Problem 12 - Precalculus : Building Concepts and Connections

Question

Find the smallest nomnegation angle betwoen the vectors \(\mathbf{v}\) and \(\mathbf{w} .\) Round your anstoer to the nearest tenth of a degree. $$\mathbf{v}=\langle 2,-4\rangle, \mathbf{w}=\langle 6,3\rangle$$

Step-by-Step Solution

Answer

The smallest non-negative angle between the vectors is 90 degrees.

Step 1: Calculate the dot product of the two vectors

The dot product of two vectors can be calculated by multiplying the corresponding components of the vectors and then summing these products. So, for vectors \(\mathbf{v}\) = \(<2,-4>\) and \(\mathbf{w}\) = \(<6,3>\), the dot product is \(2*6 + (-4*3) = 12 - 12 = 0\).

Step 2: Find the magnitude of each vector

The magnitude of a vector is calculated by squaring each component, summing these, and then taking the square root. For vectors \(\mathbf{v}\) = \(<2,-4>\) and \(\mathbf{w}\) = \(<6,3>\), the magnitudes are \(\|\mathbf{v}\|\) = \(\sqrt{2^2 + (-4)^2} = \sqrt{20}\) and \(\|\mathbf{w}\|\) = \(\sqrt{6^2 + 3^2} = \sqrt{45}\).

Step 3: Find cosine of the angle

The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them, i.e. \(\mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\|\|\mathbf{w}\|cos(\Theta)\). From Step 1, the dot product is 0, so 0 = \(\sqrt{20}\) * \(\sqrt{45}\) * \(cos(\Theta)\). Solving this for cos(\Theta), we get \(cos(\Theta)\) = \(0/(\sqrt{20}*\sqrt{45})\) = 0.

Step 4: Find non-negative angle in degrees

The angle \(\Theta\) can be found by taking the arccos (inverse cosine) of \(cos(\Theta)\). Using a calculator: \(\Theta\)=arccos(0) = 90 degrees. This is the smallest non-negative value for the angle between the vectors as angle values go from 0 to 180 degrees in arccos function.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way of multiplying two vectors that results in a scalar (a single number). To calculate the dot product, you multiply the corresponding components of the vectors and sum these products. In our example, consider vectors \( \mathbf{v} = \langle 2, -4 \rangle \) and \( \mathbf{w} = \langle 6, 3 \rangle \). The dot product is calculated as:

\( 2 \times 6 + (-4) \times 3 \)

This results in \( 12 - 12 = 0 \). In this case, the dot product being zero means the vectors are perpendicular, making the angle between them 90 degrees. Understanding the dot product is crucial, as it reveals how much one vector aligns with another. If the dot product is positive, the angle is acute (less than 90 degrees). If negative, it's obtuse (greater than 90 degrees).

Magnitude of a Vector

The magnitude of a vector is its "length" or "size" in the space. It is calculated quite simply using the Pythagorean Theorem. Given a vector \( \mathbf{v} = \langle x, y \rangle \), its magnitude \( \| \mathbf{v} \| \) is calculated as:

\( \sqrt{x^2 + y^2} \)

For \( \mathbf{v} = \langle 2, -4 \rangle \), the magnitude is:

\( \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \)

Similarly, for \( \mathbf{w} = \langle 6, 3 \rangle \), the magnitude is:

\( \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \)

The magnitude of vectors helps us understand how long or short vectors are, which is vital when comparing or combining them. Knowing how to find it is a fundamental skill when dealing with vector calculations.

Angle Between Vectors

The angle between two vectors can tell us a lot about their relationship. The cosine of this angle is found using the dot product divided by the product of the vectors' magnitudes. Mathematically, this is expressed as: \[ \cos(\Theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \cdot \|\mathbf{w}\|} \] In our example with vectors \( \mathbf{v} \) and \( \mathbf{w} \), the dot product \( \mathbf{v} \cdot \mathbf{w} = 0 \), and the magnitudes are \( \sqrt{20} \) and \( \sqrt{45} \) respectively. Substituting these values, we have: \[ \cos(\Theta) = \frac{0}{\sqrt{20} \cdot \sqrt{45}} = 0 \] From this, the angle \( \Theta \) can be found using the arccos function:

\( \Theta = \arccos(0) = 90^{\circ} \)

This means the vectors are perpendicular to each other, an essential observation in vector analysis. Understanding how to find the angle between vectors can be critical in applications like physics, computer graphics, and more.