Relation between direction cosines - l² + m² + n² = 1: Worked examples

Before we begin, one reminder of every symbol we use, in plain words:

Everything below rests on the boxed identity from the parent note: $l^{2} + m^{2} + n^{2} = 1.$

The scenario matrix

Think of this as a checklist. Every worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	All positive components	The "easy" baseline	Ex 1
B	A zero component	One axis is perpendicular ( $cos 9 0^{\circ} = 0$ )	Ex 2
C	Negative component(s)	Line leans away from an axis; $cos > 9 0^{\circ}$	Ex 3
D	The $\pm$ ambiguity	Two opposite directions share one line	Ex 4
E	Impossible angle-triple	Identity acts as a filter — no line exists	Ex 5
F	Degenerate / equal angles	Symmetric line (equal leans), the "diagonal"	Ex 6
G	Word problem (real world)	Translate physical setup → ratios	Ex 7
H	Exam twist (find a missing angle)	Back-solve one cosine from the other two	Ex 8
I	Exactly along an axis / limiting value	Line collapses onto an axis	Ex 9, Ex 10

Read the matrix once. Now we fill every cell.

Figure — Relation between direction cosines - l² + m² + n² = 1

Figure 1 — the master picture. A red unit arrow floats among the three black axes. Its three dashed shadows on the

x

y

z

-axes have lengths

l

m

n

. Squaring and adding those shadow-lengths always rebuilds the arrow's length

1

: this is the identity

l^{2} + m^{2} + n^{2} = 1

in one image. We invoke this picture directly in Ex 1 below — the three fractions

\frac{2}{7}, \frac{3}{7}, \frac{6}{7}

are exactly the three shadow-lengths, and re-squaring them rebuilds the arrow.

Cell A — all positive components

Worked example Ex 1 — the baseline

Find the direction cosines of the vector $(2, 3, 6)$ .

Forecast: all three numbers point the "positive" way, so guess: are all three cosines positive and less than $1$ ? And does the biggest component ( $6$ ) give the biggest cosine?

Step 1 — length. $2^{2} + 3^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7.$ Why this step? Direction cosines are components of a unit vector, so we must first know how long the raw vector is before shrinking it (see Unit Vectors).

Step 2 — divide each component by the length. $l = \frac{2}{7}, m = \frac{3}{7}, n = \frac{6}{7} .$ Why this step? Dividing a vector by its own length always yields a length- $1$ vector; its components are then automatically direction cosines. In the language of Figure 1, these three fractions are precisely the lengths of the red arrow's three dashed shadows on the axes.

Verify: $(\frac{2}{7})^{2} + (\frac{3}{7})^{2} + (\frac{6}{7})^{2} = \frac{4 + 9 + 36}{49} = \frac{49}{49} = 1.$ ✓ Squaring the three shadow-lengths and adding rebuilds the arrow's length $1$ — exactly what Figure 1 promised. And $\frac{6}{7}$ (largest) matches the largest component. Forecast confirmed.

Cell B — a zero component

Worked example Ex 2 — a right angle to one axis

Find the direction cosines of the line joining $(1, - 1, 2)$ and $(1, 2, 6)$ .

Forecast: the two points share the same $x$ -coordinate ( $1$ and $1$ ). The line does not move in $x$ at all. What angle does a line make with an axis it never leans toward? Guess $9 0^{\circ}$ , so $l = 0$ .

Step 1 — direction ratios. Subtract coordinates: $(1 - 1, 2 - (- 1), 6 - 2) = (0, 3, 4) .$ Why this step? The vector from one point to the other is the line's direction.

Step 2 — length. $0^{2} + 3^{2} + 4^{2} = 25 = 5.$ Why this step? We need the raw vector's length before shrinking it to a unit vector; this uses the Distance Formula in 3D.

Step 3 — cosines. $l = \frac{0}{5} = 0, m = \frac{3}{5}, n = \frac{4}{5} .$ Why this step? Same normalisation as before; the zero simply survives division.

Verify: $0 + \frac{9}{25} + \frac{16}{25} = 1.$ ✓ And $l = 0 = cos 9 0^{\circ}$ — the line is exactly perpendicular to the $x$ -axis, as forecast.

Cell C — a negative component

Worked example Ex 3 — leaning away

Find the direction cosines of $(- 1, 2, - 2)$ and describe each angle.

Forecast: two components are negative. A negative cosine means an obtuse angle (more than $9 0^{\circ}$ ) with that axis. Guess: $α$ and $γ$ are obtuse, $β$ is acute.

Step 1 — length. $(- 1)^{2} + 2^{2} + (- 2)^{2} = 1 + 4 + 4 = 9 = 3.$ Why: magnitude ignores sign (squares), so the length is a plain positive $3$ .

