Projection of vectors: Worked examples — Maths deep dive

The scenario matrix

Every projection problem falls into exactly one of these cells. The last column names the example that covers it.

#	Case class	What is special	Sign of $a \cdot b$	Covered by
C1	Acute angle ( $0^{\circ} < θ < 9 0^{\circ}$ )	shadow points along $b$	positive	Ex 1
C2	Right angle ( $θ = 9 0^{\circ}$ )	shadow is zero	zero	Ex 2
C3	Obtuse angle ( $9 0^{\circ} < θ < 18 0^{\circ}$ )	shadow points backward	negative	Ex 3
C4	Non-unit / long $b$	must divide by $b \cdot b$ , not $	\vec b	$
C5	Degenerate: $b = 0$	division by zero — undefined	$0/0$	Ex 5
C5′	Degenerate: $a = 0$	nothing to project — result is $0$	zero	Ex 5b
C6	Collinear ( $θ = 0^{\circ}$ or $18 0^{\circ}$ )	shadow = whole vector, $\pm$	$\pm	\vec a
C7	3-D decomposition	split into parallel + perpendicular	any	Ex 7
C8	Real-world word problem (ramp / force)	attach units, meaning	positive	Ex 8
C9	Exam twist (unknown solved from a projection)	work backward	given	Ex 9

We now walk them all.

Building blocks we will reuse (defined once, from zero)

Figure s01 below draws exactly this setup: the magenta arrow $a$ , the violet arrow $b$ , and the orange shadow dropped straight down onto $b$ 's line. The dashed navy line is the perpendicular "sun ray" that defines where the shadow ends — keep it in mind, every example is a version of this one picture.

Ex 1 — Acute angle (cell C1)

Worked example Sharp angle: shadow points forward

$a = (4, 3)$ , $b = (2, 0)$ . Find the scalar and vector projections of $a$ onto $b$ .

Forecast: $b$ points along $+ x$ , and $a$ leans up-and-right. Guess: the shadow is the $x$ -part of $a$ , so scalar $= 4$ and vector $= (4, 0)$ . Let's confirm. (This is precisely the geometry in figure s01 — the orange shadow along $b$ .)

Step 1. $a \cdot b = 4 (2) + 3 (0) = 8$ . Why this step? The dot product is the numerator of both formulas; compute it first.

Step 2. $b \cdot b = 2^{2} + 0^{2} = 4$ , so $∣ b ∣ = 4 = 2$ . Why this step? Scalar needs $∣ b ∣$ ; vector needs $b \cdot b$ . Get both now.

Step 3. Scalar projection $= \frac{8}{2} = 4$ . Why this step? Apply $\frac{a \cdot b}{∣ b ∣}$ . Positive ⇒ shadow lies along $b$ (acute angle, cell C1).

Step 4. Vector projection $= \frac{8}{4} (2, 0) = 2 (2, 0) = (4, 0)$ . Why this step? Apply $\frac{a \cdot b}{b \cdot b} b$ . Direction is $b$ 's, length is the shadow.

Verify: $(4, 0)$ lies on the $x$ -axis (correct direction), and its length $4$ equals the scalar projection. The $y$ -part $3$ vanished — exactly the shadow idea. ✓

Ex 2 — Right angle (cell C2)

Worked example Perpendicular: the shadow collapses

$a = (0, 5)$ , $b = (3, 0)$ . Project $a$ onto $b$ .

Forecast: $a$ points straight up, $b$ straight right — they are perpendicular. A vertical stick under a straight-down sun casts no horizontal shadow. Guess: everything is zero.

Step 1. $a \cdot b = 0 (3) + 5 (0) = 0$ . Why this step? A zero dot product is the definition of perpendicular ( $cos 9 0^{\circ} = 0$ ).

