The Dot Product as a Projection

The dot product of two vectors has two equivalent definitions, and reconciling them is one of the most beautiful results in vector algebra.

Algebraic: $\vec A\cdot\vec B = A_x B_x + A_y B_y$

Geometric: $\vec A\cdot\vec B = |\vec A|\,|\vec B|\cos\theta$

The geometric form has a wonderful interpretation: $|\vec B|\cos\theta$ is the length of the projection of $\vec B$ onto $\vec A$ (i.e., the "shadow" of $\vec B$ on the line through $\vec A$). Multiplied by $|\vec A|$, the dot product measures how much $\vec B$ goes in the direction of $\vec A$.

Step-by-Step Solution

Given: $\vec A = (3, 4)$ and $\vec B = (2, 1)$.

Find: The dot product, the angle between them, and the projection of $\vec B$ onto $\vec A$.

---

Step 1 — Compute the dot product algebraically.

$\vec A\cdot\vec B = (3)(2) + (4)(1) = 6 + 4 = 10$

Step 2 — Compute the magnitudes.

$|\vec A| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$

$|\vec B| = \sqrt{2^{2} + 1^{2}} = \sqrt{5} \approx 2.2361$

Step 3 — Use the geometric form to find the angle.

$\cos\theta = \dfrac{\vec A\cdot\vec B}{|\vec A|\,|\vec B|} = \dfrac{10}{5 \cdot \sqrt{5}} = \dfrac{10}{5\sqrt{5}} = \dfrac{2}{\sqrt{5}} \approx 0.8944$

$\theta = \arccos(0.8944) \approx 26.57°$

Step 4 — Compute the projection of $\vec B$ onto $\vec A$.

The scalar projection is:

$\text{proj}_{\vec A}\vec B = |\vec B|\cos\theta = \sqrt{5} \cdot \dfrac{2}{\sqrt{5}} = 2$

So the "shadow" of $\vec B$ on the line through $\vec A$ has length 2.

Multiplying by the unit vector $\hat A = \vec A/|\vec A| = (3/5, 4/5) = (0.6, 0.8)$ gives the vector projection:

$\vec{\text{proj}}_{\vec A}\vec B = 2 \cdot (0.6, 0.8) = (1.2, 1.6)$

Step 5 — Verify the equivalence of the two formulas.

$|\vec A|\,|\vec B|\cos\theta = 5 \cdot \sqrt{5} \cdot 0.8944 = 5 \cdot 2 = 10 \;\;\checkmark$

Both give the same answer: $\vec A\cdot\vec B = 10$.

Step 6 — Special cases.

• $\vec A\cdot\vec B = 0$ ⇔ $\theta = 90°$ → vectors are perpendicular.
• $\vec A\cdot\vec B > 0$ ⇔ $\theta < 90°$ → vectors point in "similar" directions.
• $\vec A\cdot\vec B < 0$ ⇔ $\theta > 90°$ → vectors point in "opposite" directions.

---

Answer:

$\boxed{\;\vec A\cdot\vec B = 10,\quad \theta \approx 26.57°,\quad \vec{\text{proj}}_{\vec A}\vec B = (1.2,\, 1.6)\;}$

The dot product 10 is the product of $|\vec A| = 5$ and the scalar projection of $\vec B$ onto $\vec A$ (which is 2). Geometrically, the dot product measures how much $\vec B$ extends in the direction of $\vec A$.

Try It

• Adjust the components of B to see how the projection changes.
• When the angle reaches 90°, the dot product hits zero (perpendicular vectors).
• When $\vec B$ aligns perfectly with $\vec A$, the projection equals $|\vec B|$ and the dot product is maximum.
• The HUD shows both forms of the dot product and confirms they always agree.