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Find the magnitude of each vector and the angle \(\theta, 0^{\circ} \leq \theta<360^{\circ}\), that the vector makes with the positive \(x\)-axis. $$\mathbf{U}=\langle 3,3\rangle$$

Short Answer

Expert verified
The magnitude is \( 3\sqrt{2} \) and the angle \( \theta \) is \( 45^{\circ} \).

Step by step solution

01

Calculating the Magnitude

The magnitude of a vector \( \mathbf{U} = \langle a, b \rangle \) can be calculated using the formula: \( \| \mathbf{U} \| = \sqrt{a^2 + b^2} \). Here, \( a = 3 \) and \( b = 3 \), so we have: \[ \| \mathbf{U} \| = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}. \]
02

Finding the Angle with the Positive X-Axis

To find the angle \( \theta \) that \( \mathbf{U} \) makes with the positive \( x \)-axis, we can use the inverse tangent function. The formula is: \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \), where \( a = 3 \) and \( b = 3 \). So, \( \theta = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}(1) = 45^{\circ}. \) Since both components of the vector are positive, the vector is in the first quadrant, and \( \theta = 45^{\circ} \) is already the correct angle.

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Key Concepts

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

Magnitude of a Vector
The magnitude of a vector is similar to the length of a line segment. Imagine you are measuring a straight path between two points. The magnitude tells you how long this path is.
In mathematics, especially when dealing with vectors, the magnitude is a crucial concept. If you are given a vector \( \mathbf{U} = \langle a, b \rangle \), its magnitude \( \| \mathbf{U} \| \) is calculated with the formula \( \| \mathbf{U} \| = \sqrt{a^2 + b^2} \).
This formula comes from the Pythagorean theorem from geometry. It helps us find the length of the hypotenuse of a right triangle, where \( a \) and \( b \) are the triangle's sides.
  • **Example:** For the vector \( \mathbf{U} = \langle 3, 3 \rangle \), plug the values into the formula to get \( \| \mathbf{U} \| = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \).
    This gives us the vector's magnitude, which is \( 3\sqrt{2} \).
This measurement helps us understand just how long the vector is, giving us a clearer picture of its 'size' or 'strength' in mathematical terms.
Angle with X-Axis
Vectors not only have magnitude, but they also have direction. The direction of a vector is often described by the angle it makes with the positive x-axis.
Understanding this angle is crucial for positioning a vector correctly on a graph. The angle helps identify which quadrant the vector lies in
To find this angle \( \theta \), for a vector \( \mathbf{U} = \langle a, b \rangle \), you use \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
  • **Example:** With \( \mathbf{U} = \langle 3, 3 \rangle \), you calculate \( \theta = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}(1) = 45^{\circ} \).
    The result, 45 degrees, tells us that the vector is heading halfway between the x-axis and y-axis.
This means that \( \mathbf{U} \) points northeast from the origin. Any time \( a \) and \( b \) are equal and positive, the vector will make a 45-degree angle here in the first quadrant.
Inverse Tangent Function
The inverse tangent function, often represented as \( \tan^{-1} \) or \( \arctan \), is a great tool in trigonometry. Its job is to find the angle whose tangent is a given number.
This function is beneficial when you know a ratio of sides in a right triangle and want to figure out the angle related to that ratio.
  • **Formula:** When you need to find an angle \( \theta \), you can use \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
  • **Example Usage:** If you have \( a = 3 \) and \( b = 3 \) for the vector \( \mathbf{U} = \langle 3, 3 \rangle \), then \( \tan^{-1}(1) \) results in a 45-degree angle.
    This is because an equal ratio \( \frac{3}{3} = 1\) leads to an angle that bisects the angle between the x and y axes in a coordinate system.
Remember: This function only provides the principal value, often limited to the range \( -90^{\circ} \) to \( 90^{\circ} \) or \( 0^{\circ} \) to \( 180^{\circ} \) in some settings. Thus, recognizing the quadrant of your vector is essential to avoid errors using \( \tan^{-1} \).

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