Vectors in ℝⁿ — operations, geometric interpretation: Exercises

A reminder of what each notation means, so you never hit an unexplained symbol:

Definition Symbol cheat-sheet

$v = (v_{1}, \dots, v_{n})$ — a vector: an ordered list of numbers, also an arrow from the origin.
$0 = (0, \dots, 0)$ — the zero vector: every component is $0$ ; as an arrow it has zero length and points nowhere (no direction).
$u + v$ — add component-wise (one axis at a time).
$- v = (- 1) v$ — the negation: flip every component's sign (same length, opposite direction).
$u - v$ — subtraction: this is defined as adding the negation, $u + (- v) = (u_{1} - v_{1}, \dots, u_{n} - v_{n})$ . So it is still component-wise; geometrically it is the arrow from the tip of $v$ to the tip of $u$ .
$c v$ — scale: multiply every component by the number $c$ (stretch by $∣ c ∣$ , flip if $c < 0$ ).
$∥ v ∥$ — norm / length: the straight-line distance of the arrow's tip from the origin.
$\hat{v} = v /∥ v ∥$ — the unit vector: same direction, length exactly $1$ .
$u \cdot v$ — dot product: a single number measuring how much two arrows point the same way.
$θ$ — the angle between two arrows, recovered from $cos θ = \frac{u \cdot v}{∥ u ∥∥ v ∥}$ .
$u ⊥ v$ — "perpendicular", true exactly when $u \cdot v = 0$ .

Level 1 — Recognition

These check that you can name and execute the two operations without thinking about geometry yet.

Recall Solution 1.1

WHAT: add matching components. WHY: each axis is independent, so we combine them one at a time. $u + v = (2 + (- 3), - 1 + 5, 4 + 1) = (- 1, 4, 5) .$

Recall Solution 1.2

WHAT: multiply every component by $3$ . WHY: scaling stretches the whole arrow uniformly. $3 u = (3 \cdot 2, 3 \cdot (- 1), 3 \cdot 4) = (6, - 3, 12) .$

Recall Solution 1.3

WHAT: apply Pythagoras. WHY: the arrow is the hypotenuse of a right triangle with legs $6$ and $8$ . $∥ w ∥ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10.$

Level 2 — Application

Now combine the operations and produce quantities you'll actually use.

Recall Solution 2.1

WHAT: scale each vector, then subtract. WHY: subtraction is just adding the negative scaling ( $u - v = u + (- v)$ , from the cheat-sheet). $2 u = (2, 4), \frac{1}{2} v = (2, - 1) .$ $2 u - \frac{1}{2} v = (2 - 2, 4 - (- 1)) = (0, 5) .$

Recall Solution 2.2

WHAT: divide by the length. WHY: dividing by $∥ a ∥$ rescales to length $1$ without rotating. First check $a \neq = 0$ so the division is legal: here $a = (3, - 4) \neq = 0$ . ✓ $∥ a ∥ = 3^{2} + (- 4)^{2} = 25 = 5, \hat{a} = \frac{1}{5} (3, - 4) = (0.6, - 0.8) .$ Check: $0. 6^{2} + (- 0.8)^{2} = 0.36 + 0.64 = 1 = 1.$ ✓

Recall Solution 2.3

WHAT: multiply matching components and sum. WHY: the dot product weighs how much the two arrows align. $u \cdot v = 2 \cdot 4 + (- 1) \cdot 0 + 3 \cdot (- 2) = 8 + 0 - 6 = 2.$

Level 3 — Analysis

Here the numbers must be interpreted: angles, perpendicularity, and what the sign of a dot product tells you.

The figure below shows Exercise 3.1. What to observe: the black arrow $b = (1, 0)$ lies flat along the horizontal $x$ -axis, while the red arrow $a = (1, 1)$ climbs the diagonal. The small red arc between them is the angle $θ$ . Notice the red arrow sits exactly halfway between the $x$ -axis and straight up — that visual "halfway" is precisely the $4 5^{\circ}$ we compute.

