Problem 8 - Precalculus Mathematics for Calculus

Question

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle 1,-1\rangle$$

Step-by-Step Solution

Answer

(a) -12; (b) 180 degrees.

Step 1: Calculate the dot product

To find the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\), use the formula: \[ \mathbf{u} \cdot \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 \]Substituting given values, we have:\[ (-6) \cdot 1 + 6 \cdot (-1) = -6 - 6 = -12 \]Therefore, the dot product is \(\mathbf{u} \cdot \mathbf{v} = -12\).

Step 2: Calculate the magnitudes of the vectors

The magnitude of a vector \(\mathbf{u} = \langle u_1, u_2 \rangle\) is given by:\[ ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2} \]For \(\mathbf{u} = \langle -6, 6 \rangle\):\[ ||\mathbf{u}|| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]For \(\mathbf{v} = \langle 1, -1 \rangle\):\[ ||\mathbf{v}|| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

Step 3: Calculate the angle between the vectors

To find the angle \(\theta\) between the vectors, use the formula:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \]Substitute the computed values:\[ \cos(\theta) = \frac{-12}{6\sqrt{2} \cdot \sqrt{2}} = \frac{-12}{12} = -1 \]Thus, \(\theta = \cos^{-1}(-1) = 180^\circ\).Therefore, the angle between \(\mathbf{u}\) and \(\mathbf{v}\) is 180 degrees.

Key Concepts

Dot Product

The dot product, also known as scalar product, is a way to multiply vectors to get a scalar (a single number). Imagine two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \). To find their dot product, we use the formula:

• \( \mathbf{u} \cdot \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 \)

The dot product offers insight into the relationship between the vectors. If it’s zero, the vectors are perpendicular. In our case, with vectors \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle 1, -1 \rangle \), the dot product is calculated as:

• \( (-6) \cdot 1 + 6 \cdot (-1) = -6 - 6 = -12 \)

This result, \(-12\), indicates that \( \mathbf{u} \) and \( \mathbf{v} \) have a certain directional relationship, which we'll see plays a role in calculating the angle between them.

Vector Magnitude

The magnitude of a vector represents its length. For a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \), the magnitude, denoted as \( ||\mathbf{u}|| \), is found using the formula:

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

This calculation is essentially the Pythagorean theorem extended to vectors.
For instance, for vector \( \mathbf{u} = \langle -6, 6 \rangle \), we calculate the magnitude:

• \( ||\mathbf{u}|| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \)

Similarly, for vector \( \mathbf{v} = \langle 1, -1 \rangle \), the magnitude is:

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

Knowing these magnitudes helps provide scale when measuring the direction with which our vectors interact.

Angle Between Vectors

The angle between two vectors is found using the cosine of the angle formula. This relies on both the dot product and the magnitudes we've already calculated. The formula is:

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

Once you have the cosine value, you can find the angle \( \theta \) by applying the inverse cosine function. When we plug our calculated values for vectors \( \mathbf{u} \) and \( \mathbf{v} \), we find:

• \( \cos(\theta) = \frac{-12}{6\sqrt{2} \cdot \sqrt{2}} = \frac{-12}{12} = -1 \)

Using \( \cos^{-1}(-1) \), the angle \( \theta \) is found to be \( 180^\circ \).
This means that our vectors are pointing in exactly opposite directions, as a \( 180^\circ \) angle indicates a straight line in opposing directions, emphasizing their directional relationship as shown by the dot product.