Step 2. $b \cdot b = 3^{2} + 0^{2} = 9$ , so $∣ b ∣ = 9 = 3$ . Why this step? Even though we suspect a zero answer, write out the denominators explicitly so nothing is hidden: scalar will use $∣ b ∣ = 3$ , vector will use $b \cdot b = 9$ .

Step 3. Scalar projection $= \frac{0}{3} = 0$ . Why this step? Length of shadow is genuinely zero.

Step 4. Vector projection $= \frac{0}{9} (3, 0) = (0, 0)$ . Why this step? Zero length × any direction = the zero vector.

Verify: The perpendicular part $a_{⊥} = a - 0 = (0, 5) = a$ — i.e. all of $a$ is perpendicular to $b$ , which matches the picture. ✓

Ex 3 — Obtuse angle (cell C3)

Worked example The shadow falls backward

$a = (- 3, 1)$ , $b = (4, 0)$ . Project $a$ onto $b$ .

Forecast: $a$ leans to the left, $b$ points right — the angle between them is obtuse. Guess: a negative scalar, and a vector pointing left. Figure s02 shows this exact case — watch the orange shadow land behind the origin.

Step 1. $a \cdot b = (- 3) (4) + 1 (0) = - 12$ . Why this step? A negative dot product means $cos θ < 0$ , i.e. $θ > 9 0^{\circ}$ (cell C3). The sign is already telling the story.

Step 2. $b \cdot b = 4^{2} + 0^{2} = 16$ , so $∣ b ∣ = 16 = 4$ . Why this step? Scalar divides by $∣ b ∣ = 4$ ; vector divides by $b \cdot b = 16$ .

Step 3. Scalar projection $= \frac{- 12}{4} = - 3$ . Why this step? The minus encodes "shadow on the back side"; do not drop it.

Step 4. Vector projection $= \frac{- 12}{16} (4, 0) = - \frac{3}{4} (4, 0) = (- 3, 0)$ . Why this step? Negative scalar × direction $(4, 0)$ flips it to point along $- x$ , opposite $b$ .

Verify: $(- 3, 0)$ points against $b$ and has length $3 = ∣ - 3 ∣$ = magnitude of the scalar projection. Sign and picture agree. ✓

Ex 4 — Non-unit, long $b$ (cell C4)

Worked example Making

$b$ longer must NOT change the answer $a = (2, 3)$ , $b = (0, 10)$ . Project onto $b$ , then re-do it with $b^{'} = (0, 1)$ (same direction, shorter). Do the vector answers match?

Forecast: Both $b$ vectors point straight up. The shadow of $a$ on the vertical direction should be its $y$ -part, $(0, 3)$ , regardless of how long we drew $b$ . Guess: identical vector projections.

Step 1 (with $b = (0, 10)$ ). $a \cdot b = 2 (0) + 3 (10) = 30$ , and $b \cdot b = 0 + 100 = 100$ . Why this step? Both grow with $b$ 's length; watch how they cancel.

Step 2. Vector projection $= \frac{30}{100} (0, 10) = 0.3 (0, 10) = (0, 3)$ . Why this step? The $b \cdot b$ in the denominator kills the extra length from $b$ . This is exactly why we divide by $b \cdot b$ and not $∣ b ∣$ for the vector form.

Step 3 (with $b^{'} = (0, 1)$ ). $a \cdot b^{'} = 3$ , $b^{'} \cdot b^{'} = 1$ , projection $= \frac{3}{1} (0, 1) = (0, 3)$ . Why this step? Confirms scale-independence.

Verify: Both vector projections are $(0, 3)$ — exactly the same, so making $b$ ten times longer changed nothing. (The scalar projections also both work out to $3$ here, since $∣ b ∣ = 10$ with $a \cdot b = 30$ gives $30/10 = 3$ , and $3/1 = 3$ ; both directions are the same, so their measured lengths agree too.) Length-invariance confirmed. ✓

Ex 5 — Degenerate: projecting onto the zero vector (cell C5)

Ex 5b — Degenerate: projecting the zero vector (cell C5′)

Worked example The OTHER zero case

$a = (0, 0)$ , $b = (3, 4)$ . Project $a$ onto $b$ .

