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Find the magnitude and direction angle for each vector. $$\langle 8 \sqrt{2},-8 \sqrt{2}\rangle$$

Short Answer

Expert verified
Magnitude: 16, Direction Angle: 315° or -45°

Step by step solution

01

- Calculate the Magnitude

To find the magnitude of the vector, use the formula: \[ ||\textbf{v}|| = \sqrt{(x^2 + y^2)} \] Here, \( x = 8 \sqrt{2} \) and \( y = -8 \sqrt{2} \). Substitute these values into the formula: \[ ||\textbf{v}|| = \sqrt{(8 \sqrt{2})^2 + (-8 \sqrt{2})^2} \] Simplify the expression: \[ ||\textbf{v}|| = \sqrt{(64 \cdot 2) + (64 \cdot 2)} \ ||\textbf{v}|| = \sqrt{128 + 128} \ ||\textbf{v}|| = \sqrt{256} \ ||\textbf{v}|| = 16 \]
02

- Calculate the Direction Angle

To find the direction angle of the vector, use the formula: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]where \( x = 8 \sqrt{2} \) and \( y = -8 \sqrt{2} \). Substitute these values into the formula: \[ \theta = \tan^{-1}\left(\frac{-8 \sqrt{2}}{8 \sqrt{2}}\right) \] Simplify the expression: \[ \theta = \tan^{-1}(-1) \] Find the angle where the tangent is -1, which is: \[ \theta = -45^\circ \] or equivalently \[ 315^\circ \] So, the direction angle \( \theta \) is \( 315^\circ \) or \( -45^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Magnitude
To understand vector magnitude, think of it as the length of the vector. Imagine you're drawing a straight line from one point to another in a coordinate plane. Magnitude tells you how long that straight line is. To find it, use the formula:

\[ ||\textbf{v}|| = \sqrt{(x^2 + y^2)} \]

Here, \(x\) and \(y\) are the components of the vector. For our vector \( \langle 8 \sqrt{2}, -8 \sqrt{2} \rangle \), we substitute these values into our formula:

\[ ||\textbf{v}|| = \sqrt{(8 \sqrt{2})^2 + (-8 \sqrt{2})^2} \]

This simplifies to:

\[ ||\textbf{v}|| = \sqrt{128 + 128} = \sqrt{256} = 16 \]

The magnitude of this vector is 16. This means the 'length' of the line we drew from the origin to the point is 16 units long.
Direction Angle
The direction angle of a vector tells us the exact angle it makes with the positive x-axis. It's like knowing which direction your vector is pointing. For our vector, we use the formula for direction angle:

\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]

Substituting the values, we get:

\[ \theta = \tan^{-1}\left(\frac{-8 \sqrt{2}}{8 \sqrt{2}}\right) \]
\[ \theta = \tan^{-1}(-1) \]

The arctangent of -1 corresponds to an angle of \(-45^\circ\) or 315°. The angle \(-45^\circ\) shows the vector pointing downwards and to the right. While 315° is the same angle measured clockwise from the positive x-axis. Both angles are equivalent and tell us the exact pointing direction of our vector.
Trigonometry
Trigonometry plays a crucial role in understanding vectors, as it deals with the relationships between the angles and sides of triangles. When dealing with vectors, common trigonometric functions like sine, cosine, and tangent help us find direction and magnitude.

- Tangent (\tan \theta) relates the opposite and adjacent sides of a right triangle.
- Sine (\sin\theta) relates the opposite side to the hypotenuse.
- Cosine (\cos\theta) relates the adjacent side to the hypotenuse.

For the direction angle formula:

\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]

We solved for the angle using the inverse tangent function. This is a great example of how trigonometry helps us determine the angle of direction for a vector, simplifying the process of understanding its orientation in the coordinate plane.

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