Step 2 — cosines. $l = - \frac{1}{3}, m = \frac{2}{3}, n = - \frac{2}{3} .$ Why this step? We divide each component by the length $3$ to normalise; the minus signs carry through because dividing by a positive number keeps the sign, and a negative direction cosine is exactly how the algebra records "leaning away."

Step 3 — read the angles.

$cos α = - \frac{1}{3} \Rightarrow α = cos^{- 1} (- \frac{1}{3}) \approx 109. 5^{\circ}$ (obtuse).
$cos β = \frac{2}{3} \Rightarrow β = cos^{- 1} (\frac{2}{3}) \approx 48. 2^{\circ}$ (acute).
$cos γ = - \frac{2}{3} \Rightarrow γ = cos^{- 1} (- \frac{2}{3}) \approx 131. 8^{\circ}$ (obtuse). Why: the sign of a cosine tells us which side of the $9 0^{\circ}$ mark the angle lies.

Verify: $\frac{1}{9} + \frac{4}{9} + \frac{4}{9} = \frac{9}{9} = 1.$ ✓ Signs confirm the forecast.

Figure 2 — negative cosine means obtuse angle. In the

x

–

z

plane, the red line-direction arrow points up and to the left, so its dashed shadow on the

x

-axis lands on the negative half. The black arc marks the angle

α

, which is now larger than

9 0^{\circ}

. This is why

l = cos α < 0

Cell D — the $\pm$ ambiguity

Worked example Ex 4 — one line, two directions

A line has direction ratios $(1, 1, 1)$ . Write BOTH sets of direction cosines and explain why there are two.

Forecast: a line is like a road with no arrow — you can walk it either way. So guess two answers, equal but opposite in sign.

Step 1 — length. $1^{2} + 1^{2} + 1^{2} = 3 .$ Why this step? Direction cosines are components of a unit vector, so before anything else we measure how long the raw ratio vector $(1, 1, 1)$ is; only then can we shrink it to length $1$ .

Step 2 — cosines (one orientation). $l = m = n = \frac{1}{3} .$ Why this step? Dividing each component by the length $3$ produces a unit vector; its identical components are the direction cosines for one of the two ways to walk the line.

Step 3 — the opposite orientation. Reverse the vector to $(- 1, - 1, - 1)$ : $l = m = n = - \frac{1}{3} .$ Why this step? A line (not a ray) has no preferred direction; both unit vectors $\pm \frac{1}{3} (1, 1, 1)$ lie along it.

Verify: $3 \cdot (\frac{1}{3})^{2} = 3 \cdot \frac{1}{3} = 1.$ ✓ Both sign choices pass, because squaring erases the sign.

Cell E — an impossible angle-triple

Worked example Ex 5 — the identity as a filter

Can a line make $6 0^{\circ}$ with the $x$ -axis, $6 0^{\circ}$ with the $y$ -axis, and $6 0^{\circ}$ with the $z$ -axis?

Forecast: three equal leans of $6 0^{\circ}$ ... $cos 6 0^{\circ} = \frac{1}{2}$ , and $3 \times \frac{1}{4} = \frac{3}{4} \neq = 1$ . Guess: impossible.

Step 1 — compute the sum of squares. $cos^{2} 6 0^{\circ} + cos^{2} 6 0^{\circ} + cos^{2} 6 0^{\circ} = 3 \cdot (\frac{1}{2})^{2} = 3 \cdot \frac{1}{4} = \frac{3}{4} .$ Why this step? Any real line must satisfy $l^{2} + m^{2} + n^{2} = 1$ . If a proposed triple gives anything else, no line can produce it.

Step 2 — compare with $1$ . $\frac{3}{4} \neq = 1.$ Why this step? The identity is a strict equality — the sum must be exactly $1$ . Landing on $\frac{3}{4}$ instead is the test failing, which is our whole verdict.

Conclusion: No such line exists. The identity rejects it.

Verify: contrast with the valid symmetric case $cos^{- 1} \frac{1}{3} \approx 54. 7^{\circ}$ (Ex 6), where $3 \cdot \frac{1}{3} = 1$ . ✓ So the true "equal-angle" line leans $54. 7^{\circ}$ , not $6 0^{\circ}$ .

Cell F — degenerate / equal angles (the diagonal)

Worked example Ex 6 — the perfectly symmetric line

Find the direction cosines of the line making equal angles with all three axes. What is that common angle?

Forecast: by symmetry $l = m = n$ . Plug into the identity and solve.

Step 1 — set them equal. Let $l = m = n = t$ . Why this step? "Equal angles" means equal cosines; naming the common value $t$ turns three unknowns into one.