Figure — Vectors in ℝⁿ — operations, geometric interpretation

Recall Solution 3.1

WHAT: use $cos θ = \frac{a \cdot b}{∥ a ∥∥ b ∥}$ . WHY this tool: the dot product is the only operation that packages the angle into arithmetic — we rearrange its geometric form to isolate $θ$ . $a \cdot b = 1 \cdot 1 + 1 \cdot 0 = 1, ∥ a ∥ = 2, ∥ b ∥ = 1.$ $cos θ = \frac{1}{2 \cdot 1} = \frac{1}{2} \Rightarrow θ = 4 5^{\circ} .$ WHAT IT LOOKS LIKE: exactly the red arrow of the figure above — halfway between flat and vertical, hence $4 5^{\circ}$ .

Recall Solution 3.2

WHAT: compute the dot product and read its sign. WHY: perpendicularity is exactly the statement $u \cdot v = 0$ , and the sign otherwise reports whether the angle is acute or obtuse. $u \cdot v = 3 \cdot 4 + (- 2) \cdot 6 = 12 - 12 = 0.$ Zero $\Rightarrow$ perpendicular ( $u ⊥ v$ ). Because $cos θ = \frac{u \cdot v}{∥ u ∥∥ v ∥}$ and the norms are positive, the sign of the dot product is the sign of $cos θ$ :

Dot $> 0$ : $cos θ > 0$ , so $θ < 9 0^{\circ}$ — arrows lean the same way.
Dot $= 0$ : $θ = 9 0^{\circ}$ — a right angle.
Dot $< 0$ : $cos θ < 0$ , so $θ > 9 0^{\circ}$ — arrows lean apart (obtuse).

Recall Solution 3.3

WHAT: apply the same $cos θ$ formula, now expecting a negative cosine. WHY: the dot product came out negative, which by the sign rule of 3.2 forces an obtuse angle — we still solve for $θ$ the same way. $p \cdot q = 1 \cdot (- 1) + 0 + 0 = - 1, ∥ p ∥ = 1, ∥ q ∥ = (- 1)^{2} + 1^{2} = 2 .$ $cos θ = \frac{- 1}{1 \cdot 2} = - \frac{1}{2} \Rightarrow θ = 13 5^{\circ} .$ The negative dot product forces an obtuse angle — the arrows point into different half-planes.

Level 4 — Synthesis

Combine addition, norms, and dot products in one chain of reasoning.

The figure below illustrates Exercise 4.1. What to observe: the four black arrows form a parallelogram with sides $u$ and $v$ . The red solid arrow is one diagonal $u + v$ ; the red dashed segment is the other diagonal $u - v$ (running between the tips of $u$ and $v$ ). The parallelogram law says: square those two red diagonals, add them, and you get exactly twice the squared sides — watch the red objects, they are the whole point.

Recall Solution 4.1

WHAT: compute both diagonals of the parallelogram and both sides. WHY: the law says the two diagonals' squared lengths together equal twice the two sides' squared lengths. $u + v = (4, 1) \Rightarrow ∥ u + v ∥^{2} = 16 + 1 = 17.$ $u - v = (- 2, 3) \Rightarrow ∥ u - v ∥^{2} = 4 + 9 = 13.$ Left side: $17 + 13 = 30$ . $∥ u ∥^{2} = 1 + 4 = 5, ∥ v ∥^{2} = 9 + 1 = 10 \Rightarrow 2 (5) + 2 (10) = 30.$ Both sides equal $30$ . ✓ WHAT IT LOOKS LIKE: the two red diagonals of the parallelogram in the figure above, $u + v$ (short, solid) and $u - v$ (dashed), balance against its four equal black sides.

Recall Solution 4.2

WHAT: set the dot product to zero. WHY: perpendicular $⟺$ dot $= 0$ . $a \cdot b = 3 t + 2 t = 5 t = 0 \Rightarrow t = 0.$ So $a = (0, 2)$ (straight up) and $b = (3, 0)$ (straight right) — indeed a right angle. Only $t = 0$ works.

