Direction cosines and direction ratios: Worked examples

Figure — Direction cosines and direction ratios

The scenario matrix

Every direction-cosine problem falls into one of these case classes. The last column names the example that nails it.

#	Case class	What makes it tricky	Covered by
C1	All-positive DRs	The "easy" baseline	Ex 1
C2	Mixed signs (a negative component)	Sign must survive normalising	Ex 2
C3	A zero component	Line lies in a coordinate plane / axis angle $= 9 0^{\circ}$	Ex 3
C4	Degenerate input (two identical points)	DRs all zero — line undefined	Ex 4
C5	Obtuse angle between lines (negative $cos θ$ )	Must report the actual angle	Ex 5
C6	Perpendicular / parallel test	Dot $= 0$ vs proportional DRs	Ex 6
C7	Missing DC from two known angles	Sign ambiguity → two lines	Ex 7
C8	Word problem (real direction)	Translating words into DRs	Ex 8
C9	Exam twist: DRs given as $λ$ -expressions	Solve for the unknown	Ex 9

Work through all nine and you've touched every cell.

Example 1 — C1: all-positive baseline

Example 2 — C2: a negative component

Worked example Statement

Find the DCs of the line through direction $(2, - 3, 6)$ .

Forecast: One component is negative ( $- 3$ ), so exactly one DC should come out negative — the sign is a real geometric fact (the line leans away from the positive $y$ -axis, so its angle $β$ with that axis is obtuse). This is precisely the situation drawn in figure s02.

Step 1. Length $= 2^{2} + (- 3)^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7$ . Why this step? Squaring kills the sign, so the length is always positive — but we keep the original signs on top.

Step 2. $l = \frac{2}{7}, m = - \frac{3}{7}, n = \frac{6}{7}$ . Why this step? Divide each signed DR by $7$ . The minus stays because $cos β < 0$ means $β > 9 0^{\circ}$ — look at the teal dashed projection in figure s02, which lands on the negative $y$ side.

Verify: $\frac{4}{49} + \frac{9}{49} + \frac{36}{49} = \frac{49}{49} = 1$ ✓. Since $m = - \frac{3}{7} < 0$ , $β = cos^{- 1} (- \frac{3}{7}) \approx 115. 4^{\circ}$ — obtuse, as forecast and as the figure shows.

Example 3 — C3: a zero component

Example 4 — C4: the degenerate case

Example 5 — C5: obtuse angle between two lines

Worked example Statement

Find the angle between the lines with DRs $(1, 1, 2)$ and $(2, - 1, - 1)$ .

Forecast: Some components disagree in sign, so the dot product might be negative → an obtuse angle. Watch for it.

Step 1. Numerator $= 1 \cdot 2 + 1 \cdot (- 1) + 2 \cdot (- 1) = 2 - 1 - 2 = - 1$ . Why this step? This is the dot product $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$ — the top of the cosine formula.

Step 2. Magnitudes: $1 + 1 + 4 = 6$ and $4 + 1 + 1 = 6$ . Why this step? We must divide the dot product by the product of lengths to turn it into a cosine.

Step 3. $cos θ = \frac{- 1}{6 \cdot 6} = \frac{- 1}{6}$ . Why this step? Apply the DR angle formula. The negative cosine means $θ > 9 0^{\circ}$ .

Step 4. $θ = cos^{- 1} (- \frac{1}{6}) \approx 99. 6^{\circ}$ . Why this step? $cos^{- 1}$ of a negative number lands in the second quadrant — an obtuse angle, as forecast.

Verify: $cos (99. 6^{\circ}) \approx - 0.1667 = - \frac{1}{6}$ ✓.

Example 6 — C6: perpendicular and parallel tests

Worked example Statement

(a) Are the lines $(2, 1, - 2)$ and $(1, 2, 2)$ perpendicular? (b) Are $(1, - 2, 3)$ and $(- 2, 4, - 6)$ parallel?

Forecast: (a) mixed signs — could dot to zero. (b) the second triple looks like $- 2$ times the first — smells parallel.

Step 1 (a). Dot $= 2 \cdot 1 + 1 \cdot 2 + (- 2) \cdot 2 = 2 + 2 - 4 = 0$ . Why this step? Two lines are perpendicular exactly when $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ — no need to compute the angle. So the lines ARE perpendicular.

Step 2 (b). Check proportionality: $\frac{1}{- 2} = \frac{- 2}{4} = \frac{3}{- 6} = - \frac{1}{2}$ . Why this step? Parallel lines have proportional DRs: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ . All three ratios equal $- \frac{1}{2}$ , so the lines ARE parallel (pointing in opposite directions along the same line).

Verify (a): dot $= 0$ , so $cos θ = 0$ , $θ = 9 0^{\circ}$ ✓. Verify (b): $(- 2, 4, - 6) = - 2 \cdot (1, - 2, 3)$ exactly ✓ — a genuine scalar multiple.

Example 7 — C7: missing DC from two angles

Worked example Statement

A line makes $α = 4 5^{\circ}$ with the $x$ -axis and $γ = 6 0^{\circ}$ with the $z$ -axis. Find $β$ , its angle with the $y$ -axis.

Forecast: With two DCs fixed, the identity $l^{2} + m^{2} + n^{2} = 1$ pins $m^{2}$ — but $m = \pm m^{2}$ , so expect two answers (two lines).

