Chapter 7: Problem 16
Find the magnitude and direction angle of the given vector. $$\mathbf{u}=\langle-5,-5\rangle$$
Short Answer
Expert verified
Magnitude: \( 5\sqrt{2} \), Direction: \( 225^\circ \).
Step by step solution
01
Understanding the Problem
We need to find the magnitude and direction of the vector \( \mathbf{u} = \langle -5, -5 \rangle \). The magnitude is the length of the vector, while the direction angle is the angle it makes with the positive x-axis.
02
Calculate the Magnitude
To find the magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \), use the formula: \[ \| \mathbf{u} \| = \sqrt{a^2 + b^2} \]Substitute \( a = -5 \) and \( b = -5 \):\[ \| \mathbf{u} \| = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]
03
Calculate the Direction Angle
The direction angle \( \theta \) of a vector \( \mathbf{u} = \langle a, b \rangle \) is calculated using the tangent function: \[ \theta = \tan^{-1} \left( \frac{b}{a} \right) \]Since this vector is in the third quadrant, we find:\[ \theta = \tan^{-1} \left( \frac{-5}{-5} \right) = \tan^{-1}(1) = 45^\circ \]However, in the third quadrant, the angle relative to the positive x-axis is \( 180^\circ + 45^\circ = 225^\circ \).
04
Final Commentary
The vector \( \mathbf{u} = \langle -5, -5 \rangle \) has a magnitude of \( 5\sqrt{2} \) and a direction angle of \( 225^\circ \) measured counterclockwise from the positive x-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnitude of a Vector
Vectors have both magnitude and direction. The magnitude of a vector, sometimes referred to as its 'length', quantifies how long or large the vector is.
For a vector \( \mathbf{u} = \langle a, b \rangle \), its magnitude is calculated using the formula:
In the case of the vector \( \mathbf{u} = \langle -5, -5 \rangle \), substituting these values into the formula gives us \( \| \mathbf{u} \| = \sqrt{(-5)^2 + (-5)^2} = 5\sqrt{2} \).
Therefore, the vector's magnitude is \( 5\sqrt{2} \). This tells us how far the vector extends in space from its initial point.
For a vector \( \mathbf{u} = \langle a, b \rangle \), its magnitude is calculated using the formula:
- \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \)
In the case of the vector \( \mathbf{u} = \langle -5, -5 \rangle \), substituting these values into the formula gives us \( \| \mathbf{u} \| = \sqrt{(-5)^2 + (-5)^2} = 5\sqrt{2} \).
Therefore, the vector's magnitude is \( 5\sqrt{2} \). This tells us how far the vector extends in space from its initial point.
Direction Angle
The direction angle of a vector is crucial as it describes the vector's orientation. Calculating the direction angle involves the tangent function.
The direction angle \( \theta \) of a vector \( \mathbf{u} = \langle a, b \rangle \) is often found using:
For the vector \( \mathbf{u} = \langle -5, -5 \rangle \), we calculate \( \theta = \tan^{-1}(1) = 45^\circ \). However, because vectors in the third quadrant (where both \( a \) and \( b \) are negative) point away from the positive axes, the actual angle is \( 225^\circ \).
This means the vector points further than halfway around the circle.
The direction angle \( \theta \) of a vector \( \mathbf{u} = \langle a, b \rangle \) is often found using:
- \( \theta = \tan^{-1} \left( \frac{b}{a} \right) \)
For the vector \( \mathbf{u} = \langle -5, -5 \rangle \), we calculate \( \theta = \tan^{-1}(1) = 45^\circ \). However, because vectors in the third quadrant (where both \( a \) and \( b \) are negative) point away from the positive axes, the actual angle is \( 225^\circ \).
This means the vector points further than halfway around the circle.
Third Quadrant Vector
Vectors can lie in any of the four quadrants on a Cartesian coordinate plane. These quadrants determine specific characteristics of vectors' direction angles.
The third quadrant, where our vector \( \mathbf{u} = \langle -5, -5 \rangle \) is located, is defined by both x and y components being negative. This information alters how we interpret the direction angle.
For vectors in the third quadrant, the angle computed from tangent is adjusted by adding \( 180^\circ \) because they lie beyond the second quadrant.
Thus, for our vector, while the angle based on tangent is \( 45^\circ \), the actual direction angle is \( 225^\circ \) after adding these additional degrees.
This adjustment ensures the angle aligns with the vector's true path from the origin.
The third quadrant, where our vector \( \mathbf{u} = \langle -5, -5 \rangle \) is located, is defined by both x and y components being negative. This information alters how we interpret the direction angle.
For vectors in the third quadrant, the angle computed from tangent is adjusted by adding \( 180^\circ \) because they lie beyond the second quadrant.
Thus, for our vector, while the angle based on tangent is \( 45^\circ \), the actual direction angle is \( 225^\circ \) after adding these additional degrees.
This adjustment ensures the angle aligns with the vector's true path from the origin.
Tangent Function
The tangent function is a fundamental trigonometric function that relates angles to side ratios in right triangles.
It's crucial in vector analysis for finding the direction angle, as it provides a relationship between the opposite (y) and adjacent (x) components.
The formula \( \tan \theta = \frac{b}{a} \) is used to derive \( \theta \). This equation tells us that \( \theta \) is the angle where its tangent equals the ratio of \( b \) (the y-component) to \( a \) (the x-component).
For example, in our vector \( \mathbf{u} = \langle -5, -5 \rangle \), the tangent function gives \( \tan^{-1}(1) = 45^\circ \). Yet, as previously noted, this must be corrected for the vector's position in the third quadrant, resulting in \( 225^\circ \).
The tangent function, thus, is instrumental in vector direction calculations, providing a bridge between algebraic expressions and geometric direction.
It's crucial in vector analysis for finding the direction angle, as it provides a relationship between the opposite (y) and adjacent (x) components.
The formula \( \tan \theta = \frac{b}{a} \) is used to derive \( \theta \). This equation tells us that \( \theta \) is the angle where its tangent equals the ratio of \( b \) (the y-component) to \( a \) (the x-component).
For example, in our vector \( \mathbf{u} = \langle -5, -5 \rangle \), the tangent function gives \( \tan^{-1}(1) = 45^\circ \). Yet, as previously noted, this must be corrected for the vector's position in the third quadrant, resulting in \( 225^\circ \).
The tangent function, thus, is instrumental in vector direction calculations, providing a bridge between algebraic expressions and geometric direction.