Lesson 10.8: Vectors

New Concept

This lesson introduces vectors—quantities with both magnitude and direction. You will learn to represent them geometrically, perform algebraic operations, and use their components to solve problems involving forces and motion, like calculating a plane's ground speed.

What’s next

Coming up, you'll tackle interactive examples and practice cards to master vector operations and their real-world applications.

Property

A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at $(0,0)$ and is identified by its terminal point $(a, b)$. If the initial point of a vector $\overrightarrow{CD}$ is $C(x_1, y_1)$ and the terminal point is $D(x_2, y_2)$, then the position vector is found by calculating $\overrightarrow{AB} = (x_2 - x_1, y_2 - y_1) = (a, b)$.

Examples

• For a vector with initial point $P(1, 2)$ and terminal point $Q(5, 5)$, the position vector is $\mathbf{v} = (5 - 1, 5 - 2) = (4, 3)$.

• For a vector with initial point $A(-2, 3)$ and terminal point $B(1, -1)$, the position vector is $\mathbf{v} = (1 - (-2), -1 - 3) = (3, -4)$.

Property

Given a position vector $\mathbf{v} = (a, b)$, the magnitude is found by $|\mathbf{v}| = \sqrt{a^2 + b^2}$. The direction is equal to the angle formed with the $x$-axis, or with the $y$-axis, depending on the application. For a position vector, the direction is found by $\tan \theta = \frac{b}{a} \implies \theta = \tan^{-1}\left(\frac{b}{a}\right)$.

Examples

• For $\mathbf{v} = (3, 4)$, the magnitude is $|\mathbf{v}| = \sqrt{3^2 + 4^2} = 5$. The direction is $\theta = \tan^{-1}(\frac{4}{3}) \approx 53.1^\circ$.

• For $\mathbf{u} = (-5, 12)$, the magnitude is $|\mathbf{u}| = \sqrt{(-5)^2 + 12^2} = 13$. The direction is in the second quadrant, so $\theta = \tan^{-1}(\frac{12}{-5}) + 180^\circ \approx 112.6^\circ$.

Property

The sum of two vectors $\mathbf{u}$ and $\mathbf{v}$, or vector addition, produces a third vector $\mathbf{u} + \mathbf{v}$, the resultant vector. To find $\mathbf{u} + \mathbf{v}$ where $\mathbf{u}=(a,b)$ and $\mathbf{v}=(c,d)$, we calculate $(a+c, b+d)$.
Scalar multiplication involves the product of a vector and a scalar, $k$. Each component of the vector is multiplied by the scalar: $k\mathbf{v} = (ka, kb)$.

Examples

• Given $\mathbf{u} = (1, 6)$ and $\mathbf{v} = (-3, 2)$, their sum is $\mathbf{u} + \mathbf{v} = (1 + (-3), 6 + 2) = (-2, 8)$.

• Given $\mathbf{v} = (4, -5)$, multiplying by the scalar $k=2$ gives $2\mathbf{v} = (2 \cdot 4, 2 \cdot (-5)) = (8, -10)$.

Property

If $\mathbf{v}$ is a nonzero vector, then $\frac{\mathbf{v}}{|\mathbf{v}|}$ is a unit vector in the direction of $\mathbf{v}$. Any vector divided by its magnitude is a unit vector. The horizontal unit vector is written as $\mathbf{i} = (1, 0)$ and is directed along the positive horizontal axis. The vertical unit vector is written as $\mathbf{j} = (0, 1)$ and is directed along the positive vertical axis.

Examples

• Find the unit vector for $\mathbf{v} = (5, -12)$. First, find the magnitude: $|\mathbf{v}| = \sqrt{5^2 + (-12)^2} = 13$. The unit vector is $\mathbf{u} = (\frac{5}{13}, -\frac{12}{13})$.

• Find the unit vector for $\mathbf{w} = (-7, 24)$. The magnitude is $|\mathbf{w}| = \sqrt{(-7)^2 + 24^2} = 25$. The unit vector is $\mathbf{u} = (-\frac{7}{25}, \frac{24}{25})$.

Property

Given a vector $\mathbf{v}$ with initial point $P = (x_1, y_1)$ and terminal point $Q = (x_2, y_2)$, $\mathbf{v}$ is written as $\mathbf{v} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j}$. Given $\mathbf{v} = a\mathbf{i} + b\mathbf{j}$ and $\mathbf{u} = c\mathbf{i} + d\mathbf{j}$, then $\mathbf{v} + \mathbf{u} = (a + c)\mathbf{i} + (b + d)\mathbf{j}$ and $\mathbf{v} - \mathbf{u} = (a - c)\mathbf{i} + (b - d)\mathbf{j}$.

Examples

• Given $\mathbf{v} = 2\mathbf{i} + 5\mathbf{j}$ and $\mathbf{u} = 4\mathbf{i} - 2\mathbf{j}$, their sum is $\mathbf{v} + \mathbf{u} = (2+4)\mathbf{i} + (5-2)\mathbf{j} = 6\mathbf{i} + 3\mathbf{j}$.

• Given $\mathbf{v} = -\mathbf{i} + 3\mathbf{j}$ and $\mathbf{u} = 2\mathbf{i} + 7\mathbf{j}$, their difference is $\mathbf{v} - \mathbf{u} = (-1-2)\mathbf{i} + (3-7)\mathbf{j} = -3\mathbf{i} - 4\mathbf{j}$.

Property

The dot product of two vectors $\mathbf{v} = (a, b)$ and $\mathbf{u} = (c, d)$ is the sum of the product of the horizontal components and the product of the vertical components. The formula is $\mathbf{v} \cdot \mathbf{u} = ac + bd$. To find the angle $\theta$ between the two vectors, use the formula $\cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|}$.

Examples

• Find the dot product of $\mathbf{v} = (3, 5)$ and $\mathbf{u} = (2, -4)$. $\mathbf{v} \cdot \mathbf{u} = (3)(2) + (5)(-4) = 6 - 20 = -14$.

• Find the dot product of $\mathbf{a} = 7\mathbf{i} - 2\mathbf{j}$ and $\mathbf{b} = \mathbf{i} + 3\mathbf{j}$. $\mathbf{a} \cdot \mathbf{b} = (7)(1) + (-2)(3) = 7 - 6 = 1$.