Problem 9 - Algebra and Trigonometry

Question

Find (a) \(\mathbf{u} \cdot \mathbf{v}\) and (b) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$ \mathbf{u}=\langle 3,-2\rangle, \quad \mathbf{v}=\langle 1,2\rangle $$

Step-by-Step Solution

Answer

(a) The dot product is -1; (b) the angle between them is approximately 97 degrees.

Step 1: Write the formula for the dot product

The dot product of two vectors \(\mathbf{u} = \langle u_1, u_2 \rangle\) and \(\mathbf{v} = \langle v_1, v_2 \rangle\) is given by the formula \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2\).

Step 2: Calculate the dot product

Substitute the components of \(\mathbf{u} = \langle 3, -2 \rangle\) and \(\mathbf{v} = \langle 1, 2 \rangle\) into the dot product formula: \(\mathbf{u} \cdot \mathbf{v} = 3 \times 1 + (-2) \times 2\). Simplify to get: \(\mathbf{u} \cdot \mathbf{v} = 3 - 4 = -1\).

Step 3: Calculate magnitudes of vectors

The magnitude of a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\) is given by \(\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2}\). Calculate \(\|\mathbf{u}\| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\) and \(\|\mathbf{v}\| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\).

Step 4: Write the formula for the cosine of the angle

The cosine of the angle \(\theta\) between two vectors can be found using the formula \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\).

Step 5: Calculate the cosine of the angle

Substitute the known values into the formula: \(\cos(\theta) = \frac{-1}{\sqrt{13} \times \sqrt{5}}\). Simplify this to \(\cos(\theta) = \frac{-1}{\sqrt{65}}\).

Step 6: Find the angle using the inverse cosine

Find \(\theta\) by taking the inverse cosine: \(\theta = \cos^{-1}\left(\frac{-1}{\sqrt{65}}\right)\). Calculate this to find \(\theta \approx 97\) degrees.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is similar to the length of a line segment in geometry. For any vector, say \( \mathbf{u} = \langle u_1, u_2 \rangle \), its magnitude is calculated using the Pythagorean theorem. It is denoted by \( \| \mathbf{u} \| \) and is given by the formula:

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

This formula effectively breaks down the vector into two parts: a horizontal component \(u_1\) and a vertical component \(u_2\).
By squaring each component, adding them together, and then taking the square root of the result, we get the magnitude of the vector. For example, for the vector \( \mathbf{u} = \langle 3, -2 \rangle \), the magnitude is:

• \( \| \mathbf{u} \| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)

Understanding the magnitude is crucial because it tells us how "long" or significant a vector is in space.
It doesn't matter whether the vector components are positive or negative since the square of any real number is positive.

Cosine of the Angle Between Vectors

The cosine of the angle between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) can tell us about their direction. Cosine, in this context, relates directly to the dot product and the magnitudes of the vectors.
To find the cosine of the angle \( \theta \), use the formula:

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

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is a result of multiplying corresponding components of \( \mathbf{u} \) and \( \mathbf{v} \) and then adding them:

• For \( \mathbf{u} = \langle 3, -2 \rangle \) and \( \mathbf{v} = \langle 1, 2 \rangle \), \( \mathbf{u} \cdot \mathbf{v} = 3 \times 1 + (-2) \times 2 = 3 - 4 = -1 \).

This dot product tells us how much \( \mathbf{u} \) and \( \mathbf{v} \) are aligned or opposed to each other. If cosine is zero, vectors are perpendicular; if it's positive, they angle less than 90 degrees; if negative, they angle more than 90 degrees.

Inverse Cosine

The inverse cosine function, often denoted as \( \cos^{-1} \) or \( \text{acos} \), is used to find the angle when given the cosine value. Once you have \( \cos(\theta) \) from the vectors, \( \theta \) can be found by:

• \( \theta = \cos^{-1}\left(\frac{-1}{\sqrt{65}}\right) \)

This step converts the ratio obtained in the previous step back into an angle, expressed in degrees or radians. To use \( \cos^{-1} \):

• Ensure the input value is between -1 and 1, as this range is valid for cosine.

In practical use, you'll typically utilize a calculator set to degree mode to find the angle in degrees. For instance, given the cosine value \( \frac{-1}{\sqrt{65}} \), the inverse cosine yields \( \theta \approx 97 \) degrees.
This measure of the angle helps in understanding the spatial relationship between the two vectors.