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Chapter 9: Problem 46

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=\sqrt{3}, \quad \theta=300^{\circ}$$

Short Answer

Expert verified
The vector is \(\frac{\sqrt{3}}{2}\mathbf{i} - \frac{3}{2}\mathbf{j}\).

Step by step solution

01

Convert the Angle

First, we need to convert the given angle from degrees to radians. Since there are \(360^\circ\) in a circle and \(2\pi\) radians, we convert \(\theta = 300^\circ\) to radians by using the formula:\[\theta_{\text{radians}} = \frac{300 \times \pi}{180} = \frac{5\pi}{3}\]
02

Determine the Horizontal Component

The horizontal component of the vector can be found using the cosine function. If the angle \(\theta\) is with respect to the positive x-axis, the formula is:\[v_x = |\mathbf{v}| \cdot \cos(\theta) = \sqrt{3} \cdot \cos\left(\frac{5\pi}{3}\right)\]Knowing that \(\cos\left(300^\circ\right) = \frac{1}{2}\), the horizontal component is:\[v_x = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}\]
03

Determine the Vertical Component

The vertical component is determined using the sine function. The formula is:\[v_y = |\mathbf{v}| \cdot \sin(\theta) = \sqrt{3} \cdot \sin\left(\frac{5\pi}{3}\right)\]Knowing that \(\sin\left(300^\circ\right) = -\frac{\sqrt{3}}{2}\), the vertical component is:\[v_y = \sqrt{3} \cdot -\frac{\sqrt{3}}{2} = -\frac{3}{2}\]
04

Express the Vector in Terms of i and j

The vector \(\mathbf{v}\) can be written using the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\), which represent the horizontal and vertical components, respectively. Thus, the vector is given by:\[\mathbf{v} = v_x \cdot \mathbf{i} + v_y \cdot \mathbf{j} = \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j}\]
05

Final Step: Write the Complete Vector

The vector using components \(\mathbf{i}\) and \(\mathbf{j}\) is:\[\mathbf{v} = \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j}\]

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Key Concepts

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

Understanding the Horizontal Component
The horizontal component of a vector represents its projection along the x-axis. Imagine you are shining a light on the vector and it casts a shadow on the horizontal line; that shadow is the horizontal component. In mathematical terms, this component can be found using the cosine function, which relates the angle of the vector to its horizontal projection.

When you have a vector with a known angle \( \theta \) and magnitude \( |\mathbf{v}| \), you employ the formula:
  • \( v_x = |\mathbf{v}| \cdot \cos(\theta) \)
This formula tells you how much of the vector's length is in the horizontal direction.

Taking our example from the exercise, where the angle \( \theta = 300^{\circ} \), the horizontal component is calculated using \( \cos(300^{\circ}) = \frac{1}{2} \). This results in \( v_x = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \). So, the horizontal component describes how far the vector extends across the horizontal axis.
Breaking Down the Vertical Component
The vertical component of a vector is akin to its shadow on the vertical axis, revealing the vector鈥檚 influence in the up and down direction. This component is crucial when the vector extends in a direction other than purely horizontal or along the diagonal.

To determine the vertical component, the sine function is used. The formula is quite similar to that used in finding the horizontal component:
  • \( v_y = |\mathbf{v}| \cdot \sin(\theta) \)
This formula gives us the segment of the vector that is moving up or down.

Using the earlier context with \( \theta = 300^{\circ} \) and knowing \( \sin(300^{\circ}) = -\frac{\sqrt{3}}{2} \), we find the vertical component: \( v_y = \sqrt{3} \cdot -\frac{\sqrt{3}}{2} = -\frac{3}{2} \). Therefore, the vertical component is negative here, indicating that it moves downward, below the x-axis.
Utilizing Unit Vectors i and j
Unit vectors are incredibly useful in vector mathematics. They provide a standardized way to express vectors in a plane, using components of length 1 in specific directions. The vector \( \mathbf{i} \) refers to the unit vector along the x-axis (horizontal), while \( \mathbf{j} \) refers to the unit vector along the y-axis (vertical).

By using unit vectors, any vector \( \mathbf{v} \) can be expressed as a combination of \( \mathbf{i} \) and \( \mathbf{j} \). The general expression is:
  • \( \mathbf{v} = v_x \cdot \mathbf{i} + v_y \cdot \mathbf{j} \)
This form is straightforward and shows the horizontal and vertical components clearly.

For the specific vector from the exercise, the components are \( v_x = \frac{\sqrt{3}}{2} \) and \( v_y = -\frac{3}{2} \). Thus, the vector \( \mathbf{v} \) is expressed as \( \mathbf{v} = \frac{\sqrt{3}}{2} \mathbf{i} - \frac{3}{2} \mathbf{j} \). This notation provides an intuitive understanding of the vector's direction and magnitude in relation to the axes.

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Most popular questions from this chapter

Show that the vectors proj, \(\mathbf{u}\) and \(\mathbf{u}-\) proj, \(\mathbf{u}\) are orthogonal.

Let a \(=\langle 2,2,2\rangle\) \(\mathbf{b}=\langle- 2,-2,0\rangle,\) and \(\mathbf{r}=\langle x, y, z\rangle\) (a) Show that the vector equation \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, by expanding the dot product and simplifying the resulting algebraic equation. (b) Find the center and radius of the sphere. (c) Interpret the result of part (a) geometrically, using the fact that the dot product of two vectors is 0 only if the vectors are perpendicular. endpoints of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{r},\) noting that the end. points of a and b are the endpoints of a diameter and the endpoint of \(\mathbf{r} \text { is an arbitrary point on the sphere. }]\) (d) Using your observations from part (a), find a vector equation for the sphere in which the points \((0,1,3)\) and \((2,-1,4)\) form the endpoints of a diameter. Simplify the vector equation to obtain an algebraic equation for the sphere. What are its center and radius?

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\langle 3,4\rangle$$

A tetrahedron is a solid with four triangular faces, four vertices, and six edges, as shown in the figure. In a regular tetrahedron, the edges are all of the same length. Consider the tetrahedron with vertices \(A(1,0,0), B(0,1,0), C(0,0,1),\) and \(D(1,1,1)\) (a) Show that the tetrahedron is regular. (b) The center of the tetrahedron is the point \(E\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\) (the "average" of the vertices). Find the angle between the vectors that join the center to any two of the vertices (for instance, \(\langle A E B\) ). This angle is called the central angle of the tetrahedron. NOTE: In a molecule of methane \(\left(\mathrm{CH}_{4}\right)\) the four hydrogen atoms form the vertices of a regular tetrahedron with the carbon atom at the center. In this case chemists refer to the central angle as the bond angle. In the figure, the tetrahedron in the exercise is shown, with the vertices labeled \(H\) for hydrogen, and the center labeled \(C\) for carbon. (figure cannot copy)

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=\langle 10,-1\rangle, \quad \mathbf{v}=\langle- 2,-2\rangle$$
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