Problem 37 - Precalculus with Limits

Question

In Exercises 31-40, find the angle \(\theta\) between the vectors. \(\mathbf{u} = 5\mathbf{i} + 5\mathbf{j}\) \(\mathbf{v} = -6\mathbf{i} + 6\mathbf{j}\)

Step-by-Step Solution

Answer

\(\theta = 90^\circ\)

Step 1: Compute the dot product

To compute the dot product, \(\mathbf{u}\cdot\mathbf{v}\), you need to multiply corresponding components of each vector and then add those products: \(\mathbf{u}\cdot\mathbf{v}= (5*-6)+(5*6)= -30+30=0\)

Step 2: Compute the magnitudes

Next, the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\) need to be computed. The formula for the magnitude is \(|\mathbf{u}| = \sqrt{x^2 + y^2}\) where x, y are the components of the vector. For \(\mathbf{u}\), it is \(|\mathbf{u}| = \sqrt{5^2+5^2} = \sqrt{50}\) and for \(\mathbf{v}\), it is \(|\mathbf{v}| = \sqrt{(-6)^2+6^2} = \sqrt{72}\)

Step 3: Compute the angle

Finally, the angle \(\theta\) can be computed by substituting the values in the formula \(\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos{\theta}\). This formula, when rearranged for \(\theta\), is \(\theta = \arccos (\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|})\). This yields \(\arccos (\frac{0}{\sqrt{50}*\sqrt{72}}) = \arccos (0) = 90^\circ\).

Key Concepts

Dot Product

Understanding the dot product is key to finding the angle between two vectors. It is a scalar representation of two vectors' magnitude and the cosine of the angle between them.
The dot product, denoted as \( \mathbf{u} \cdot \mathbf{v} \), is calculated by multiplying the corresponding components of vectors \( \mathbf{u} \) and \( \mathbf{v} \) and then adding those products together. In our case, \( \mathbf{u} \cdot \mathbf{v} = (5 \times -6) + (5 \times 6) = -30 + 30 = 0 \).
A dot product of zero signifies that vectors are perpendicular to each other as the cosine of 90 degrees is zero. This operation is crucial for many applications, such as projecting one vector onto another or determining whether two vectors are orthogonal.

Vector Magnitudes

The magnitude of a vector is a measure of its length or size, which is an integral component when working with vectors in various mathematical and physical contexts. To calculate a vector's magnitude, we use the formula \( |\mathbf{v}| = \sqrt{x^2 + y^2} \) for two-dimensional vectors, where \(x \) and \(y \) are the vector's horizontal and vertical components, respectively.
In our exercise, the magnitudes for \( \mathbf{u} \) and \( \mathbf{v} \) are \( |\mathbf{u}| = \sqrt{5^2 + 5^2} = \sqrt{50} \) and \( |\mathbf{v}| = \sqrt{(-6)^2 + 6^2} = \sqrt{72} \).
Keep in mind that the magnitude is always a positive number or zero, and it geometrically represents the vector's 'length' from its starting point to its end point when drawn on a coordinate plane.

Arccos Function

The arccos function, or inverse cosine function, allows us to find the angle whose cosine is a given number. It is denoted as \( \arccos \) and is ultimately the key to unlocking the angle between vectors when we have their dot product and magnitudes.
In our scenario, once we've obtained the dot product (which is 0) and calculated the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \) (which are \( \sqrt{50} \) and \( \sqrt{72} \) respectively), we use the arccos function to find the angle \( \theta \) by the formula \( \theta = \arccos (\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}) \). We end up with \( \arccos (0) \) which equals 90 degrees, indicating that the vectors are orthogonal to each other (i.e., they form a right angle).
The range of the arccos function is from 0 to 180 degrees, suitable for all possible angles between two vectors in two-dimensional space.