Forecast: This time $b$ is a perfectly good direction, but $a$ is nothing — a point at the origin. A zero-length stick casts a zero-length shadow. Guess: scalar $0$ , vector $0$ (and this time it really is defined).

Step 1. $a \cdot b = 0 (3) + 0 (4) = 0$ . Why this step? Any dot product with the zero vector is $0$ .

Step 2. $b \cdot b = 3^{2} + 4^{2} = 25$ , so $∣ b ∣ = 25 = 5$ — a nonzero denominator. Why this step? Because $b \neq = 0$ , the division is legal (unlike Ex 5).

Step 3. Scalar $= \frac{0}{5} = 0$ ; vector $= \frac{0}{25} (3, 4) = (0, 0)$ . Why this step? $0/25$ is a genuine zero, not $0/0$ . The answer exists and equals the zero vector.

Verify: Contrast with Ex 5: here the denominators $5$ and $25$ are nonzero, so we divide legally and get a well-defined $0$ . Zero input $a$ → zero output; zero direction $b$ → undefined. Two very different degenerate cases. ✓

Ex 6 — Collinear vectors (cell C6)

Worked example Same line, both directions

(a) $a = (6, 0)$ , $b = (2, 0)$ (same direction, $θ = 0^{\circ}$ ). (b) $a = (6, 0)$ , $b = (- 2, 0)$ (opposite direction, $θ = 18 0^{\circ}$ ).

Forecast: $a$ already lies on the line of $b$ . Its shadow should be its whole self, $(6, 0)$ , in both cases — because the vector projection lands back on the $a$ line either way. The scalar should flip sign in (b).

Step 1 (a). $a \cdot b = 12$ , $b \cdot b = 4$ so $∣ b ∣ = 2$ . Vector $= \frac{12}{4} (2, 0) = (6, 0)$ . Scalar $= \frac{12}{2} = 6$ . Why this step? $θ = 0^{\circ}$ , $cos 0^{\circ} = 1$ , so scalar $= ∣ a ∣ = 6$ : the shadow is the full length.

Step 2 (b). $a \cdot b = - 12$ , $b \cdot b = 4$ so $∣ b ∣ = 2$ . Vector $= \frac{- 12}{4} (- 2, 0) = (6, 0)$ . Scalar $= \frac{- 12}{2} = - 6$ . Why this step? $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ , so scalar $= - ∣ a ∣ = - 6$ ; but the negative scalar × negative direction $(- 2, 0)$ swings the vector back to $(6, 0)$ .

Verify: In both cases the vector projection is $(6, 0) = a$ (fully recovered, as it must be for a collinear vector). Only the scalar carries the direction sign. ✓

Ex 7 — 3-D orthogonal decomposition (cell C7)

Ex 8 — Word problem: force along a ramp (cell C8)

Worked example How much force pushes along the slope?

A box on a ramp feels force $F = (0, - 10)$ N (gravity, pointing straight down). The ramp direction is $b = (4, 3)$ (rightward and up the incline). How much of the force acts along the ramp, and what is the vector along it?

Forecast: Gravity points down, the ramp goes up-right, so gravity's component along $b$ opposes the up-ramp direction — expect a negative scalar (force pulls the box down the slope). Figure s03 draws the box, the down-pointing gravity, and its orange shadow sliding down the ramp. Links to Work done by a force, where $W = F \cdot d$ is exactly this projection idea.

Step 1. $F \cdot b = 0 (4) + (- 10) (3) = - 30$ . Why this step? The dot product is the signed amount of force in $b$ 's direction.

Step 2. $b \cdot b = 4^{2} + 3^{2} = 25$ , so $∣ b ∣ = 25 = 5$ . Why this step? We need $∣ b ∣$ for the scalar (in newtons) and $b \cdot b$ for the vector.

