Dot Product

Question 1

What is another name for the dot product and why?

Answer 1

The scalar product — because the result is a scalar (a number), not a vector.

Question 2

What is the geometric definition of the dot product of vectors u and v?

Answer 2

u · v = |u||v|cosθ, where θ is the angle between the vectors.

Question 3

What is the range of θ in the dot product, and why?

Answer 3

0 ≤ θ ≤ π. It is always the smallest possible angle between the two vectors.

Question 4

Why can you not write the dot product as “uv”?

Answer 4

“uv” has no meaning for vectors. You must write u · v explicitly — the dot is required to indicate this specific operation.

Question 5

What is the rearranged formula for finding the angle between two vectors?

Answer 5

cosθ = (u · v) / (|u||v|)

Question 6

Property 1 — When are two non-zero vectors orthogonal (perpendicular)?

Answer 6

If and only if u · v = 0. Since cosθ = 0 when θ = 90°, the dot product equals zero for perpendicular vectors.

Question 7

Why do the unit vectors satisfy i·j = j·k = i·k = 0?

Answer 7

Because i, j, k point along the x, y, z axes — they are all perpendicular to each other, so the angle between any two is 90°, giving cos90° = 0.

Question 8

Property 2 — Is the dot product commutative? Show why

Answer 8

Yes. u · v = |u||v|cosθ and v · u = |v||u|cosθ.

Since multiplication of scalars is commutative, both are equal. So u · v = v · u.

Question 9

Property 3 — What is u · u equal to, and why?

Answer 9

u · u = |u|².

The angle between a vector and itself is θ = 0, so u · u = |u||u|cos0° = |u|² × 1 = |u|².

This is useful for finding magnitude: |u| = √(u · u).

Question 10

Property 4 — What is the associative property of the dot product with a scalar k?

Answer 10

(ku) · v = k(u · v) = u · (kv).

Scaling one vector by k just scales the dot product by k, regardless of which vector it’s applied to.

Question 11

Property 5 — What is the distributive property of the dot product?

Answer 11

u · (v + w) = u · v + u · w.

The dot product distributes over vector addition, just like multiplication distributes over addition.

Question 12

What is the dot product formula in component form for u = (u1, u2) and v = (v1, v2)?

Answer 12

u · v = u1v1 + u2v2.

In 3D: u · v = u1v1 + u2v2 + u3v3. Multiply matching components and add.

Question 13

Proof — why does the component form u·v = u1v1 + u2v2 work?

Answer 13

Write u = u1i + u2j and v = v1i + v2j.

Expand: u·v = u1v1(i·i) + u1v2(i·j) + u2v1(j·i) + u2v2(j·j).

Since i·i = j·j = 1 and i·j = j·i = 0, this simplifies to u1v1 + u2v2. ✓

Question 14

Problem type 1 — How do you find the dot product of two vectors in component form? e.g. u = (3, −2, 1), v = (4, 5, −3)

Answer 14

Multiply matching components and add: u · v = (3)(4) + (−2)(5) + (1)(−3) = 12 − 10 − 3 = −1.

Question 15

Problem type 2 — How do you find the angle between two vectors in component form? e.g. u = (3, −2, 1), v = (4, 5, −3)

Answer 15

Step 1: find u · v = −1 (see above).

Step 2: |u| = √(9+4+1) = √14, |v| = √(16+25+9) = √50.

Step 3: cosθ = −1/(√14 · √50) = −1/√700. Step 4: θ = cos⁻¹(−1/√700) ≈ 92.2°.

Question 16

What is i·i, j·j, and k·k, and why?

Answer 16

All equal 1. Each is a unit vector (magnitude = 1), and the angle with itself is 0°. So using u·u = |u|²: i·i = 1² = 1, and same for j·j and k·k.

Question 17

What is i·j, i·k, and j·k, and why?

Answer 17

All equal 0. The x, y, and z axes are perpendicular to each other, so the angle between any two different unit vectors is 90°. Since cos90° = 0, the dot product is 1 × 1 × 0 = 0.

Question 18

How do you find a vector perpendicular to u = (a, b) using the dot product?

Answer 18

Set v = (x, y) and solve u · v = 0.
This gives ax + by = 0.
Let y = t (any scalar), then x = −bt/a.
So v = t(−b, a) — infinitely many solutions, all scalar multiples of each other.

Question 19

What is the shortcut for finding a perpendicular vector to u = (a, b) in 2D?

Answer 19

Swap the components and negate one: v = (−b, a). Always verify with u · v = a(−b) + b(a) = 0 ✓

Question 20

Why are there infinitely many vectors perpendicular to a given 2D vector?

Answer 20

Because the condition u · v = 0 gives one equation with two unknowns. One variable is free — letting it equal a scalar t gives a whole family of solutions: v = t(−b, a). Every scalar multiple is also perpendicular.

Question 21

Problem type - Find all vectors perpendicular to u = (2, 3)

Answer 21

Set v = (a, b) and solve u · v = 0 → 2a + 3b = 0.

Let b = t → a = −3t/2.

So v = t(−3, 2) for any scalar t. Verify: (2)(−3) + (3)(2) = 0

Question 22

In parallelogram ABCD, given vertices A, B and D, how do you find the coordinates of C?

Answer 22

C = B + D − A. Rearranged from the rule that opposite sides are equal: AB = DC.

Question 23

How do you find the interior angles of a parallelogram given its vertices?

Answer 23

Pick a vertex, form two vectors along its adjacent sides, then use cosθ = (u · v) / (|u||v|).

The opposite angle = 180° − θ since adjacent interior angles in a parallelogram are supplementary.

Question 24

How do you find the angles between the diagonals of a parallelogram?

Answer 24

Form the two diagonal vectors AC = C − A and BD = D − B, then use the dot product formula to find θ.

The second angle is simply 180° − θ.

Question 25

Find a vector perpendicular to u = (1, 2, 3)

Answer 25

Set v = (a, b, c) and solve u · v = 0 → a + 2b + 3c = 0, so a = −2b − 3c. Freely choose any values for b and c, then solve for a. e.g. let b = 1, c = 1 → a = −2(1) − 3(1) = −5, giving v = (−5, 1, 1). Verify: (1)(−5) + (2)(1) + (3)(1) = 0 ✓