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Chapter 6: Problem 12

Sketch each vector as a position vector and find its magnitude. $$\mathbf{v}=-5 \mathbf{j}$$

Short Answer

Expert verified
The position vector is a vertical arrow pointing downwards from the origin with a length proportional to 5. The magnitude of vector v is 5.

Step by step solution

01

Sketch the Vector v as a Position Vector

Position vector v represented by -5j is a vertical vector pointing downwards on the y-axis because -j represents the negative direction on the vertical axis. To plot this, you can start at the origin and draw a relatively long arrow pointing downwards. The length of the arrow should be proportional to the magnitude of the vector (5 in this case)
02

Compute the Magnitude of Vector v

The magnitude of a vector is its length. Since vector v is only in the j direction with no i component (x-direction component), the length of this vector corresponds to the absolute value of its scalar multiple, which is |-5| = 5

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Key Concepts

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

Position Vector
When talking about vectors in the context of position, a position vector specifically describes a point's location relative to an origin in a coordinate system. In our example, the vector \( \mathbf{v} = -5 \mathbf{j} \) is a position vector because it defines a point along the y-axis, starting from the origin and pointing downwards.
This downward direction comes from the vector's negative j-component. Plotting this position vector starts at the origin
  • The origin is typically represented by the coordinates (0,0) in a 2D space.
  • The vector direction is then determined by its components; here, \(-5j\) indicates movement exclusively along the negative y-axis.
For visualization, a position vector can effectively communicate both direction and magnitude, helping us interpret spatial data in a geometric way.
Magnitude of a Vector
The magnitude of a vector is essentially its length when represented in space. Calculating this is crucial for understanding the vector's scale and potential effect. In this exercise, despite \( \mathbf{v} = -5\mathbf{j} \) having only a component in one direction, its magnitude or length is still a straightforward computation.
To find it, we rely on the absolute value of its scalar multiplier:
  • Since the vector only stretches along the negative y-axis, its length is simply the absolute value of -5.
  • This results in a magnitude of 5, indicating the vector reaches a distance of 5 units from the origin.
It's important to understand that the magnitude is void of direction information, focusing solely on the vector's extent in the space.
Vector Components
Vector components are foundational in understanding the precise effect of a vector in a coordinate system. Each component represents projection or influence along corresponding axes.
In our case with \( \mathbf{v} = -5 \mathbf{j} \), here are the insights about its components:
  • The vector lacks an x-component (or i-component), indicating no horizontal shift from the origin, which simplifies the representation to vertical movement only.
  • The presence of only the j-component (-5) signifies downward movement along the y-axis, describing the vector's complete impact in two-dimensional space.
By understanding vector components, we can break down complex movements into simpler directional influences, facilitating problem-solving and analysis in physics and engineering contexts.

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Most popular questions from this chapter

The rectangular coordinates of a point are given. Find polar coordinates of each point. Express \(\theta\) in radians. $$(5,0)$$

Convert each rectangular equation to a polar equation that expresses \(r\) in terms of \(\theta\) $$x^{2}=6 y$$

Polar coordinates of a point are given. Find the rectangular coordinates of each point. $$\left(-4, \frac{\pi}{2}\right)$$

The rectangular coordinates of a point are given. Find polar coordinates of each point. Express \(\theta\) in radians. $$(2,-2)$$

Exercises \(112-114\) will help you prepare for the material covered in the next section. In each exercise, use a calculator to complete the table of coordinates. Where necessary, round to two decimal places. Then plot the resulting points, \((r, \theta),\) using a polar coordinate system. $$\begin{array}{c|c|c|c|c|c|c|c|c|c} \boldsymbol{\theta} & \boldsymbol{0} & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2 \pi}{3} & \frac{3 \pi}{4} & \frac{5 \pi}{6} & \pi \\ \hline r=4 \sin 2 \theta & & & & & & & \\ \hline \end{array}$$
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