Problem 38 - Contemporary Precalculus

Question

Find the magnitude and direction angle of the vector \(\boldsymbol{v}\). $$\mathbf{v}=\langle 4,5\rangle$$

Step-by-Step Solution

Answer

Answer: The magnitude of the vector \(\boldsymbol{v}\) is \(\sqrt{41}\), and its direction angle is approximately \(51.34^{\circ}\).

Step 1: Calculate the magnitude of the vector

To find the magnitude of the vector \(\boldsymbol{v}=\langle 4,5\rangle\), we can use the Pythagorean theorem: $$ \begin{aligned} ||\boldsymbol{v}|| &= \sqrt{(4)^2 + (5)^2} \\ &= \sqrt{16 + 25} \\ &= \sqrt{41} \end{aligned} $$ So, the magnitude of the vector \(\boldsymbol{v}\) is \(\sqrt{41}\).

Step 2: Find the direction angle of the vector

In order to find the direction angle \(\theta\) of the vector \(\boldsymbol{v}\), we can use the tangent function since we know the components of the vector: $$\tan(\theta) = \frac{5}{4}$$ To find the angle, we will use the inverse tangent function (or arctangent), denoted by \(\tan^{-1}\): $$\theta = \tan^{-1}\left(\frac{5}{4}\right)$$ Now we can use a calculator to find the arctangent of \(\frac{5}{4}\), to get: $$\theta \approx 51.34^{\circ}$$ The direction angle of the vector \(\boldsymbol{v}\) is approximately \(51.34^{\circ}\). To summarize, the magnitude of vector \(\boldsymbol{v}=\langle 4,5\rangle\) is \(\sqrt{41}\), and its direction angle is approximately \(51.34^{\circ}\).

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle used to calculate the magnitude (or length) of a vector. When you have a vector like \( \mathbf{v} = \langle 4,5 \rangle \), imagine it as forming a right triangle. The components \(4\) and \(5\) can be thought of as the lengths of the triangle's legs. The magnitude of the vector is the hypotenuse.
Here's how it works: the Pythagorean theorem states that the square of the hypotenuse (\(c\)) is equal to the sum of the squares of the legs (\(a\) and \(b\)). Mathematically, it's expressed as:

• \(c^2 = a^2 + b^2\)

Using the theorem, you can calculate the vector magnitude \(||\mathbf{v}||\) as follows:

• \(||\mathbf{v}|| = \sqrt{(4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41}\)

This method is excellent for finding the actual length of any vector in a 2D space, using its horizontal and vertical components. It essentially helps you 'measure' the vector.

Direction angle

The direction angle of a vector tells you the direction in which the vector is pointing. For a vector \( \mathbf{v} = \langle 4, 5 \rangle \), we need to determine this angle with respect to the positive x-axis.
The direction of a vector in two dimensions is given by the angle \(\theta\), which can be calculated using the relationship provided by the tangent function:

• \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)
• Here, 'opposite' refers to the y-component (5), and 'adjacent' refers to the x-component (4).

By knowing the direction angle, you can understand how the vector orientates itself around a coordinate plane. This is useful in both physics and engineering for determining the influence of forces.

Inverse tangent function

The inverse tangent function, also known as arctangent, is key for finding the direction angle of a vector. After determining the ratio of the vector components using tangent, you apply the inverse tangent to find the actual angle.
In our example of vector \( \mathbf{v} = \langle 4, 5 \rangle \):

• We have \(\tan(\theta) = \frac{5}{4}\).
• Applying the inverse tangent function gives us \(\theta = \tan^{-1}\left(\frac{5}{4}\right)\).

This function does the reverse of the tangent function: while tangent gives you a ratio from an angle, inverse tangent provides you the angle from a ratio. After calculation, this results in a direction angle, \(\theta \approx 51.34^{\circ}\), placing our vector in the first quadrant of the coordinate plane. Understanding the inverse tangent function is crucial for translating between geometric figures and numerical values.