Problem 35 - Precalculus with Limits

Question

In Exercises 31-40, find the angle \(\theta\) between the vectors. \(\mathbf{u} = 2\mathbf{i} - \mathbf{j}\) \(\mathbf{v} = 6\mathbf{i} + 4\mathbf{j}\)

Step-by-Step Solution

Answer

The angle \(\theta\) between the vectors \(\mathbf{u} = 2\mathbf{i} - \mathbf{j}\) and \(\mathbf{v} = 6\mathbf{i} + 4\mathbf{j}\) is \(\cos^{-1}(\frac{8}{\sqrt{5}\sqrt{52}})\) degrees. This value needs to be calculated using a calculator, typically resulting in an approximate numerical value.

Step 1: Calculate the dot product of u and v

The dot product of two vectors \(\mathbf{u} = a_{1}\mathbf{i} + a_{2}\mathbf{j}\) and \(\mathbf{v} = b_{1}\mathbf{i} + b_{2}\mathbf{j}\) is given by the formula \(\mathbf{u} \cdot \mathbf{v} = a_{1}b_{1} + a_{2}b_{2}\). Applying this to the given vectors, we get \(\mathbf{u} \cdot \mathbf{v} = 2(6) - 1(4) = 8\).

Step 2: Find the magnitudes of u and v

The magnitude of a vector \(\mathbf{w} = c_{1}\mathbf{i} + c_{2}\mathbf{j}\) is given by the formula \(||\mathbf{w}|| = \sqrt{c_{1}^{2} + c_{2}^{2}}\). Applying this to \(\mathbf{u}\) and \(\mathbf{v}\), we get \(||\mathbf{u}|| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\) and \(||\mathbf{v}|| = \sqrt{6^{2} + 4^{2}} = \sqrt{52}\).

Step 3: Substitute in the cosine theta formula

We can substitute the values obtained in steps 1 and 2 into the formula \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||}\) to get the cos of the angle. Hence, \(\cos(\theta) = \frac{8}{\sqrt{5}\sqrt{52}}\). Now, to find the angle, calculate the cosine inverse of the obtained value, i.e., \(\theta = \cos^{-1}(\frac{8}{\sqrt{5}\sqrt{52}})\).

Step 4: Calculate the angle

Finally, calculating the cosine inverse from the previous step, we obtain the angle between the two vectors. Note that the angle must be calculated in degrees.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a crucial operation in vector algebra that combines two vectors to yield a scalar quantity. It's calculated by multiplying corresponding components of the vectors and then summing those products.

For example, if we have vectors \(\mathbf{u} = a_{1}\mathbf{i} + a_{2}\mathbf{j}\) and \(\mathbf{v} = b_{1}\mathbf{i} + b_{2}\mathbf{j}\), the dot product is given by \(\mathbf{u} \cdot \mathbf{v} = a_{1}b_{1} + a_{2}b_{2}\). This concept plays an integral role in determining the angle between two vectors as it reflects how much one vector extends in the direction of another.

Vector Magnitude

The magnitude of a vector—often thought of as its 'length'—measures how long the vector is. For a two-dimensional vector expressed as \(\mathbf{w} = c_{1}\mathbf{i} + c_{2}\mathbf{j}\), the magnitude is calculated using the Pythagorean theorem which results in the formula \(||\mathbf{w}|| = \sqrt{c_{1}^{2} + c_{2}^{2}}\).

In practical terms, if you picture a vector as an arrow, the magnitude is the distance from the tail of the arrow to its point. Knowing the magnitude of vectors is essential when calculating the angle between them, as it's part of the formula used to determine this angle.

Cosine Theta Formula

The cosine theta formula is the key to finding the angle between two vectors. It ties together their dot product and magnitudes. The formula states that \(\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||}\), where \(\theta\) is the angle formed between the two vectors.

By rearranging this formula, we can solve for \(\theta\) when given the vectors' components. It's a way of measuring how 'aligned' two vectors are: a cosine of 1 means they are perfectly aligned, 0 means they are perpendicular, and -1 means they are oppositely aligned. This formula is particularly helpful in physics and engineering, where angles between directions of forces, velocities, or fields are crucial.

Inverse Trigonometric Functions

Once we have the cosine of an angle from the cosine theta formula, we use inverse trigonometric functions to find the angle itself. For instance, the inverse cosine function, denoted as \(\cos^{-1}\) or arccosine, reverses what the cosine function does. If we have the value of \(\cos(\theta)\), we apply the inverse to get \(\theta = \cos^{-1}(\cos(\theta))\).

It is worth noting that inverse trigonometric functions can return angles in different units, such as radians or degrees. In the context of our problem, we are particularly interested in degrees because it is the most common unit for measuring angles in everyday scenarios. Inverse trigonometric functions are vital in fields that require angle computation, such as navigation, architecture, and computer graphics.