0) + s e$ . Then $s$ is a signed distance because the direction vector has unit length.

Angle problems are direction problems

For an angle, remove the anchor part of each line equation and retain its nonzero direction vector. Compute a dot product before doing unnecessary algebra with the points. A zero result settles perpendicularity immediately. Otherwise, divide its absolute value by the vector lengths to obtain the smaller angle.

Example — A perpendicularity pattern

Problem
Find the angle between directions $(a, b, c)$ and $(b - c, c - a, a - b)$ whenever both triples define directions.

1.Expand their dot product: $a (b - c) + b (c - a) + c (a - b) = ab - a c + b c - ab + a c - b c = 0$ . Every term cancels.
2.If both vectors are nonzero, the dot-product test gives a right angle: $θ = 9 0^{\circ}$ . This conclusion holds without knowing the individual values of $a, b, c$ .
3.Check the assumption: when $a = b = c$ , the second triple is $(0, 0, 0)$ and has no direction. In that case the proposed angle is undefined, so the right-angle conclusion requires the stated nonzero condition.

Example — Use a right angle to determine a parameter

Problem
The lines $(x - 1) / (- 3) = (y - 2) / (2 k) = (z - 3) /2$ and $(x - 1) / (3 k) = (y - 1) /1 = (z - 6) / (- 5)$ have perpendicular directions. Find $k$ .

1.The numerators are already coordinate differences, so read directions $(- 3, 2 k, 2)$ and $(3 k, 1, - 5)$ .
2.Set their dot product to zero: $(- 3) (3 k) + (2 k) (1) + 2 (- 5) = 0$ . This simplifies to $- 9 k + 2 k - 10 = 0$ .
3.Thus $- 7 k = 10$ and $k = - 10/7$ . Substitution gives dot product zero, and the value is nonzero, so the displayed fractions involving $k$ remain defined.

A common perpendicular direction

The cross product plays two related roles. Its zero value detects parallel directions. When it is nonzero, it supplies a vector perpendicular to both given directions. Attach that vector to a specified anchor if you are asked to construct a line with a direction perpendicular to two given lines.

Example — Construct a direction perpendicular to two lines

Problem
Find a vector equation through $(1, 2, - 4)$ with direction perpendicular to both $(x - 8) /3 = (y + 19) / (- 16) = (z - 10) /7$ and $(x - 15) /3 = (y - 29) /8 = (z - 5) / (- 5)$ .

1.Read the given directions $u = (3, - 16, 7)$ and $v = (3, 8, - 5)$ . Their anchors do not affect the required direction.
2.Compute $u \times v = (80 - 56, 21 + 15, 24 + 48) = (24, 36, 72) = 12 (2, 3, 6)$ .
3.Use the simpler direction $(2, 3, 6)$ . Through the required point, the line is $r = (1, 2, - 4) + t (2, 3, 6)$ .
4.Check: $(2, 3, 6) \cdot (3, - 16, 7) = 6 - 48 + 42 = 0$ and $(2, 3, 6) \cdot (3, 8, - 5) = 6 + 24 - 30 = 0$ . These verify perpendicular directions; they do not assert that the constructed line intersects both given lines.

Distance problems also need location

Two parallel lines have the same direction but can be far apart, so directions alone cannot determine distance. The missing information is the displacement between a point on one line and a point on the other. Projecting or taking a cross product removes the parts of this connector that merely run along the lines.

First compute $n = u \times v$ . If it is nonzero, take the absolute dot product of the anchor difference with $n$ and divide by $∣ n ∣$ . If it is zero, use a common nonzero direction and the parallel-distance formula instead. This branch in the method is essential: the nonparallel formula would divide by zero for parallel lines.

Example — A complete skew-distance solution

Problem
Find the shortest distance between $r = (6, 2, 2) + t (1, - 2, 2)$ and $r = (- 4, 0, - 1) + s (3, - 2, - 2)$ .

1.The anchor difference is $Δ a = (- 10, - 2, - 3)$ . The directions have cross product $(1, - 2, 2) \times (3, - 2, - 2) = (8, 8, 4)$ .
2.This normal has length $64 + 64 + 16 = 12$ , so the directions are nonparallel.
3.The scalar triple product is $(- 10, - 2, - 3) \cdot (8, 8, 4) = - 80 - 16 - 12 = - 108$ .
4.Take the absolute value and divide: $d = 108/12 = 9$ units. The positive gap confirms that the nonparallel lines are skew. The negative scalar triple product only describes orientation; distance itself is nonnegative.

Example — Distinguish parallel lines from one coincident line

Problem
Compare $r = (1, 0, 0) + t (1, 1, 0)$ with $r = (0, 1, 2) + s (2, 2, 0)$ and find their distance.

