Problem 7 - Precalculus

Question

Find the magnitude and direction angle of the given vector. $$\mathbf{u}=\langle 3,8\rangle$$

Step-by-Step Solution

Answer

Magnitude is \( \sqrt{73} \), and the direction angle is approximately \( 69.44^\circ \).

Step 1: Understanding the Vector

We have a vector \( \mathbf{u} = \langle 3, 8 \rangle \), which means it has an x-component of 3 and a y-component of 8. Our goal is to find the magnitude and the direction angle of this vector.

Step 2: Calculate the Magnitude

The magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \) is given by the formula:\[||\mathbf{u}|| = \sqrt{a^2 + b^2}\]Substitute the x and y components of the vector:\[||\mathbf{u}|| = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}\]Thus, the magnitude of the vector is \( \sqrt{73} \).

Step 3: Calculate the Direction Angle

The direction angle \( \theta \) for a vector \( \mathbf{u} = \langle a, b \rangle \) is determined using the tangent function:\[\tan \theta = \frac{b}{a}\]Substitute in our values:\[\tan \theta = \frac{8}{3}\]To find \( \theta \), take the inverse tangent:\[\theta = \tan^{-1}\left(\frac{8}{3}\right)\]Using a calculator, \( \theta \approx 69.44^\circ \). Thus, the direction angle of the vector is approximately \( 69.44^\circ \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector gives us the length or size of the vector. This is an important concept because it tells us how long the vector is, similar to measuring the length of a line segment. When we are given a vector, such as \( \mathbf{u} = \langle 3, 8 \rangle \), it has two components: one in the x-direction (3) and one in the y-direction (8).
To find the magnitude, we use the formula:

• \( ||\mathbf{u}|| = \sqrt{a^2 + b^2} \)

In this equation, \( a \) and \( b \) are the components of the vector, which are 3 and 8 in our example. When you substitute these values into the formula, you calculate:

• \( ||\mathbf{u}|| = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73} \)

Therefore, the magnitude of the vector \( \mathbf{u} \) is \( \sqrt{73} \). This tells us that the vector is \( \sqrt{73} \) units long.

Direction Angle of a Vector

The direction angle of a vector provides the angle the vector makes with the positive x-axis. This is a crucial piece of information because it describes the vector's direction.For the vector \( \mathbf{u} = \langle 3, 8 \rangle \), to find this angle, we use the tangent function. The formula to find the direction angle \( \theta \) is:

• \( \tan \theta = \frac{b}{a} \)

Here, \( a = 3 \) and \( b = 8 \) are the components of the vector. Substitute these values:

• \( \tan \theta = \frac{8}{3} \)

To find the angle \( \theta \), we need to solve for \( \theta \) by taking the inverse tangent. This gives us:

• \( \theta = \tan^{-1}\left(\frac{8}{3}\right) \)

Using a calculator, \( \theta \approx 69.44^\circ \). Thus, the direction angle of the vector \( \mathbf{u} \) is approximately \( 69.44^\circ \), meaning it points in this direction relative to the positive x-axis.

Inverse Tangent Function

The inverse tangent function, also written as \( \tan^{-1} \) or \( \arctan \), is a valuable tool in finding angles when the tangent value is known. This happens specifically when you know the ratio of the opposite side to the adjacent side of a right triangle, which is the definition of the tangent in trigonometry.
In vector calculations, the inverse tangent helps determine the direction angle of the vector. If you have a vector \( \mathbf{u} = \langle a, b \rangle \), the direction angle \( \theta \) can be found using:

• \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)

This function finds the angle \( \theta \) whose tangent is \( \frac{b}{a} \). For example, if a vector has components 3 and 8, as in \( \tan \theta = \frac{8}{3} \), you apply the inverse tangent to get:

• \( \theta = \tan^{-1}\left(\frac{8}{3}\right) \)

Using a calculator, this results in an angle of approximately \( 69.44^\circ \). This measurement helps understand in which direction the vector is pointing, relative to a flat horizonal line, or the positive x-axis.