Vectors

Unit 4 · Functions Involving Parameters, Vectors, and Matrices

What AP Precalc asks here

A 2D vector $\vec v = (a, b)$ has magnitude $|\vec v| = \sqrt{a^2 + b^2}$ and direction angle $\theta = \arctan(b/a)$ (adjusted for quadrant). Vector arithmetic is component-wise. The dot product $\vec u \cdot \vec v = u_1 v_1 + u_2 v_2$ tests perpendicularity (zero means perpendicular) and gives the cosine of the angle between vectors via $\cos\theta = (\vec u \cdot \vec v) / (|\vec u| |\vec v|)$ .

Vector operations

Magnitude

|\vec v| = \sqrt{a^2 + b^2}

Sum

\vec u + \vec v = (u_1 + v_1, u_2 + v_2)

Scalar multiplication

k \vec v = (ka, kb)

Dot product

\vec u \cdot \vec v = u_1 v_1 + u_2 v_2

AP Tip: Dot product = 0 ⇔ vectors are perpendicular. This is the fastest perpendicularity test.

Type 1

Magnitude and direction

Apply Pythagorean formula for magnitude; use arctan with quadrant adjustment for direction.

Example 1

What is the magnitude of

\vec v = (3, 4)

? (A) 5 (B) 7 (C) 12 (D) 25

Practice more of this type— AI-generated · always-new problems

Generate Problems →

Type 2

Add, subtract, scalar multiply

Operate component-wise.

Example 2

Compute

2(1, -3) + (4, 5)

. (A)

(5, 2)

(B)

(6, -1)

(C)

(6, 2)

(D)

(10, -10)

Practice more of this type— AI-generated · always-new problems

Generate Problems →

Type 3

Dot product

Multiply components pairwise and add.

Example 3

Are

\vec u = (2, 3)

and

\vec v = (-3, 2)

perpendicular? (A) Yes, because

\vec u \cdot \vec v = 0

(B) No, because

\vec u \cdot \vec v = 5

Practice more of this type— AI-generated · always-new problems

Generate Problems →