This tutorial explains the dot product (scalar product) of two vectors, its properties, geometric meaning, and applications such as finding angles and determining whether vectors are orthogonal.
Let \[ \mathbf{v} = \langle v_1, v_2 \rangle \quad \text{and} \quad \mathbf{u} = \langle u_1, u_2 \rangle \] be two vectors in the plane. The dot product of v and u is defined by:
\[ \mathbf{v} \cdot \mathbf{u} = v_1u_1 + v_2u_2 \]Important: The result of a dot product is a scalar (a real number), not a vector.
Given: \[ \mathbf{v} = \langle -2, 3 \rangle, \quad \mathbf{u} = \langle 4, 6 \rangle \] find \( \mathbf{v} \cdot \mathbf{u} \).
Apply the definition of the dot product:
\[ \mathbf{v} \cdot \mathbf{u} = (-2)(4) + (3)(6) \] \[ = -8 + 18 = 10 \]The dot product of the two vectors is 10.
Let: \[ \mathbf{v} = \langle 3, -4 \rangle \]
Method 1: Definition
\[ \|\mathbf{v}\| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5 \]Method 2: Dot Product Property
\[ \mathbf{v} \cdot \mathbf{v} = 3^2 + (-4)^2 = 25 \] \[ \|\mathbf{v}\| = \sqrt{25} = 5 \]Both methods give the same magnitude, confirming the property.
If \( \theta \) is the angle between vectors \( \mathbf{v} \) and \( \mathbf{u} \), then:
\[ \mathbf{v} \cdot \mathbf{u} = \|\mathbf{v}\| \, \|\mathbf{u}\| \cos \theta \]
This formula allows us to compute the angle between two vectors and determine whether they are orthogonal.
If \( \cos \theta = 0 \), then \( \theta = \frac{\pi}{2} \) and the vectors are perpendicular.
Show that: \[ \mathbf{v} = \langle 3, -4 \rangle, \quad \mathbf{u} = \langle 4, 3 \rangle \] are orthogonal.
Since the dot product is zero, the vectors are orthogonal.
Find the angle between: \[ \mathbf{v} = \langle 1, 1 \rangle, \quad \mathbf{u} = \langle -4, 3 \rangle \]