Problem 34 - Precalculus Mathematics for Calculus

Question

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\langle 4,0,2\rangle, \quad \mathbf{v}=\langle 2,-1,0\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 36.87 degrees.

Step 1: Dot Product

Calculate the dot product of the two vectors \(\mathbf{u}\) and \(\mathbf{v}\). The formula for the dot product \( \mathbf{u} \cdot \mathbf{v} \) is:\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\]Substituting the coordinates of \(\mathbf{u}\) and \(\mathbf{v}\):\[4 \cdot 2 + 0 \cdot (-1) + 2 \cdot 0 = 8 + 0 + 0 = 8\]

Step 2: Magnitudes of Vectors

Calculate the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) using the formula:\[||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2}\]\[||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]For \(\mathbf{u}:\)\[||\mathbf{u}|| = \sqrt{4^2 + 0^2 + 2^2} = \sqrt{16 + 0 + 4} = \sqrt{20}\]For \(\mathbf{v}:\)\[||\mathbf{v}|| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5}\]

Step 3: Calculate Cosine of the Angle

Use the formula for the cosine of the angle \(\theta\) between two vectors:\[\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||}\]Substituting the known values:\[\cos(\theta) = \frac{8}{\sqrt{20} \cdot \sqrt{5}} = \frac{8}{\sqrt{100}} = \frac{8}{10} = 0.8\]

Step 4: Angle in Degrees

Convert the angle from its cosine form to degrees using the inverse cosine function:\[\theta = \cos^{-1}(0.8)\]Using a calculator, find \(\theta\).\[\theta \approx 36.87^\circ\]

Key Concepts

Dot Product

Understanding the concept of the dot product helps to determine how much one vector extends in the direction of another vector. For two vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the dot product is computed using the formula:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \)

It involves multiplying each corresponding component of the vectors and adding the results.
The dot product is a scalar and not a vector, meaning the result is a single number, not another vector.
For example, the calculation \( 4 \times 2 + 0 \times (-1) + 2 \times 0 = 8 + 0 + 0 = 8 \) for vectors \( \mathbf{u} = \langle 4,0,2 \rangle \) and \( \mathbf{v} = \langle 2,-1,0 \rangle \) shows their dot product is 8. This outcome signifies how aligned the vectors are. If the dot product is positive, as in this instance, it indicates that both vectors point in generally the same direction.

Magnitude of a Vector

The magnitude, or length, of a vector provides a measure of how long the vector is. For a vector \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \), the magnitude is calculated using the formula:

• \( ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2} \)

This involves taking the square root of the sum of the squares of its components. It gives a positive number that represents the vector's length in the vector space.
For instance, for \( \mathbf{u} = \langle 4,0,2 \rangle \), we find:

• \( ||\mathbf{u}|| = \sqrt{4^2 + 0^2 + 2^2} = \sqrt{20} \)

Similarly, for \( \mathbf{v} = \langle 2,-1,0 \rangle \), we compute:

• \( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5} \)

Understanding magnitudes is crucial when comparing vector lengths or normalizing vectors to have a length of 1.

Cosine of the Angle Between Vectors

The cosine of the angle between vectors provides insight into the orientation of these vectors with respect to each other. To find the cosine of the angle \(\theta\) between vectors \(\mathbf{u}\) and \(\mathbf{v}\), use the formula:

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

This involves dividing the dot product by the product of the magnitudes of the vectors.
This calculation tells us whether vectors are oriented in the same direction, opposite, or perpendicular. A cosine value of 1 signifies identical direction, whereas -1 indicates opposite directions. Any value between will suggest a certain degree of angular difference.
In our problem, substituting the known values, we have:

• \( \cos(\theta) = \frac{8}{\sqrt{20} \cdot \sqrt{5}} = \frac{8}{\sqrt{100}} = 0.8 \)

Using the inverse cosine function, you can then convert this into degrees, which gives \( \theta \approx 36.87^\circ \). This tells us the angle between the vectors is moderately acute.