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(a) In unit-vector notation, what is the sum \(\vec{a}+\vec{b}\) if \(\vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}+(3.0 \mathrm{~m}) \hat{\mathrm{k}}\) and \(\vec{b}=(-13.0 \mathrm{~m}) \hat{\mathrm{i}}+(7.0 \mathrm{~m}) \hat{\mathrm{k}} ?\) What are the (b) magnitude and (c) direction of \(\vec{a}+\vec{b}\) ?

Short Answer

Expert verified
\(\vec{a} + \vec{b} = (-9.0 \mathrm{\ m}) \hat{\mathrm{i}} + (10.0 \mathrm{\ m}) \hat{\mathrm{k}} \), with magnitude \( \approx 13.45 \mathrm{\ m} \) and direction \( \approx -48.01^\circ \).

Step by step solution

01

Identify the Given Vectors

We are given two vectors \( \vec{a} = (4.0 \mathrm{\ m}) \hat{\mathrm{i}} + (3.0 \mathrm{\ m}) \hat{\mathrm{k}} \) and \( \vec{b} = (-13.0 \mathrm{\ m}) \hat{\mathrm{i}} + (7.0 \mathrm{\ m}) \hat{\mathrm{k}} \). These vectors are expressed in unit-vector notation. We need to add these vectors first.
02

Calculate the Sum of the Vectors

Add the corresponding components of each vector: for the \( \hat{\mathrm{i}} \) components, \( 4.0 \mathrm{\ m} + (-13.0 \mathrm{\ m}) = -9.0 \mathrm{\ m} \). For the \( \hat{\mathrm{k}} \) components, \( 3.0 \mathrm{\ m} + 7.0 \mathrm{\ m} = 10.0 \mathrm{\ m} \). Thus, the sum vector \( \vec{a} + \vec{b} = (-9.0 \mathrm{\ m}) \hat{\mathrm{i}} + (10.0 \mathrm{\ m}) \hat{\mathrm{k}} \).
03

Determine the Magnitude of the Sum Vector

The magnitude of a vector \( \vec{v} = (v_x \hat{\mathrm{i}} + v_y \hat{\mathrm{k}}) \) is given by \( \|\vec{v}\| = \sqrt{v_x^2 + v_y^2} \). Substituting for \( \vec{a} + \vec{b} \), we get: \[ \| \vec{a} + \vec{b} \| = \sqrt{(-9.0)^2 + (10.0)^2} = \sqrt{81 + 100} = \sqrt{181} \approx 13.45 \mathrm{\ m}. \]
04

Find the Direction of the Sum Vector

The direction \( \theta \) of a vector in the \( \hat{\mathrm{i}}\)-\( \hat{\mathrm{k}} \) plane can be found using \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \). Performing the calculation: \( \theta = \tan^{-1}\left(\frac{10.0}{-9.0}\right) \approx \tan^{-1}(-1.11) \approx -48.01^\circ \). This angle is measured from the negative \( \hat{\mathrm{i}} \) axis towards the positive \( \hat{\mathrm{k}} \) axis.

