Unit 8 · Topic 4.8 · Optional

Vectors

Represent quantities with magnitude and direction, operate on their components, and use trigonometry and the dot product to connect algebraic and geometric information.

Learning Goals

Find vector components from initial and terminal points.
Determine magnitude and direction from component form.
Add, subtract, and scalar-multiply vectors algebraically and geometrically.
Construct unit vectors and use \(\mathbf i,\mathbf j\) notation.
Use the dot product to find angles and identify perpendicular vectors.
Interpret resultant vectors in geometric and applied settings.

1. Scalars and Vectors

A scalar records magnitude only. A vector records both magnitude and direction.

Quantity	Type	Reason
18 seconds	Scalar	Time has no spatial direction.
12 meters east	Vector	Both size and direction are specified.
60 kilometers per hour	Scalar speed	Only rate magnitude is given.
60 kilometers per hour northwest	Vector velocity	Rate and direction are given.

Common vector notations are \(\mathbf v\), \(\vec v\), and \(\langle a,b\rangle\).

2. A Directed Line Segment

The vector arrow's horizontal and vertical components form a right triangle.

The tail is the initial point and the arrowhead is the terminal point. The segment length is the vector's magnitude, while the arrow identifies its direction.

3. Component Form from Two Points

For initial point \(P(x_1,y_1)\) and terminal point \(Q(x_2,y_2)\), subtract initial coordinates from terminal coordinates:

\[\overrightarrow{PQ}=\langle x_2-x_1,\ y_2-y_1\rangle.\]

If \(P=(-3,2)\) and \(Q=(4,6)\), then \(\overrightarrow{PQ}=\langle7,4\rangle\): move 7 units right and 4 units up.

4. Position Vectors, Equality, and the Zero Vector

Translating a vector without rotating or resizing it does not change the vector. Thus, the arrow from \((-3,2)\) to \((4,6)\) is equal to the position vector from \((0,0)\) to \((7,4)\).

Idea	Component condition
\(\mathbf u=\mathbf v\)	Corresponding components are equal.
Opposite vectors	\(-\mathbf v\) has equal magnitude and reverse direction.
Zero vector	\(\mathbf 0=\langle0,0\rangle\), with magnitude 0 and no unique direction.

5. Magnitude from the Pythagorean Theorem

For \(\mathbf v=\langle a,b\rangle\),

\[\|\mathbf v\|=\sqrt{a^2+b^2}.\]

For \(\mathbf v=\langle3,4\rangle\), \(\|\mathbf v\|=5\). Magnitude is a nonnegative scalar, not another vector.

6. Direction Angle and Trigonometric Components

If a nonzero vector has magnitude r and direction angle \(\theta\), measured counterclockwise from the positive x-axis, then

\[\mathbf v=\langle r\cos\theta,\ r\sin\theta\rangle.\]

Conversely, \(\tan\theta=b/a\), but inverse tangent alone may return an angle in the wrong quadrant. Use the signs of a and b, or an atan2 calculation, to select \(0\le\theta<2\pi\).

For \(\langle-3,3\rangle\), the reference angle is \(\pi/4\), but the vector lies in Quadrant II, so \(\theta=3\pi/4\).

7. Vector Addition

Add corresponding components:

\[\langle a,b\rangle+\langle c,d\rangle=\langle a+c,b+d\rangle.\]

Geometrically, place the second vector's tail at the first vector's head. The resultant points from the first tail to the final head. Equivalently, the two vectors form adjacent sides of a parallelogram and their sum is the diagonal.

For \(\mathbf u=\langle2,5\rangle\) and \(\mathbf v=\langle-1,3\rangle\), \(\mathbf u+\mathbf v=\langle1,8\rangle\).

8. Vector Subtraction

\[\mathbf u-\mathbf v=\mathbf u+(-\mathbf v).\]

In components, \(\langle a,b\rangle-\langle c,d\rangle=\langle a-c,b-d\rangle\). When u and v share a tail, \(\mathbf u-\mathbf v\) points from the head of v to the head of u.

