/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Find the magnitude of the vector... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the magnitude of the vector a and the smallest positive angle \(\boldsymbol{\theta}\) from the positive \(\boldsymbol{x}\) -axis to the vector \(O P\) that corresponds to a. $$\mathbf{a}=\langle 0,-5\rangle$$

Short Answer

Expert verified
Magnitude is 5, angle is \(270^\circ\).

Step by step solution

01

Identifying the Components of the Vector

The vector \( \mathbf{a} \) has components \( \langle 0, -5 \rangle \). This means that the vector starts at the origin (0,0) and ends at the point (0,-5). The \(x\)-component is 0 and the \(y\)-component is -5.
02

Calculating the Magnitude of the Vector

The magnitude of a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is calculated using the formula \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \). Substitute \(a_1 = 0\) and \(a_2 = -5\) to find:\[\| \mathbf{a} \| = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5\]
03

Finding the Direction Angle

To find the smallest positive angle \( \theta \) from the positive \( x \)-axis, we use the tangent function such that \( \tan(\theta) = \frac{a_2}{a_1} \). Here, \(a_1 = 0\) and \(a_2 = -5\). Since the vector points directly downward, it is on the negative \(y\)-axis:\[\theta = 270^\circ\] because the vector is perpendicular to the \(x\)-axis and points in the downward direction.

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Key Concepts

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

Magnitude of a Vector
Understanding the magnitude of a vector is essential in vector mathematics. The magnitude, or length, of a vector represents how long the vector is in a geometric space. It's like finding the length of a segment in a line, but for vectors in a coordinate plane.
To calculate the magnitude of a vector with components \( \mathbf{a} = \langle a_1, a_2 \rangle \), you use the Pythagorean theorem. The formula is:
  • \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \)
For the vector \( \mathbf{a} = \langle 0, -5 \rangle \), we identify \( a_1 = 0 \) and \( a_2 = -5 \). Plugging these into the formula gives us:
  • \( \| \mathbf{a} \| = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5 \)
Thus, the magnitude of the vector is 5 units. This method works for any two-dimensional vector, helping to determine its length in the plane.
Direction Angle of a Vector
The direction angle of a vector is the angle it forms with the positive x-axis, measured counterclockwise. This angle helps understand the vector's orientation in space.
For a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), you can find this angle \( \theta \) using the tangent function:
  • \( \tan(\theta) = \frac{a_2}{a_1} \)
In the case of \( \mathbf{a} = \langle 0, -5 \rangle \), since \( a_1 = 0 \), the direction points perpendicular to the x-axis.
The positive x-axis direction when the vector points downwards is:
  • \( \theta = 270^\circ \)
This angle indicates the vector is directed downward along the negative y-axis.
Vector Components
A vector in two dimensions can be expressed with components, which are essentially the projections of the vector along the x- and y-axes.
For any vector \( \mathbf{a} \) expressed as \( \langle a_1, a_2 \rangle \):
  • \( a_1 \) is the x-component. It shows how far the vector moves horizontally.
  • \( a_2 \) is the y-component, indicating vertical movement.
For example, the vector \( \mathbf{a} = \langle 0, -5 \rangle \) has:
  • x-component = 0, signifying no horizontal change;
  • y-component = -5, meaning the vector moves 5 units in the negative y direction.
Understanding these components helps visualize and calculate other vector properties, such as direction and magnitude, easily.

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