Introduction
In mathematics, vectors are quantities that have both magnitude and direction. They are commonly represented as arrows in space, with the length of the arrow indicating the magnitude and the direction indicating the direction of the vector. Vectors are crucial in various fields such as physics, engineering, and computer science. In this topic, we will explore the fundamental concepts of vectors, including addition, subtraction, scalar multiplication, and dot product.
Vector Representation
A vector in two-dimensional space can be represented as $\vec{v} = \begin{bmatrix} x \ y \end{bmatrix}$, where $x$ and $y$ are the horizontal and vertical components of the vector, respectively. The magnitude of the vector can be calculated using the formula:
$$|\vec{v}| = \sqrt{x^2 + y^2}$$
Example:
Given two vectors $\vec{a} = \begin{bmatrix} 3 \ 4 \end{bmatrix}$ and $\vec{b} = \begin{bmatrix} -2 \ 5 \end{bmatrix}$, find the magnitude of each vector.
Solution: For vector $\vec{a}$: $$|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
For vector $\vec{b}$: $$|\vec{b}| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$$
Vector Addition and Subtraction
Adding or subtracting vectors involves combining their respective horizontal and vertical components.
Addition:
If $\vec{u} = \begin{bmatrix} x_1 \ y_1 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} x_2 \ y_2 \end{bmatrix}$, then the sum $\vec{u} + \vec{v}$ is calculated as: $$\vec{u} + \vec{v} = \begin{bmatrix} x_1 + x_2 \ y_1 + y_2 \end{bmatrix}$$
Subtraction:
Similarly, the difference of two vectors $\vec{u} - \vec{v}$ is given by: $$\vec{u} - \vec{v} = \begin{bmatrix} x_1 - x_2 \ y_1 - y_2 \end{bmatrix}$$
Example:
Given vectors $\vec{p} = \begin{bmatrix} 2 \ -1 \end{bmatrix}$ and $\vec{q} = \begin{bmatrix} -3 \ 4 \end{bmatrix}$, find $\vec{p} + \vec{q}$ and $\vec{p} - \vec{q}$.
Solution: For vector addition: $$\vec{p} + \vec{q} = \begin{bmatrix} 2 \ -1 \end{bmatrix} + \begin{bmatrix} -3 \ 4 \end{bmatrix} = \begin{bmatrix} 2 + (-3) \ -1 + 4 \end{bmatrix} = \begin{bmatrix} -1 \ 3 \end{bmatrix}$$
For vector subtraction: $$\vec{p} - \vec{q} = \begin{bmatrix} 2 \ -1 \end{bmatrix} - \begin{bmatrix} -3 \ 4 \end{bmatrix} = \begin{bmatrix} 2 - (-3) \ -1 - 4 \end{bmatrix} = \begin{bmatrix} 5 \ -5 \end{bmatrix}$$
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a single number).
If $\vec{v} = \begin{bmatrix} x \ y \end{bmatrix}$ and $k$ is a scalar, then the scalar product $k\vec{v}$ is: $$k\vec{v} = \begin{bmatrix} kx \ ky \end{bmatrix}$$
Example:
Given vector $\vec{r} = \begin{bmatrix} 3 \ -2 \end{bmatrix}$ and scalar $k = 4$, find $k\vec{r}$.
Solution: $$4\vec{r} = 4 \times \begin{bmatrix} 3 \ -2 \end{bmatrix} = \begin{bmatrix} 4 \times 3 \ 4 \times (-2) \end{bmatrix} = \begin{bmatrix} 12 \ -8 \end{bmatrix}$$
Dot Product
The dot product of two vectors $\vec{u} = \begin{bmatrix} x_1 \ y_1 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} x_2 \ y_2 \end{bmatrix}$ is given by: $$\vec{u} \cdot \vec{v} = x_1x_2 + y_1y_2$$
Example:
Given vectors $\vec{m} = \begin{bmatrix} 2 \ 5 \end{bmatrix}$ and $\vec{n} = \begin{bmatrix} -1 \ 3 \end{bmatrix}$, find $\vec{m} \cdot \vec{n}$.
Solution: $$\vec{m} \cdot \vec{n} = 2 \times (-1) + 5 \times 3 = -2 + 15 = 13$$
Common Mistakes
- Mixing up the order of vector components when performing vector operations.
- Forgetting to consider the direction when adding or subtracting vectors.
- Incorrectly calculating the magnitude or dot product of vectors.
Key Points
- Vectors have both magnitude and direction.
- Vectors can be added, subtracted, multiplied by a scalar, and have a dot product.
- The magnitude of a vector is calculated using the Pythagorean theorem.
Practice Questions
- Given vectors $\vec{u} = \begin{bmatrix} 1 \ 3 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} -2 \ 4 \end{bmatrix}$, find $\vec{u} + \vec{v}$.
Solution: $$\vec{u} + \vec{v} = \begin{bmatrix} 1 \ 3 \end{bmatrix} + \begin{bmatrix} -2 \ 4 \end{bmatrix} = \begin{bmatrix} 1 + (-2) \ 3 + 4 \end{bmatrix} = \begin{bmatrix} -1 \ 7 \end{bmatrix}$$
- If $\vec{a} = \begin{bmatrix} 4 \ -1 \end{bmatrix}$ and $\vec{b} = \begin{bmatrix} 2 \ 3 \end{bmatrix}$, calculate $\vec{a} - \vec{b}$.
Solution: $$\vec{a} - \vec{b} = \begin{bmatrix} 4 \ -1 \end{bmatrix} - \begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} 4 - 2 \ -1 - 3 \end{bmatrix} = \begin{bmatrix} 2 \ -4 \end{bmatrix}$$
- Determine the magnitude of the vector $\vec{c} = \begin{bmatrix} -3 \ 4 \end{bmatrix}$.
Solution: $$|\vec{c}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
- Find $3\vec{b}$ where $\vec{b} = \begin{bmatrix} -1 \ 2 \end{bmatrix}$.
Solution: $$3\vec{b} = 3 \times \begin{bmatrix} -1 \ 2 \end{bmatrix} = \begin{bmatrix} 3 \times (-1) \ 3 \times 2 \end{bmatrix} = \begin{bmatrix} -3 \ 6 \end{bmatrix}$$
- Calculate the dot product of vectors $\vec{x} = \begin{bmatrix} 2 \ 1 \end{bmatrix}$ and $\vec{y} = \begin{bmatrix} -3 \ 4 \end{bmatrix}$.
Solution: $$\vec{x} \cdot \vec{y} = 2 \times (-3) + 1 \times 4 = -6 + 4 = -2$$