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

Short Answer

Expert verified
The sum is \[ \vec{a} + \vec{b} = (-9.0 \text{ m}) \hat{i} + (10.0 \text{ m}) \hat{j} \] \[ |\vec{a} + \vec{b}| \approx 13.45 \text{ m} \] \[ \theta \approx 131.6^{\circ} \] relative to \(\hat{i}\).

Step by step solution

01

- Write the given vectors

The given vectors are: \[ \vec{a} = (4.0 \text{ m}) \hat{i} + (3.0 \text{ m}) \hat{j} \] \[ \vec{b} = (-13.0 \text{ m}) \hat{i} + (7.0 \text{ m}) \hat{j} \]
02

- Add the vectors

Add the corresponding components of \(\vec{a}\) and \(\vec{b}\): \[ \vec{a} + \vec{b} = \left( (4.0 \text{ m}) + (-13.0 \text{ m}) \right) \hat{i} + \left( (3.0 \text{ m}) + (7.0 \text{ m}) \right) \hat{j} \] This gives: \[ \vec{a} + \vec{b} = (-9.0 \text{ m}) \hat{i} + (10.0 \text{ m}) \hat{j} \]
03

- Calculate the magnitude

Use the Pythagorean theorem to find the magnitude of \(\vec{a} + \vec{b}\): \[ |\vec{a} + \vec{b}| = \sqrt{ (-9.0 \text{ m})^2 + (10.0 \text{ m})^2 } \] \[ = \sqrt{ 81 + 100 } \] \[ = \sqrt{ 181 } \text{ m} \] \[ \approx 13.45 \text{ m} \]
04

- Calculate the direction

Use the tangent function to find the direction of \(\vec{a} + \vec{b}\) relative to \(\hat{i}\): \[ \theta = \tan^{-1} \left( \frac{10.0 \text{ m}}{-9.0 \text{ m}} \right) \] \[ \theta = \tan^{-1} \left( -\frac{10}{9} \right) \] \[ \theta \approx -48.4^{\circ} \] The negative angle indicates the direction is below the x-axis. To express as an angle from the positive x-axis: \[ \theta = 180^{\circ} - 48.4^{\circ} \] \[ \theta \approx 131.6^{\circ} \]

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Key Concepts

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

Unit Vectors
Unit vectors are basic building blocks in vector mathematics. They have a magnitude of 1 and indicate direction along a specific axis.
In 2D space, you often see two unit vectors: \(\hat{i}\) for the x-axis and \(\hat{j}\) for the y-axis.
  • \(\hat{i}\) points in the horizontal direction.
  • \(\hat{j}\) points in the vertical direction.
For example, a vector \(\vec{a} = 4\hat{i} + 3\hat{j}\) means it moves 4 units in the x-direction and 3 units in the y-direction. By using unit vectors, describing and manipulating vectors becomes straightforward and systematic.
Magnitude Calculation
Magnitude represents the length or size of a vector.
To calculate the magnitude of a vector, you can use the Pythagorean theorem.
For a vector \(\vec{v} = a\hat{i} + b\hat{j}\), its magnitude \(|\vec{v}|\) is computed as follows:
\[|\vec{v}| = \sqrt{a^2 + b^2}\]
In our example, we have a resultant vector \(\vec{a} + \vec{b} = -9 \hat{i} + 10 \hat{j}\).
So, the magnitude is:
\[|\vec{a} + \vec{b}| = \sqrt{(-9)^2 + (10)^2}\]
\[|\vec{a} + \vec{b}| = \sqrt{81 + 100}\]\
\[|\vec{a} + \vec{b}| \approx 13.45 \text{ meters}\]
This is the distance from the origin to the point \((-9, 10)\) in 2D space.
Direction Determination
Direction tells us where the vector points relative to a reference axis.
To find the direction of a vector, we use trigonometric functions.
For a vector \(\vec{v} = a \hat{i} + b \hat{j}\), we can determine the angle \(\theta\) relative to the positive x-axis using the tangent function:
\[\theta = \tan^{-1} \left( \frac{b}{a} \right)\]
In the example given, the resultant vector is \(\vec{a} + \vec{b} = -9 \hat{i} + 10 \hat{j}\):
\[\theta = \tan^{-1} \left( \frac{10}{-9} \right)\]
\[\theta \approx -48.4^{\circ}\]
The negative sign indicates the direction is below the x-axis.
To express this in terms of the positive x-axis, we add 180°:
\[\theta = 180^{\circ} - 48.4^{\circ}\]\[\theta \approx 131.6^{\circ}\]
So, our vector points at an angle of approximately 131.6° from the positive x-axis, going counterclockwise.

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Most popular questions from this chapter

A golfer takes three putts to get the ball into the hole. The first putt displaces the ball \(3.66 \mathrm{~m}\) north, the second \(1.83 \mathrm{~m}\) southeast, and the third \(0.91 \mathrm{~m}\) southwest. What are (a) the magnitude and (b) the direction of the displacement needed to get the ball into the hole on the first putt?

Prove that two vectors must have equal magnitudes if their sum is perpendicular to their difference.

Oasis \(B\) is \(25 \mathrm{~km}\) due east of oasis \(A\). Starting from oasis \(A\), a camel walks \(24 \mathrm{~km}\) in a direction \(15^{\circ}\) south of east and then walks \(8.0 \mathrm{~km}\) due north. How far is the camel then from oasis \(B\) ?

You are to make four straight-line moves over a flat desert floor, starting at the origin of an \(x y\) coordinate system and ending at the \(x y\) coordinates \((-140 \mathrm{~m}, 30 \mathrm{~m})\). The \(x\) -component and \(y\) -component of your moves are the following, respectively, in meters: (20 and 60), then \(\left(b_{x}\right.\) and \(-70\) ), then \(\left(-20\right.\) and \(\left.c_{y}\right)\), then \((-60\) and \(-70)\). What are (a) component \(b_{x}\) and (b) component \(c_{y}\) ? What are (c) the magnitude and (d) the angle (relative to the positive direction of the \(x\) axis) of the overall displacement?

For the vectors $$\vec{a}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}} \text { and } \vec{b}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}+(-2.0 \mathrm{~m}) \hat{\mathrm{j}},$$ give \(\vec{a}+\vec{b}\) in (a) unit-vector notation, and as (b) a magnitude and (c) an angle (relative to \(\hat{\mathrm{i}}\) ). Now give \(\vec{b}-\vec{a}\) in (d) unit-vector notation, and as (e) a magnitude and (f) an angle.

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