1.The directions are proportional, so choose the common direction $u = (1, 1, 0)$ . The anchor difference is $(- 1, 1, 2)$ .
2.Its cross product with the common direction is $u \times (- 1, 1, 2) = (2, - 2, 2)$ , whose length is $12$ .
3.Divide by $∣ u ∣ = 2$ to obtain $d = 6$ . The nonzero distance shows that the lines are distinct and parallel, rather than coincident.

Match each vector to its job

Normalize a direction to obtain direction cosines. Use two directions in the angle formula. Use their cross product for a common normal. Use the anchor difference when measuring a gap. Swapping these roles is a common source of a plausible-looking but incorrect answer. A final check should also confirm that every required direction is nonzero.

A strategy for a mixed question

Translate the information into anchors and directions before selecting a formula. If the question asks for a line, construct it and recover the given point or points as a check. If it asks for an angle, use only directions. If it asks for distance, classify the directions first and then include the anchor difference.

A diagram helps decide what the calculation should represent, but a perspective drawing cannot prove that two spatial lines intersect. Use shared coordinate values for an intersection claim. Likewise, orthogonal directions need not imply a zero gap. Keep the three decisions separate: what the lines are, how they point, and whether they meet.

Check the complete chapter

These questions mix representations and methods. Before calculating, identify which pieces of information matter and which can be ignored. The correct choice should survive a change of anchor along the same line and a rescaling of any nonzero direction vector.

Quick check

Which condition distinguishes direction cosines from general direction ratios?

Quick check

On $r = a + t (b - a)$ , which parameter gives the second anchor $b$ ?

Quick check

A line has direction $(0, 0, 5)$ and passes through $(2, - 3, 1)$ . Which coordinates remain fixed?

Quick check

Which information is sufficient for the smaller angle between two lines?

Quick check

Two nonzero directions have zero cross product and nonzero dot product. What is their smaller line angle?

Quick check

For nonparallel directions, what does a nonzero scalar triple product with the anchor difference tell you?

Quick check

Why is an absolute value used in a scalar-triple-product distance?

Quick check

The directions are $(a, b, c)$ and $(b - c, c - a, a - b)$ . If $a = b = c = 1$ , what is wrong with asking their angle?

Quick check

Which situation guarantees zero shortest distance?

Quick check

For parallel lines, which expression gives their distance using common direction $u$ and anchor difference $Δ a$ ?

Practice Problems

Find direction cosines in the orientation of $(1, - 2, 2)$ , and give the reversed orientation.
For $A = (2, - 1, 3)$ and $B = (4, 3, 1)$ , find a unit direction from $A$ to $B$ and an equation of their line.
Write a line through $(4, - 2, 1)$ parallel to $(0, 2, - 1)$ without using any zero denominator.
Find the smaller angle between directions $(1, 2, 2)$ and $(2, - 1, 0)$ .
Find $a$ if $(1, a, 1)$ is perpendicular to $(2, 1, - 3)$ .
Find the shortest distance between $r = t (1, 2, 0)$ and $r = (0, 0, 3) + s (2, - 1, 0)$ . Classify the lines.
Find the distance between $r = (0, 0, 0) + t (1, 1, 0)$ and $r = (0, 0, 4) + s (- 2, - 2, 0)$ .
Decide whether $(2/3, 1/3, 2/3)$ is a direction-cosine triple and explain the test.
Show that $r = (1, 0, 2) + t (1, 1, - 1)$ and $r = (2, 1, 1) + s (1, - 1, 0)$ intersect, and state their distance.
Construct a line through the origin with direction perpendicular to both $(1, 1, 0)$ and $(0, 1, 1)$ .
The directions of two lines are $(1, 0, 0)$ and $(0, 1, 0)$ , and their anchors are $(0, 0, 0)$ and $(0, 0, h)$ . Find their distance and decide when they intersect.
Explain why normalizing either direction vector changes neither the line angle nor the shortest-distance formula. Assume the direction is nonzero.

The vector length is $3$ , so use $(1/3, - 2/3, 2/3)$ . Reverse all signs to obtain $(- 1/3, 2/3, - 2/3)$ .

Key Takeaways

• Coordinates locate points; nonzero vectors supply directions; line equations combine both. • Direction cosines are normalized direction ratios, and two points supply ratios by subtraction. • Angles compare directions through dot and cross products, independently of anchors. • Distances require the anchor difference as well as directions; choose the parallel or nonparallel formula first. • Check zero components, signs, common parameters, and nonzero-vector assumptions throughout a solution.

Previous · Lesson 4

Shortest Distance Between Lines

End of chapter

One picture behind the chapter

Directions, signs, and zero components

Angle problems are direction problems

A common perpendicular direction

Distance problems also need location

A strategy for a mixed question

Check the complete chapter

Quiz

Practice Problems

Key Takeaways