Recall Solution 4.3

WHAT: use $∥ c u ∥ = ∣ c ∣ ∥ u ∥$ . WHY: scaling by $c$ multiplies length by $∣ c ∣$ (the absolute value, since flipping doesn't change length). $∥ u ∥ = 5, ∣ c ∣ \cdot 5 = 15 \Rightarrow ∣ c ∣ = 3 \Rightarrow c = 3 or c = - 3.$ Both are valid: $c = 3$ keeps the direction, $c = - 3$ reverses it — same length either way.

Level 5 — Mastery

One extended problem weaving the whole chapter together.

Recall Solution 5.1

WHAT we're doing: splitting $u$ into a piece along $v$ and a piece square to it — a first taste of Projections and Orthogonal Decomposition. WHY this works: the parallel piece is a scaling of $v$ , and the right scaling is picked so the leftover is perpendicular. The dot product is the exact tool that measures "how much of $u$ lies along $v$ ."

Step 0 — legality check. We are about to divide by $v \cdot v = ∥ v ∥^{2}$ , so we need $v \neq = 0$ . Here $v = (1, 3) \neq = 0$ , so the projection is well-defined. (Projecting onto $0$ would be meaningless — $0$ points nowhere.)

Step 1 — the parallel part. Write $p = k v$ for some scalar $k$ (parallel means "a scaled copy"). The formula that makes the remainder perpendicular is $k = \frac{u \cdot v}{v \cdot v} .$ Compute: $u \cdot v = 2 \cdot 1 + 1 \cdot 3 = 5, v \cdot v = 1^{2} + 3^{2} = 10.$ $k = \frac{5}{10} = \frac{1}{2}, p = \frac{1}{2} (1, 3) = (\frac{1}{2}, \frac{3}{2}) .$

Step 2 — the perpendicular part. Whatever is left over (using subtraction $u - p = u + (- p)$ ): $q = u - p = (2 - \frac{1}{2}, 1 - \frac{3}{2}) = (\frac{3}{2}, - \frac{1}{2}) .$

Step 3 — (b) verify perpendicularity. $q \cdot v = \frac{3}{2} \cdot 1 + (- \frac{1}{2}) \cdot 3 = \frac{3}{2} - \frac{3}{2} = 0. ✓$ And $p + q = (\frac{1}{2} + \frac{3}{2}, \frac{3}{2} - \frac{1}{2}) = (2, 1) = u .$ ✓

WHY the choice of $k$ was forced: we insisted the leftover $q = u - k v$ be perpendicular to $v$ , i.e. $q \cdot v = 0$ . Expand that requirement using the dot product's distributive rule: $(u - k v) \cdot v = u \cdot v - k (v \cdot v) = 0.$ Solving the single unknown $k$ gives $k = \frac{u \cdot v}{v \cdot v}$ — this is the only value that makes the leftover square to $v$ . Any other $k$ would leave a nonzero dot product, so $q$ would not be perpendicular. (Note $v \cdot v \neq = 0$ because $v \neq = 0$ — the same legality check as Step 0.)

Recall Solution 5.2

WHY this should hold: $p$ and $q$ meet at a right angle, so they form a right triangle with $u$ as the hypotenuse — Pythagoras applies. $∥ u ∥^{2} = 2^{2} + 1^{2} = 5.$ $∥ p ∥^{2} = (\frac{1}{2})^{2} + (\frac{3}{2})^{2} = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = 2.5.$ $∥ q ∥^{2} = (\frac{3}{2})^{2} + (- \frac{1}{2})^{2} = \frac{9}{4} + \frac{1}{4} = 2.5.$ $∥ p ∥^{2} + ∥ q ∥^{2} = 2.5 + 2.5 = 5 = ∥ u ∥^{2} . ✓$

Connections

Projections and Orthogonal Decomposition — Exercise 5 is the seed of the full projection formula.
Dot Product and Orthogonality — the perpendicularity tests in Levels 3–5.
Norms and Distance in Rn — the length and parallelogram computations.
Vectors in ℝⁿ — operations, geometric interpretation — the parent note these exercises drill.