Step 2 — use the identity. $t^{2} + t^{2} + t^{2} = 1 \Rightarrow 3 t^{2} = 1 \Rightarrow t = \pm \frac{1}{3} .$ Why this step? Every valid direction cosine triple must obey $l^{2} + m^{2} + n^{2} = 1$ ; substituting $l = m = n = t$ turns that identity into a single equation we can solve for $t$ .

Step 3 — the common angle. $α = cos^{- 1} \frac{1}{3} \approx 54.7 4^{\circ} .$ Why: this is the famous "body diagonal" of a cube direction $(1, 1, 1)$ — Ex 4 was this same line.

Verify: $3 \cdot \frac{1}{3} = 1$ ✓ and $cos 54.7 4^{\circ} \approx 0.5774 \approx \frac{1}{3}$ ✓.

Cell G — a word problem (real world)

Worked example Ex 7 — a guy-wire on a mast

A straight cable runs from the top of a $12$ m vertical pole to a ground anchor $3$ m east and $4$ m north of the pole's base. Set up axes with $x$ = east, $y$ = north, $z$ = up. Find the cable's direction cosines (from the top down to the anchor).

Forecast: the cable drops $12$ m in $z$ (negative, going down) and moves $3, 4$ horizontally. Guess: $n$ is negative and largest in size, since most of the run is vertical.

Step 1 — build the direction vector. Top of pole is at $(0, 0, 12)$ ; anchor at $(3, 4, 0)$ . Vector top→anchor: $(3 - 0, 4 - 0, 0 - 12) = (3, 4, - 12) .$ Why this step? We translate the physical geometry into a direction ratio triple.

Step 2 — length. $3^{2} + 4^{2} + (- 12)^{2} = 9 + 16 + 144 = 169 = 13.$ Why: this is also the true cable length — $13$ m.

Step 3 — cosines. $l = \frac{3}{13}, m = \frac{4}{13}, n = - \frac{12}{13} .$ Why this step? Dividing each component of the direction ratio by the length $13$ shrinks the cable direction to a unit vector; those normalised components are the direction cosines.

Verify: $\frac{9 + 16 + 144}{169} = \frac{169}{169} = 1$ ✓. $n$ is negative and largest in size — the cable indeed heads mostly downward, as forecast. Units: length $13$ m is a real distance (Distance Formula in 3D).

Cell H — exam twist (find the missing angle)

Worked example Ex 8 — two angles given, find the third

A line makes $9 0^{\circ}$ with the $x$ -axis and $4 5^{\circ}$ with the $y$ -axis. Find the angle it makes with the $z$ -axis.

Forecast: $l = cos 9 0^{\circ} = 0$ , $m = cos 4 5^{\circ} = \frac{1}{2}$ . The identity must supply $n$ . Since $0 + \frac{1}{2} = \frac{1}{2}$ , we expect $n^{2} = \frac{1}{2}$ too, so another $4 5^{\circ}$ -ish answer.

Step 1 — turn angles into cosines. $l = 0, m = \frac{1}{2} .$ Why this step? The identity is written in cosines, so we translate first.

Step 2 — apply the identity. $n^{2} = 1 - l^{2} - m^{2} = 1 - 0 - \frac{1}{2} = \frac{1}{2} .$ Why this step? With two cosines known, the third is fixed up to sign — the coupling from the parent note.

Step 3 — solve. $n = \pm \frac{1}{2} \Rightarrow γ = cos^{- 1} (\pm \frac{1}{2}) = 4 5^{\circ} or 13 5^{\circ} .$ Why $\pm$ : the two opposite directions of the line (Cell D again).

Verify: $0 + \frac{1}{2} + \frac{1}{2} = 1$ ✓ for either sign. Both $4 5^{\circ}$ and $13 5^{\circ}$ are geometrically valid.

Cell I — exactly along an axis, and the limit toward it

Worked example Ex 9 — a line lying along the

$x$ -axis Find the direction cosines of a line with direction ratios $(1, 0, 0)$ — i.e. pointing straight along the $x$ -axis.

Forecast: the line is the $x$ -axis. So it makes $0^{\circ}$ with $x$ (perfectly aligned) and $9 0^{\circ}$ with both $y$ and $z$ . Guess: $(l, m, n) = (\pm 1, 0, 0)$ .

Step 1 — length. $1^{2} + 0^{2} + 0^{2} = 1 = 1.$ Why this step? We still normalise; here the vector already has length $1$ , so nothing shrinks.

Step 2 — cosines. $l = \frac{1}{1} = 1, m = 0, n = 0.$ Reversing the direction to $(- 1, 0, 0)$ gives $l = - 1$ . So the two orientations are $(\pm 1, 0, 0)$ . Why this step? Normalising an already-unit vector leaves it unchanged; $l = 1 = cos 0^{\circ}$ confirms the line is perfectly aligned with the $x$ -axis.