9. Scalar Multiplication

\[k\langle a,b\rangle=\langle ka,kb\rangle,\qquad \|k\mathbf v\|=|k|\,\|\mathbf v\|.\]

Scalar k	Effect
\(k>1\)	Same direction, greater magnitude
\(0<k<1\)	Same direction, smaller magnitude
\(k<0\)	Reverse direction; magnitude scales by \(\|k\|\)
\(k=0\)	Produces the zero vector

10. Unit Vectors

A unit vector has magnitude 1. For any nonzero vector v, divide by its magnitude:

\[\widehat{\mathbf v}=\frac{\mathbf v}{\|\mathbf v\|}.\]

For \(\mathbf v=\langle-5,12\rangle\), \(\|\mathbf v\|=13\), so

\[\widehat{\mathbf v}=\left\langle-\frac5{13},\frac{12}{13}\right\rangle.\]

Its magnitude is \(\sqrt{25/169+144/169}=1\). The zero vector cannot be normalized because division by zero is undefined.

11. Standard Unit Vectors

\[\mathbf i=\langle1,0\rangle,\qquad \mathbf j=\langle0,1\rangle.\]

Therefore, \(\langle a,b\rangle=a\mathbf i+b\mathbf j\). For example, \(\langle4,-3\rangle=4\mathbf i-3\mathbf j\), showing the horizontal and vertical contributions separately.

12. The Dot Product

For \(\mathbf u=\langle a,b\rangle\) and \(\mathbf v=\langle c,d\rangle\),

\[\mathbf u\cdot\mathbf v=ac+bd.\]

The result is a scalar. For \(\mathbf u=\langle2,-1\rangle\) and \(\mathbf v=\langle3,4\rangle\),

\[\mathbf u\cdot\mathbf v=2(3)+(-1)(4)=2.\]

Do not confuse a dot product with componentwise multiplication that returns a vector.

13. Angle Between Two Vectors

The algebraic and geometric forms of the dot product combine to give

\[\mathbf u\cdot\mathbf v=\|\mathbf u\|\,\|\mathbf v\|\cos\theta,\qquad \theta=\arccos\left(\frac{\mathbf u\cdot\mathbf v}{\|\mathbf u\|\,\|\mathbf v\|}\right).\]

Dot product	Angle	Geometric relation
Positive	\(0\le\theta<\pi/2\)	Acute
Zero	\(\theta=\pi/2\)	Perpendicular for nonzero vectors
Negative	\(\pi/2<\theta\le\pi\)	Obtuse

14. Resultants and the Law of Cosines

If \(\theta\) is the angle between u and v when they share a tail, then

\[\|\mathbf u+\mathbf v\|^2=\|\mathbf u\|^2+\|\mathbf v\|^2+2\|\mathbf u\|\,\|\mathbf v\|\cos\theta.\]

This follows from expanding \((\mathbf u+\mathbf v)\cdot(\mathbf u+\mathbf v)\). For perpendicular vectors of magnitudes 6 and 8, the resultant magnitude is \(\sqrt{6^2+8^2}=10\).

In a context such as wind plus an airplane's air velocity, add components first, then report the resultant magnitude with an appropriate direction and units.

15. AP Workflow and Common Errors

Identify initial and terminal points or translate the context into components.
Compute terminal minus initial for a directed segment.
Use the Pythagorean theorem for magnitude and quadrant information for direction.
Perform vector operations component by component.
Normalize only after finding the original magnitude.
Use the dot product for angles or perpendicularity.
Interpret the final magnitude, direction, and units in context.

Do not reverse initial and terminal coordinates.
Do not add magnitudes when the vectors point in different directions.
Do not let inverse tangent erase quadrant information.
Do not multiply by a negative scalar without reversing direction.
Do not divide the zero vector by its magnitude.
Do not call vectors perpendicular from a zero dot product unless both are nonzero.

Key Takeaways

Represent magnitude and direction with components and use vector addition, scalar multiplication, and dot products.
Core relationship: \(|\langle a,b\rangle|=\sqrt{a^2+b^2}\)
Error check: Do not add components when calculating magnitude.

Checkpoint · Topic 4.8

Find the component form and magnitude of the vector from \(P=(-2,5)\) to \(Q=(4,-3)\).
Find the direction angle of \(\langle-4,-4\sqrt3\rangle\) on \(0\le\theta<2\pi\).
For \(\mathbf u=\langle3,-2\rangle\) and \(\mathbf v=\langle-1,5\rangle\), calculate \(2\mathbf u-\mathbf v\).
Find and verify a unit vector in the direction of \(\langle8,-6\rangle\).
Use the dot product to find the angle between \(\langle1,2\rangle\) and \(\langle2,-1\rangle\), and interpret the result.