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Key Concepts

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

Unit-Vector Notation
When dealing with vector addition, unit-vector notation is an efficient and precise way to express vectors. This method utilizes specific symbols for directions in three-dimensional space. The most common unit vectors are:
  • \( \hat{\mathrm{i}} \) for the x-axis direction
  • \( \hat{\mathrm{j}} \) for the y-axis direction
  • \( \hat{\mathrm{k}} \) for the z-axis direction
Using unit-vector notation, we can succinctly write vectors as combinations of these unit vectors with their respective components. Take the vectors \( \vec{a} \) and \( \vec{b} \) given in the exercise:
  • \( \vec{a} = (4.0 \mathrm{\ m}) \hat{\mathrm{i}} + (3.0 \mathrm{\ m}) \hat{\mathrm{k}} \)
  • \( \vec{b} = (-13.0 \mathrm{\ m}) \hat{\mathrm{i}} + (7.0 \mathrm{\ m}) \hat{\mathrm{k}} \)
Notice how each component is paired with a unit vector, indicating its direction along the axes. To add two vectors given in this form, simply add their corresponding components. This step is fundamental for combining vectors effectively and forms the basis for more complex vector manipulations.
Magnitude of a Vector
The magnitude of a vector represents its length or size and is an essential aspect of vector analysis. It provides information about how "long" a vector is in its given space, independent of its direction. For a vector \( \vec{v} = (v_x \hat{\mathrm{i}} + v_y \hat{\mathrm{k}}) \), the magnitude is calculated using the Pythagorean theorem as follows: \[ \|\vec{v}\| = \sqrt{v_x^2 + v_y^2} \]Applying this formula to the sum vector \( \vec{a} + \vec{b} = (-9.0 \mathrm{\ m}) \hat{\mathrm{i}} + (10.0 \mathrm{\ m}) \hat{\mathrm{k}} \), we find: \[\| \vec{a} + \vec{b} \| = \sqrt{(-9.0)^2 + (10.0)^2} = \sqrt{181} \approx 13.45 \mathrm{\ m}\]The magnitude is a key feature that helps in understanding the overall effect of vector quantities in physics and engineering. It's crucial particularly in cases where force, velocity, or any other vector quantity is analyzed for its impact.
Direction of a Vector
The direction of a vector is equally important as its magnitude because it tells us where the vector is headed in space. For two-dimensional vectors lying in the \( \hat{\mathrm{i}} \)-\( \hat{\mathrm{k}} \) plane, the direction can be found using the arctangent function: \[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\] Direction is measured as an angle, often in degrees. For the vector sum \( \vec{a} + \vec{b} \), we calculate:\[\theta = \tan^{-1}\left(\frac{10.0}{-9.0}\right) \approx \tan^{-1}(-1.11) \approx -48.01^\circ\]This angle is measured from the negative \( \hat{\mathrm{i}} \) axis toward the positive \( \hat{\mathrm{k}} \) direction. It is crucial to understand the role of direction to fully grasp vectors' influence in a given context. Remember, angles can be adjusted contextually, often needing interpretation as to where they are measured from.

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Most popular questions from this chapter

A particle undergoes three successive displacements in a plane, as follows: \(\vec{d}_{1}, 4.00 \mathrm{~m}\) southwest; then \(\vec{d}_{2}, 5.00 \mathrm{~m}\) east; and finally \(\vec{d}_{3}, 6.00 \mathrm{~m}\) in a direction \(60.0^{\circ}\) north of east. Choose a coordinate system with the \(y\) axis pointing north and the \(x\) axis pointing east. What are (a) the \(x\) component and (b) the \(y\) component of \(\vec{d}_{1}\) ? What are (c) the \(x\) component and (d) the \(y\) component of \(\overrightarrow{d_{2}}\) ? What are (e) the \(x\) component and (f) the \(y\) component of \(\vec{d}_{3} ?\) Next, consider the net displacement of the particle for the three successive displacements. What are (g) the \(x\) component, (h) the \(y\) component, (i) the magnitude, and (j) the direction of the net displacement? If the particle is to return directly to the starting point, (k) how far and (l) in what direction should it move?

A vector \(\vec{a}\) of magnitude 10 units and another vector \(\vec{b}\) of magnitude 6.0 units differ in directions by \(60^{\circ} .\) Find (a) the scalar product of the two vectors and (b) the magnitude of the vector product \(\vec{a} \times \vec{b}\)

Find (a) "north cross west," (b) "down dot south," (c) "east cross up," (d) "west dot west," and (e) "south cross south." Let each "vector" have unit magnitude.

If \(\vec{d}_{1}=3 \hat{i}-2 \hat{j}+4 \hat{k}, \vec{d}_{2}=-5 \hat{i}+2 \hat{j}-\hat{k},\) then what is \(\left(\vec{d}_{1}+\vec{d}_{2}\right) \cdot\left(\vec{d}_{1} \times 4 \vec{d}_{2}\right) ?\)

Let i be directed to the east, \(\hat{j}\) be directed to the north, and \(\hat{k}\) be directed upward. What are the values of products (a) \(\hat{\mathrm{i}} \cdot \hat{\mathrm{k}},(\mathrm{b})\) \((-\hat{\mathrm{k}}) \cdot(-\hat{\mathrm{j}}),\) and \((\mathrm{c}) \hat{\mathrm{j}} \cdot(-\hat{\mathrm{j}}) ?\) What are the directions (such as east or down of products \((\mathrm{d}) \hat{\mathrm{k}} \times \hat{\mathrm{j}},(\mathrm{e})(-\hat{\mathrm{i}}) \times(-\hat{\mathrm{j}}),\) and (f) \((-\hat{\mathrm{k}}) \times(-\hat{\mathrm{j}}) ?\)

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