Step 3 — read the angles. $α = cos^{- 1} (1) = 0^{\circ}$ , $β = γ = cos^{- 1} (0) = 9 0^{\circ}$ . Why this step? Turning each cosine back into its angle via $cos^{- 1}$ lets us describe the orientation in plain degrees: $0^{\circ}$ means "runs exactly along the $x$ -axis," and $9 0^{\circ}$ means "perpendicular to the $y$ - and $z$ -axes" — confirming in words what the numbers say.

Verify: $1^{2} + 0^{2} + 0^{2} = 1$ ✓. A direction cosine can indeed hit its extreme value $\pm 1$ — this is the only case where one cosine is $\pm 1$ and the other two are forced to $0$ .

Worked example Ex 10 — the limit: swinging onto the

$z$ -axis A line makes angle $θ$ with both the $x$ - and $y$ -axes and we shrink $θ \to 9 0^{\circ}$ . What direction does the line approach, and what are its limiting direction cosines?

Forecast: if the line becomes perpendicular to both $x$ and $y$ , the only room left is straight up/down — the $z$ -axis. Guess $l, m \to 0$ and $n \to \pm 1$ .

Step 1 — write $l, m$ in terms of $θ$ . $l = m = cos θ .$ Why this step? The problem says the line makes the same angle $θ$ with both the $x$ - and $y$ -axes, and a direction cosine is precisely the cosine of that angle; so both $l$ and $m$ equal $cos θ$ .

Step 2 — force the identity. $2 cos^{2} θ + n^{2} = 1 \Rightarrow n^{2} = 1 - 2 cos^{2} θ .$ Why this step? The third cosine is always whatever the identity requires; solving for $n^{2}$ expresses the vertical lean purely in terms of $θ$ .

Step 3 — take the limit $θ \to 9 0^{\circ}$ . As $θ \to 9 0^{\circ}$ , $cos θ \to 0$ , so $n^{2} \to 1 - 0 = 1 \Rightarrow n \to \pm 1, l, m \to 0.$ Why this step? This is the degenerate limit where the direction $(0, 0, \pm 1)$ is exactly the $z$ -axis, making angle $0^{\circ}$ (or $18 0^{\circ}$ ) with it — the same "along-an-axis" case as Ex 9, now reached as a smooth limit rather than a fixed input.

Verify: limiting triple $(0, 0, \pm 1)$ : $0 + 0 + 1 = 1$ ✓. And $n = 1 = cos 0^{\circ}$ means the line is the $z$ -axis, matching the forecast.

Figure 3 — the limit toward the $z$ -axis. Several faint black arrows show the line at

θ = 5 5^{\circ}, 6 5^{\circ}, 7 5^{\circ}, 8 5^{\circ}

; as

θ

grows toward

9 0^{\circ}

they lean ever more upright. The bold red arrow is the limit

(0, 0, 1)

— the line has become the

z

-axis itself.

Fill-check: did we cover the matrix?

Recall Tick every cell

A (all positive) ::: Ex 1 B (zero component) ::: Ex 2 C (negative component) ::: Ex 3 D ( $\pm$ ambiguity) ::: Ex 4 E (impossible triple) ::: Ex 5 F (equal angles / degenerate) ::: Ex 6 G (word problem) ::: Ex 7 H (missing angle) ::: Ex 8 I (exactly along an axis / limiting value) ::: Ex 9 and Ex 10

Active Recall

Recall Rapid fire

Direction cosines of $(2, 3, 6)$ ? ::: $(\frac{2}{7}, \frac{3}{7}, \frac{6}{7})$ . What does a zero direction cosine mean? ::: The line is perpendicular ( $9 0^{\circ}$ ) to that axis. What does a negative direction cosine mean? ::: The angle with that axis is obtuse ( $> 9 0^{\circ}$ ). Direction cosines of a line along the $x$ -axis? ::: $(\pm 1, 0, 0)$ . Common angle of the equal-angle line? ::: $cos^{- 1} \frac{1}{3} \approx 54.7 4^{\circ}$ . Can a line make $6 0^{\circ}$ with all three axes? ::: No; $3 cos^{2} 6 0^{\circ} = \frac{3}{4} \neq = 1$ . Two angles $9 0^{\circ}, 4 5^{\circ}$ given — third angle? ::: $4 5^{\circ}$ or $13 5^{\circ}$ .

Connections

Direction Ratios — every example starts by writing a raw ratio triple.
Unit Vectors — dividing by length is the normalisation step.
Distance Formula in 3D — supplies the magnitude $a^{2} + b^{2} + c^{2}$ .
Dot Product — why each shadow equals a cosine.
Angle Between Two Lines — next step: combine two cosine-triples.
Equation of a Line in 3D — direction cosines fix the line's slope and hence its orientation in space.