Definition of Scalar Product
The scalar product, also known as the dot product, is a fundamental operation in vector algebra. For two vectors $\mathbf{a} = \begin{pmatrix} a_1\\a_2 \\a_3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} b_1\\b_2 \\b_3 \end{pmatrix}$ in three-dimensional space, their scalar product is defined as:
$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$
This operation takes two vectors as input and produces a scalar (hence the name) as output.
NoteThe scalar product is denoted by a dot (·) between the vectors, which is why it's also called the dot product.
Properties of the Scalar Product
The scalar product possesses several important properties that make it a versatile tool in vector calculations:
- Commutativity: $\mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v}$ This means the order of the vectors doesn't matter when calculating their scalar product.
- Distributivity over addition: $\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$ This property allows us to break down complex scalar products into simpler ones.
- Scalar multiplication: $(k\mathbf{v}) \cdot \mathbf{w} = k(\mathbf{v} \cdot \mathbf{w})$ Where $k$ is a scalar. This property shows how scalar multiplication interacts with the dot product.
- Self-dot product: $\mathbf{v} \cdot \mathbf{v} = ||\mathbf{v}||^2$ The scalar product of a vector with itself equals the square of its magnitude. This is because
$$\mathbf{v}\cdot\mathbf{v} = v_1^2 + v_2^2 + v_3^2 = \sqrt{v_1^2 + v_2^2 + v_3^2}^2 = ||\mathbf{v}||^2$$
ExampleLet $\mathbf{a} = \begin{pmatrix}1 \\2 \\3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 4\\ 5\\6 \end{pmatrix}$. Then:
$\mathbf{a} \cdot \mathbf{b} = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32$
Demonstrating commutativity: $\mathbf{b} \cdot \mathbf{a} = 4(1) + 5(2) + 6(3) = 4 + 10 + 18 = 32$
Geometric Interpretation
One of the most powerful aspects of the scalar product is its geometric interpretation. For two vectors $\mathbf{v}$ and $\mathbf{w}$, their scalar product is related to their magnitudes and the angle between them by the formula:
$$\mathbf{v} \cdot \mathbf{w} = ||\mathbf{v}||\times ||\mathbf{w}||\cos\theta$$
Where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}$, and $||\mathbf{v}||$ and $||\mathbf{w}||$ are the magnitudes of the vectors.
This formula provides a powerful way to find angles between vectors and has numerous applications in physics, particularly in calculating work done by a force.
The above figure shows a quick geometrical interpretation of the dot product. If you have two vectors, and draw a perpendicular line with one attaching it to another to make a right angle triangle, the length of the the 'adjacent' component is equal to $||\mathbf{v}||\cos(\theta)$ where $\mathbf{v}$ is the hypotenuse of this triangle.
By multiplying this length by $\mathbf{w}$ which is going to be the vector pointing in the same direction as the adjacent component of the right angle triangle, we will get $||\mathbf{w}||\,||\mathbf{v}||\cos\theta$
In physics, this can be used to represent the force in a direction from each component $x,y,z$, or how much 'power' one vector has combined with another in a particular direction.
Applications of the Dot Product
Perpendicular Vectors
One key application of the dot product is in determining whether two vectors are perpendicular (orthogonal). Two non-zero vectors $\mathbf{v}$ and $\mathbf{w}$ are perpendicular if and only if their scalar product is zero:
$$\mathbf{v} \cdot \mathbf{w} = 0 \iff \mathbf{v} \perp \mathbf{w}$$
This follows from the geometric interpretation, as $\cos 90° = 0$.
ExampleLet $\mathbf{a} = \begin{pmatrix} 1\\2 \\-1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2\\-1 \\ 2\end{pmatrix}$. Are they perpendicular?
$\mathbf{a} \cdot \mathbf{b} = 1(2) + 2(-1) + (-1)(2) = 2 - 2 - 2 = -2$
Since $\mathbf{a} \cdot \mathbf{b} \neq 0$, the vectors are not perpendicular.
Parallel Vectors
For parallel vectors, the angle between them is either 0° (same direction) or 180° (opposite directions). In either case, $|\cos\theta| = 1$, which leads to:
$$||\mathbf{v} \cdot \mathbf{w}|| = ||\mathbf{v}||\,||\mathbf{w}||$$
This equality holds if and only if the vectors are parallel.
Finding Angles Between Vectors
The scalar product formula can be rearranged to find the angle between two vectors:
$$\theta = \arccos\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}||\mathbf{w}|}\right)$$
This is particularly useful in geometry and physics problems where angles between vectors are required.
