Problem 11 - Thomas Calculus

Question

Find the angles between the vectors in Exercises \(9-12\) to the nearest hundredth of a radian. $$ \mathbf{u}=\sqrt{3} \mathbf{i}-7 \mathbf{j}, \quad \mathbf{v}=\sqrt{3} \mathbf{i}+\mathbf{j}-2 \mathbf{k} $$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 1.93 radians.

Step 1: Understand the Problem

We need to find the angle between two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), given in component form as \(\mathbf{u}=\sqrt{3} \mathbf{i}-7 \mathbf{j}\) and \(\mathbf{v}=\sqrt{3}\mathbf{i}+\mathbf{j}-2 \mathbf{k}\). The angle between two vectors is found using the dot product formula.

Step 2: Recall the Formula for the Angle Between Two Vectors

The cosine of the angle \(\theta\) between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be found using the formula:\[\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\]where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product and \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) are the magnitudes of the vectors.

Step 3: Calculate the Dot Product

Calculate the dot product \(\mathbf{u} \cdot \mathbf{v}\):\[\mathbf{u} \cdot \mathbf{v} = (\sqrt{3})(\sqrt{3}) + (-7)(1) + (0)(-2) = 3 - 7 + 0 = -4\]

Step 4: Calculate the Magnitudes

Find the magnitude of each vector:- \(\|\mathbf{u}\| = \sqrt{(\sqrt{3})^2 + (-7)^2} = \sqrt{3 + 49} = \sqrt{52}\)- \(\|\mathbf{v}\| = \sqrt{(\sqrt{3})^2 + (1)^2 + (-2)^2} = \sqrt{3 + 1 + 4} = \sqrt{8}\)

Step 5: Substitute into the Formula

Substitute the dot product and magnitudes into the cosine formula:\[\cos \theta = \frac{-4}{\sqrt{52}\sqrt{8}}\]

Step 6: Solve for the Angle

Calculate the value of \(\theta\) using the inverse cosine function:\[\theta = \cos^{-1}\left(\frac{-4}{\sqrt{52}\sqrt{8}}\right)\]

Step 7: Evaluate and Round the Result

Use a calculator to evaluate \(\theta\) to the nearest hundredth of a radian:\[\theta \approx \cos^{-1}\left(\frac{-4}{\sqrt{52}\sqrt{8}}\right) \approx 1.93\] radians.

Key Concepts

Dot Product

The dot product is a key operation in vector mathematics, essential for finding the angle between vectors. When you take the dot product of two vectors, you're essentially multiplying their corresponding components and summing up those products. The formula for the dot product of two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \) is:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

In our example, the dot product is calculated as:

• \( (\sqrt{3})(\sqrt{3}) + (-7)(1) + (0)(-2) = 3 - 7 + 0 = -4 \)

This operation helps determine how much one vector extends in the direction of the other. If the dot product is zero, the vectors are perpendicular.

Magnitude of a Vector

The magnitude of a vector represents its length or size, ignoring its direction. Calculating the magnitude involves squaring each of its components, summing them up, and then taking the square root of this sum. For a vector \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \), the magnitude is given by:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

For vector \( \mathbf{u} = \sqrt{3} \mathbf{i} - 7 \mathbf{j} \), its magnitude is:

• \( \|\mathbf{u}\| = \sqrt{(\sqrt{3})^2 + (-7)^2} = \sqrt{3 + 49} = \sqrt{52} \)

Calculating these magnitudes is crucial because they are used to normalize the dot product when finding the angle between two vectors.

Inverse Cosine Function

To find the angle \( \theta \) between two vectors, we use the inverse cosine function. This function is the mathematical way to "reverse" the cosine function and find the angle whose cosine is a given number. From the dot product formula for vectors \( \mathbf{u} \) and \( \mathbf{v} \), we know:

• \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|} \)

The inverse cosine function, expressed as \( \cos^{-1} \), then helps find \( \theta \) from:

• \( \theta = \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\right) \)

In this example, we plug in our numbers:

• \( \theta = \cos^{-1}\left(\frac{-4}{\sqrt{52}\sqrt{8}}\right) \approx 1.93 \)

This result shows how the two vectors relate directionally.

Vector Components

Vector components are the individual parts that define a vector in space. Each component corresponds to a coordinate direction, typically given as \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) for the \(x\), \(y\), and \(z\)-axes, respectively. For example, the vector \( \mathbf{u} = \sqrt{3} \mathbf{i} - 7 \mathbf{j}\) has:

• \(x\)-component: \(\sqrt{3}\)
• \(y\)-component: \(-7\)
• \(z\)-component: \(0\) (since it's not included)

Understanding vector components is crucial because they allow us to perform operations such as addition, subtraction, and finding dot products.Moreover, they help visualize the direction and shape of vectors in the geometric plane or space. Recognizing how these components come together enables students to perform a wide variety of vector-related calculations.