Problem 36 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$\mathbf{u}=-6 \mathbf{i}-3 \mathbf{j}$$ $$\mathbf{v}=-8 \mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle θ between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\Theta = cos^{-1}(\frac{36}{\sqrt{45}*\sqrt{80}}). \)

Step 1: Calculate Dot Product

The dot product of \(\mathbf{u}\) and \(\mathbf{v}\) can be calculated by multiplying their corresponding \(\mathbf{i}\) and \(\mathbf{j}\) components, then adding those products together. Hence, \(\mathbf{u}\cdot\mathbf{v} = (-6*-8) + (-3*4) = 48 -12 = 36.\)

Step 2: Calculate Magnitude

Compute the magnitude of each vector. The magnitude of a vector can be calculated by taking the square root of the sum of the squares of the components. Therefore, \(||\mathbf{u}||= \sqrt{(-6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}\) and \(||\mathbf{v}|| = \sqrt{(-8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80}\).

Step 3: Calculate Cosine of Angle

The cosine of the angle θ between two vectors can be calculated by dividing the dot product of the vectors by the product of their magnitudes. Hence, \( cos\Theta = \frac{\mathbf{u}\cdot\mathbf{v}}{{||\mathbf{u}||*||\mathbf{v}||}} = \frac{36}{\sqrt{45}*\sqrt{80}}\).

Step 4: Calculate Angle

Find the angle by taking the arccos of the value calculated in the previous step. Hence, \(\Theta = cos^{-1}(\frac{36}{\sqrt{45}*\sqrt{80}})\).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar. When finding the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\), you multiply their corresponding components and then sum those results. For instance, given \(\mathbf{u} = -6 \mathbf{i} - 3 \mathbf{j}\) and \(\mathbf{v} = -8 \mathbf{i} + 4 \mathbf{j}\), the dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as:

• First, multiply the \(\mathbf{i}\) components: \((-6) \times (-8)\)
• Next, multiply the \(\mathbf{j}\) components: \((-3) \times 4\)
• Add the results: \(48 - 12 = 36\)

The dot product tells us if the vectors are working together. If it's positive, they're generally pointing in the same direction, if negative, they oppose each other, and if zero, they are perpendicular.

Vector Magnitude

The magnitude of a vector, often referred to as its length, describes how long the vector is. It's a measure of the distance from the vector's tail to its head. To calculate the magnitude of a vector, you use the formula \(\sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the components of the vector. For the vector \(\mathbf{u} = -6 \mathbf{i} - 3 \mathbf{j}\), the magnitude is:

• Square each component: \((-6)^2 = 36\) and \((-3)^2 = 9\)
• Sum the squares: \(36 + 9 = 45\)
• Take the square root: \(\sqrt{45}\)

This process is repeated for \(\mathbf{v} = -8 \mathbf{i} + 4 \mathbf{j}\), resulting in \(\sqrt{80}\). The magnitude helps in quantifying the vector's size in space, essential for further computations like finding angles.

Cosine of Angle

The cosine of the angle between two vectors is a crucial step for finding the angle itself. The formula to find the cosine of the angle \(\Theta\) is given by:\[\cos\Theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \times ||\mathbf{v}||}\]So, from our exercise, the cosine of the angle is:

• Use the dot product from before: \(36\)
• Use the magnitudes: \(\sqrt{45}\) and \(\sqrt{80}\)
• Plug these into the formula: \(\cos\Theta = \frac{36}{\sqrt{45} \times \sqrt{80}}\)

The cosine function relates the angle to these vectors' dot product and magnitudes, offering a precise dimension of how aligned they are in space. Once you have \(\cos \Theta\), you can find the angle \(\Theta\) using the arccosine function.