Question

(a) find the dot product v \(\cdot \mathbf{w} ;\) (b) find the angle between \(\mathbf{v}\) and \(\mathbf{w} ;\) (c) state whether the vectors are parallel, orthogonal, or neither. $$ \mathbf{v}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{w}=\mathbf{i}-2 \mathbf{j} $$

Short answer

(a) \(\textbf{v} \cdot \textbf{w} = 0\), (b) \(\theta = 90^\textdegree\), (c) The vectors are orthogonal.

Step-by-step solution

- Find the Dot Product \(\textbf{v} \cdot \textbf{w}\)

The dot product of two vectors \(\textbf{v} = \textbf{a}_1\textbf{i} + \textbf{a}_2\textbf{j}\) and \(\textbf{w} = \textbf{b}_1\textbf{i} + \textbf{b}_2\textbf{j}\) is calculated as \(\textbf{v} \cdot \textbf{w} = a_1b_1 + a_2b_2\). Given \(\textbf{v} = 2\textbf{i} + \textbf{j}\) and \(\textbf{w} = \textbf{i} - 2\textbf{j}\), we get \[\textbf{v} \cdot \textbf{w} = (2)(1) + (1)(-2) = 2 - 2 = 0.\textbf{v} \cdot \textbf{w} = 0\]

- Find the Magnitudes of the Vectors

The magnitude of a vector \(\textbf{v} = \textbf{a}_1\textbf{i} + \textbf{a}_2\textbf{j}\) is given by \(|\textbf{v}| = \sqrt{a_1^2 + a_2^2}\). So, \(|\textbf{v}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \). Similarly, \(|\textbf{w}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\).

- Find the Angle Between the Vectors

The angle \(\theta\) between two vectors is given by \(\textbf{v} \cdot \textbf{w} = |\textbf{v}||\textbf{w}| \cos \theta \). We know \(\textbf{v} \cdot \textbf{w} = 0\). Therefore, \(\textbf{v} \cdot \textbf{w} = |\textbf{v}||\textbf{w}| \cos \theta = 0\). As the magnitudes are non-zero, that leads to \(\theta = \frac{\textbackslashpi}{2} = 90^\textdegree\).

- Determine if Vectors are Parallel, Orthogonal, or Neither

Since the dot product \(\textbf{v} \cdot \textbf{w} = 0\) and the angle \(\theta = 90^\textdegree\), the vectors \(\textbf{v}\) and \(\textbf{w}\) are orthogonal.

Key concepts

Vector Operations

Vector operations are fundamental to understanding many concepts in physics and mathematics. They involve mathematical manipulations of vectors, which are quantities having both magnitude and direction. The two main vector operations we deal with are vector addition and scalar multiplication. Vector addition involves adding corresponding components of the vectors. If we have \(\textbf{a} = a_1 \textbf{i} + a_2 \textbf{j}\) and \(\textbf{b} = b_1 \textbf{i} + b_2 \textbf{j}\), then their sum is \(\textbf{a} + \textbf{b} = (a_1 + b_1)\textbf{i} + (a_2 + b_2)\textbf{j}\). Scalar multiplication, on the other hand, involves multiplying each component of a vector by a scalar value. For instance, if we have a vector \(\textbf{v} = a_1\textbf{i} + a_2\textbf{j}\) and a scalar \(k\), then the scalar multiplication is represented as \(\textbf{v}' = k(a_1\textbf{i} + a_2 \textbf{j}) = (ka_1)\textbf{i} + (ka_2)\textbf{j}\). Understanding these operations is crucial for solving problems involving vectors, such as calculating the dot product or finding vector magnitudes.

Vector Magnitude

The vector magnitude, also known as the vector length or norm, is a measure of how long the vector is. It is calculated using the Pythagorean theorem. For a vector \(\textbf{v} = a_1 \textbf{i} + a_2 \textbf{j}\), the magnitude is given by \(|\textbf{v}| = \sqrt{a_1^2 + a_2^2}\). For example, let's find the magnitude of \(\textbf{v} = 2\textbf{i} + \textbf{j}\). Here we use the formula: \(|\textbf{v}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\). Similarly, for \(\textbf{w} = \textbf{i} - 2\textbf{j}\), the magnitude is \(|\textbf{w}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\). Magnitudes are always non-negative and give us an idea of the size of the vectors, irrespective of their direction.

Angle Between Vectors

To find the angle between two vectors, we use the dot product formula. The formula states that \(\textbf{v} \cdot \textbf{w} = |\textbf{v}| |\textbf{w}| \cos \theta\), where \(\theta\) is the angle between the vectors. The dot product \(\textbf{v} \cdot \textbf{w}\) is calculated as the sum of the products of their corresponding components. For \(\textbf{v} = 2\textbf{i} + \textbf{j}\) and \(\textbf{w} = \textbf{i} - 2\textbf{j}\), the dot product is: \(\textbf{v} \cdot \textbf{w} = (2)(1) + (1)(-2) = 2 - 2 = 0\). Given that \(\textbf{v} \cdot \textbf{w} = 0\) and the magnitudes we calculated earlier are non-zero, we solve the equation \(0 = \sqrt{5} \sqrt{5} \cos \theta\) to find \(\cos \theta = 0\). This implies that \(\theta = \frac{\pi}{2}\) or \(90^{\circ}\). Therefore, the angle between the vectors is 90 degrees.

Orthogonal Vectors

Orthogonal vectors are vectors that are perpendicular to each other. This perpendicularity condition is mathematically represented by their dot product being zero. If \(\textbf{v} \cdot \textbf{w} = 0\), the vectors \(\textbf{v}\) and \(\textbf{w}\) are orthogonal. From our previous solution steps, we have seen that \(2 \textbf{i} + \textbf{j} \) and \( \textbf{i} - 2 \textbf{j}\) have a dot product of zero: \( 2\cdot1 + 1\cdot(-2) = 2 - 2 = 0\). This confirms that \(\textbf{v}\) and \(\textbf{w}\) are orthogonal, meaning they meet at a right angle. Recognizing orthogonal vectors is important in many areas of mathematics and physics, including solving systems of equations and understanding physical phenomena like forces and motion.