Question

For \(\mathbf{v}=-2 \mathbf{i}-\mathbf{j}\) and \(\mathbf{w}=2 \mathbf{i}+\mathbf{j},\) find the dot product \(\mathbf{v} \cdot \mathbf{w}\) and the angle between \(\mathbf{v}\) and \(\mathbf{w}\).

Short answer

Dot product is -5. Angle between vectors is 180°.

Step-by-step solution

– Define the Dot Product Formula

The dot product of two vectors \(\textbf{v} = a_1 \textbf{i} + a_2 \textbf{j}\) and \(\textbf{w} = b_1 \textbf{i} + b_2 \textbf{j}\) is given by \(\textbf{v} \cdot \textbf{w} = a_1 b_1 + a_2 b_2\).

– Assign Components of Vectors

From the given vectors, \(\textbf{v} = -2 \textbf{i} - \textbf{j}\) and \(\textbf{w} = 2 \textbf{i} + \textbf{j}\), assign \(a_1 = -2, a_2 = -1, b_1 = 2, b_2 = 1\).

– Calculate the Dot Product

Using the dot product formula, \(\textbf{v} \cdot \textbf{w} = (-2)(2) + (-1)(1) = -4 - 1 = -5\).

– Define the Formula for the Angle Between Vectors

The angle \(\theta\) between two vectors can be found using \(\textbf{v} \cdot \textbf{w} = ||\textbf{v}|| ||\textbf{w}|| \cos(\theta) \). Rearrange to find \(\cos(\theta) = \frac{\textbf{v} \cdot \textbf{w}}{||\textbf{v}|| ||\textbf{w}||}\).

– Calculate the Magnitudes of the Vectors

Calculate the magnitudes: \(||\textbf{v}|| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\) and \(||\textbf{w}|| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}\).

– Calculate the cosine of the Angle

Substitute into the formula: \(\cos(\theta) = \frac{-5}{\sqrt{5} \sqrt{5}} = \frac{-5}{5} = -1\).

– Find the Angle

The angle whose cosine is -1 is \(\theta = 180\degree\).

Key concepts

vectors

Vectors are mathematical objects that have both magnitude and direction. They are usually represented in a coordinate system by an arrow. The length of the arrow depicts the magnitude, while the direction of the arrow represents the direction of the vector. For example, the vector \(\textbf{v} = -2 \textbf{i} - \textbf{j}\) has components in both the i (horizontal) and j (vertical) directions. Understanding vectors is crucial because they are widely used in physics and engineering to represent quantities such as force and velocity.

magnitude of vectors

The magnitude of a vector represents its length. It's calculated using the Pythagorean theorem. For a vector \(\textbf{v} = a_1 \textbf{i} + a_2 \textbf{j}\), the magnitude \(||\textbf{v}||\) can be found using the formula \[ ||\textbf{v}|| = \sqrt{a_1^2 + a_2^2}\.\] For instance, given the vector \(\textbf{v} = -2 \textbf{i} - \textbf{j}\), the magnitude is \(\sqrt{(-2)^2 + (-1)^2} = \sqrt{5}\). Similarly, for \(\textbf{w} = 2 \textbf{i} + \textbf{j}\), the magnitude is also \(\sqrt{5}\). Magnitude helps in understanding how long or strong a vector is.

angle between vectors

The angle between two vectors is a measure of their direction relative to each other. It can be determined using the dot product. The formula for finding the angle \(\theta\) between vectors \(\textbf{v}\) and \(\textbf{w}\) is: \[ \textbf{v} \cdot \textbf{w} = ||\textbf{v}|| ||\textbf{w}|| \cos(\theta).\] Rearranging to get \(\theta\), we have: \[ \cos(\theta) = \frac{\textbf{v} \cdot \textbf{w}}{||\textbf{v}|| ||\textbf{w}||}.\] For example, calculating the dot product of \(\textbf{v} = -2 \textbf{i} - \textbf{j}\) and \(\textbf{w} = 2 \textbf{i} + \textbf{j}\), we get \((\textbf{v} \cdot \textbf{w}) = -5\). Given both magnitudes as \(\sqrt{5} \), the cosine of the angle is \(-1\), meaning the angle \(\theta\) is \180\textdegree. This angle indicates that vectors \textbf{v} and \textbf{w} are pointing in exactly opposite directions.

cosine of angle

The cosine of the angle between two vectors is a crucial component in determining their relationship. The cosine function \(\text{cos}\theta\) links the dot product of the vectors with their magnitudes. If \(\text{cos}\theta \) is 1, the vectors are pointing in the same direction. If it’s -1, they are pointing in opposite directions. For vectors with an angle of 90 degrees between them, the cosine value is 0, making them perpendicular. Using the problem example, \[ \cos(\theta) = \frac{-5}{\sqrt{5} \sqrt{5}} = -1,\] tells us that \(\theta\) is 180 degrees. This value of \(-1\) confirms that the vectors \(\textbf{v}\) and \(\textbf{w}\) are directly opposite each other in direction, emphasizing the utility of the cosine function in vector analysis.