Step 3. Scalar projection $= \frac{- 30}{5} = - 6$ N. Why this step? $6$ N of gravity acts along the ramp line; the minus means it points down the slope, dragging the box down. Units: N (a force), correct.

Step 4. Vector projection $= \frac{- 30}{25} (4, 3) = - 1.2 (4, 3) = (- 4.8, - 3.6)$ N. Why this step? Direction points down-and-left along the ramp — the box slides down.

Verify: Magnitude of the vector $= (- 4.8)^{2} + (- 3.6)^{2} = 23.04 + 12.96 = 36 = 6$ N, matching the scalar magnitude. Direction $(- 4.8, - 3.6)$ is $- 1.2 \times (4, 3)$ : exactly opposite the up-ramp direction. Physically sensible. ✓

Ex 9 — Exam twist: solve for an unknown (cell C9)

Worked example Work backward from a given projection

$a = (t, 4)$ with unknown $t$ . It is given that the scalar projection of $a$ onto $b = (3, 0)$ equals $5$ . Find $t$ . Then find the value of $t$ that makes the projection zero.

Forecast: The scalar projection onto the $x$ -axis is basically $a$ 's $x$ -part. If it must be $5$ , guess $t = 5$ ; if zero, guess $t = 0$ .

Step 1. $b \cdot b = 3^{2} + 0^{2} = 9$ , so $∣ b ∣ = 3$ . Scalar projection $= \frac{a \cdot b}{∣ b ∣} = \frac{3 t + 0}{3} = t$ . Why this step? Express the given quantity in terms of the unknown before solving.

Step 2 (projection $= 5$ ). Set $t = 5$ . Why this step? The equation collapsed to $t = 5$ directly. Clean.

Step 3 (projection $= 0$ ). Set $t = 0$ . Why this step? A zero projection means $a$ is perpendicular to $b$ ; with $b$ along $x$ , that forces $a$ 's $x$ -part $= 0$ , i.e. $t = 0$ , giving $a = (0, 4)$ (straight up — perpendicular to the $x$ -axis ✓).

Verify: For $t = 5$ : $a \cdot b = 3 (5) = 15$ , scalar $= 15/3 = 5$ ✓. For $t = 0$ : $a \cdot b = 0$ , scalar $= 0$ ✓, and $(0, 4) ⊥ (3, 0)$ . ✓

Active recall

Recall Which cell is which?

Obtuse angle gives what sign of scalar projection? ::: Negative — shadow falls backward. Projecting onto $b = 0$ gives? ::: Undefined — no direction, $0/0$ . Projecting the zero vector $a = 0$ onto a nonzero $b$ gives? ::: The zero vector $(0, 0)$ — well-defined, since the denominator $b \cdot b \neq = 0$ . Right-angle projection value? ::: Exactly $0$ (a genuine zero, not undefined). Making $b$ ten times longer changes the vector projection how? ::: Not at all — $b \cdot b$ cancels the extra length. Collinear opposite vectors: scalar sign? ::: Negative ( $cos 18 0^{\circ} = - 1$ ), but the vector projection still lands on $a$ 's line. Force $(0, - 10)$ along ramp $(4, 3)$ : scalar projection? ::: $- 6$ N (down the slope). What does $comp_{b} a$ mean in words? ::: The component of $a$ along $b$ — a signed number (length of the shadow).

Connections

Projection of vectors — the parent: formulas and the shadow intuition.
Dot product — the numerator of every example here.
Unit vectors — why $b = 0$ (no $\hat{b}$ ) breaks everything in Ex 5.
Orthogonal decomposition — Ex 7's split.
Gram-Schmidt process — repeats Ex 7's move to build orthonormal bases.
Least squares regression — projection of data onto a subspace (same C4 scale-invariance).
Work done by a force — Ex 8 is a projection in physics clothing.

Worked examples — Projection of vectors