Problem 35 - Contemporary Precalculus

Question

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

Step-by-Step Solution

Answer

Answer: The magnitude of the vector is \(4\sqrt{2}\), and the direction angle is \(45^\circ\).

Step 1: Find the magnitude of the vector

To find the magnitude of the vector \(\boldsymbol{v}=\langle 4,4\rangle\), we can use the Pythagorean theorem. The magnitude is the length of the vector, denoted by \(|\boldsymbol{v}|\), and can be found using the formula: $$|\boldsymbol{v}| = \sqrt{x^2 + y^2}$$ where \(x\) is the x-component of the vector and \(y\) is the y-component. For this vector, we have \(x=4\) and \(y=4\).

Step 2: Calculate the magnitude of the vector

Using the formula from step 1, and substituting the x and y components, we obtain: $$|\boldsymbol{v}| = \sqrt{4^2 + 4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$$ So, the magnitude of the vector \(\boldsymbol{v}=\langle 4,4\rangle\) is \(4\sqrt{2}\).

Step 3: Find the direction angle of the vector

To find the direction angle of the vector (let's call it \(\theta\)), we can use the tangent inverse function (arctan) applied to the ratio of the y-component to the x-component of the vector. For this vector, both components are equal (x=y=4), so we can use the formula: $$\theta = \arctan\left(\frac{y}{x}\right)$$

Step 4: Calculate the direction angle of the vector

Using the formula from step 3, and substituting the x and y components, we obtain: $$\theta = \arctan\left(\frac{4}{4}\right) = \arctan(1)$$ Now using a calculator or a reference table, we find that \(\arctan(1)\) is equal to \(45^\circ\). Therefore, the direction angle of the vector \(\boldsymbol{v}=\langle 4,4\rangle\) is \(45^\circ\). So, the magnitude and direction angle of the vector \(\boldsymbol{v}=\langle 4,4\rangle\) are \(4\sqrt{2}\) and \(45^\circ\), respectively.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that relates the lengths of the sides of a right-angled triangle. In the context of vectors, it is instrumental in calculating the magnitude of a vector. The formula for the theorem is \(a^2 + b^2 = c^2\).However, when dealing with vectors, this theorem is adapted to find the magnitude of a vector \(\mathbf{v} = \langle x, y \rangle\). The magnitude is essentially the length of the vector from the origin to the point \(\langle x, y \rangle\). You can calculate it by using:

• \(|\mathbf{v}| = \sqrt{x^2 + y^2}\)

In our example, both the x and y components are 4. So, \(x^2 = 16\) and \(y^2 = 16\). Add them to get 32, and then take the square root to find the magnitude, \(\sqrt{32} = 4\sqrt{2}\).
This use of the Pythagorean theorem helps determine how far a vector stretches in a two-dimensional plane.

Vector Components

Understanding vector components is key in breaking down a vector into its horizontal and vertical parts. Think of a vector as an arrow. In two dimensions, each vector can be broken down into two perpendicular parts: the x-component and the y-component.For the vector \(\mathbf{v} = \langle x, y \rangle\), \(

• The x-component tells us how far the vector stretches along the horizontal axis.
• The y-component represents the length along the vertical axis.

\)In practical terms, if you overlay this vector onto a graph:

• Start at the origin and move 4 units right.
• Next, move 4 units up.

These movements present the vector directionally on the graph. Each part is like the ingredients of the vector's resulting direction and length. By knowing these components, you can better understand and calculate the vector's magnitude using the Pythagorean theorem.

Inverse Tangent Function

The inverse tangent function, often called arctan, helps find the angle a vector makes with the horizontal axis. The direction angle, \(\theta\), is a measure of this slant.To locate this angle for the vector \(\mathbf{v} = \langle 4, 4 \rangle\), use the formula:

• \(\theta = \arctan\left(\frac{y}{x}\right)\)

Since both the x and y components are 4 here, the ratio simplifies to 1. Thus, you compute:

• \(\theta = \arctan(1)\)

In this case, \(\arctan(1)\) gives an angle of \(45^\circ\). This tells us the vector points halfway between the x and y axes, forming an equal angle with both. The inverse tangent function is a handy tool for turning vector components into comprehensible angles, making it easier to visualize the vector's direction in a plane.