Step 1. $l = cos 4 5^{\circ} = \frac{1}{2}$ , so $l^{2} = \frac{1}{2}$ . And $n = cos 6 0^{\circ} = \frac{1}{2}$ , so $n^{2} = \frac{1}{4}$ . Why this step? DCs are cosines of the direction angles; square them ready for the identity.

Step 2. $m^{2} = 1 - l^{2} - n^{2} = 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ . Why this step? Rearranging $l^{2} + m^{2} + n^{2} = 1$ isolates the unknown.

Step 3. $m = \pm \frac{1}{2} \Rightarrow β = cos^{- 1} (\pm \frac{1}{2}) = 6 0^{\circ} or 12 0^{\circ}$ . Why this step? Both signs are geometrically valid — they are the two lines symmetric about the $x z$ -plane.

Verify: for $m = \frac{1}{2}$ : $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$ ✓. For $m = - \frac{1}{2}$ : $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$ ✓ (squaring erases the sign). Both are real lines.

Example 8 — C8: word problem

Worked example Statement

A drone flies from the control tower at ground point $P (1, 0, 2)$ straight to a rooftop at $Q (4, 4, 6)$ (units: metres, $z$ = height). Find the direction cosines of its flight path and the angle the path makes with the $z$ -axis.

Forecast: It climbs (positive $z$ -change), so $n > 0$ . The angle with the vertical $z$ -axis will be less than $9 0^{\circ}$ ; the angle with the ground is its complement.

Step 1. DRs $= P Q = (4 - 1, 4 - 0, 6 - 2) = (3, 4, 4)$ . Why this step? The straight flight path points along $P Q$ ; its components are direction ratios.

Step 2. Length $= 3^{2} + 4^{2} + 4^{2} = 9 + 16 + 16 = 41$ . Why this step? Needed to normalise into DCs (and it is literally the flight distance in metres, $\approx 6.40$ m).

Step 3. DCs $= (\frac{3}{41}, \frac{4}{41}, \frac{4}{41})$ . Why this step? Divide each DR by the length. All positive, so all angles are acute — matches "climbing away from origin."

Step 4. Angle with $z$ -axis: $γ = cos^{- 1} (\frac{4}{41}) \approx 51. 3^{\circ}$ . Why this step? $n = cos γ$ ; take arccos.

Verify: $\frac{9}{41} + \frac{16}{41} + \frac{16}{41} = \frac{41}{41} = 1$ ✓. And $cos^{- 1} (4/ 41) \approx 51.3 4^{\circ}$ , so the path rises $\approx 90 - 51.3 = 38. 7^{\circ}$ above the ground — a gentle climb, as forecast. (See Distance between points in 3D: the $41$ is exactly the distance $P Q$ .)

Example 9 — C9: exam twist with a parameter

Worked example Statement

The direction ratios of a line are proportional to $(λ, λ + 1, 1)$ and the line is perpendicular to the line with DRs $(1, - 2, 1)$ . Find $λ$ , then the DCs.

(Note: we say direction ratios, not cosines — by definition direction cosines must already satisfy $l^{2} + m^{2} + n^{2} = 1$ , which $(λ, λ + 1, 1)$ generally does not.)

Forecast: "Perpendicular" hands us a dot-product-equals-zero equation. One equation, one unknown $λ$ — solvable.

Step 1. Perpendicularity: $(λ) (1) + (λ + 1) (- 2) + (1) (1) = 0$ . Why this step? Two lines are perpendicular iff $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ (built in the parent). Uses Vectors and dot product.

Step 2. Expand: $λ - 2 λ - 2 + 1 = 0 \Rightarrow - λ - 1 = 0 \Rightarrow λ = - 1$ . Why this step? Just simplify and solve the linear equation.

Step 3. Substitute $λ = - 1$ : DRs $= (- 1, 0, 1)$ . Why this step? $λ + 1 = 0$ , giving a zero middle component (a C3 flavour hiding inside).

Step 4. Length $= (- 1)^{2} + 0^{2} + 1^{2} = 2$ , so DCs $= (- \frac{1}{2}, 0, \frac{1}{2})$ . Why this step? Normalise; the sign on $- \frac{1}{2}$ is kept.

Verify: Perpendicular check: $(- 1) (1) + (0) (- 2) + (1) (1) = - 1 + 0 + 1 = 0$ ✓. Unit check: $\frac{1}{2} + 0 + \frac{1}{2} = 1$ ✓.

Recall Self-test (reveal the answer after the arrow)

Which example handles a zero DR? ::: Example 3 (C3) — one component $0$ , giving a $9 0^{\circ}$ axis angle. Which example gives an obtuse angle between lines? ::: Example 5 (C5) — negative dot product, $θ \approx 99. 6^{\circ}$ . What do you report when two given points coincide? ::: Undefined — no direction exists (Example 4, C4). After finding $m^{2} = \frac{1}{4}$ , why two answers for $β$ ? ::: $m = \pm \frac{1}{2}$ ; both signs satisfy $l^{2} + m^{2} + n^{2} = 1$ (Example 7, C7). Which equation does "perpendicular" hand you in Example 9? ::: The dot product $= 0$ , i.e. $λ - 2 (λ + 1) + 1 = 0$ .

Connections

Direction cosines and direction ratios — parent; all formulas used here are derived there.
Vectors and dot product — every angle/perpendicular test is a dot product.
Equation of a line in 3D — DRs are the line's direction vector.
Distance between points in 3D — the normalising length $a^{2} + b^{2} + c^{2}$ is a distance (Ex 8).
Angle between line and plane — extends the angle idea to a plane's normal.
Equation of a plane — a plane's normal has its own DRs/DCs.