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(II) (a) Determine the magnitude and direction of the sum of the three vectors $$ \vec{\mathbf{v}}_{1}=4.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}} \cdot \vec{\mathbf{v}}_{2}=\hat{\mathbf{i}}+\hat{\mathbf{j}} $$ $$ \vec{\mathbf{v}}_{3}=-2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}} .(b) \text { Determine } \vec{\mathbf{v}}_{1}-\vec{\mathbf{v}}_{2}+\vec{\mathbf{v}}_{3} $$

Short Answer

Expert verified
The magnitude is \(3\sqrt{2} \) and direction is 315°. \(\vec{\mathbf{v}}_1 - \vec{\mathbf{v}}_2 + \vec{\mathbf{v}}_3 = \hat{\mathbf{i}} - 5\hat{\mathbf{j}} \).

Step by step solution

01

Write Down the Vectors

First, clearly note the given vectors in component form. These are:\( \vec{\mathbf{v}}_{1} = 4.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} \)\( \vec{\mathbf{v}}_{2} = \hat{\mathbf{i}} + \hat{\mathbf{j}} \)\( \vec{\mathbf{v}}_{3} = -2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}} \)
02

Compute the Sum of the Vectors

Add the vectors component-wise:\[ \vec{\mathbf{v}} = \vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 + \vec{\mathbf{v}}_3 = (4.0 + 1 - 2.0) \hat{\mathbf{i}} + (-8.0 + 1 + 4.0) \hat{\mathbf{j}} \]The sum is: \( \vec{\mathbf{v}} = 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}} \)
03

Find the Magnitude of the Sum

The magnitude of a vector \( \vec{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \) is found using:\[ |\vec{\mathbf{v}}| = \sqrt{a^2 + b^2} \]For our vector:\[ |\vec{\mathbf{v}}| = \sqrt{(3.0)^2 + (-3.0)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \]
04

Determine the Direction of the Sum

The direction (angle) \( \theta \) of a vector with components \( (a, b) \) is given by:\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]For our vector:\[ \theta = \tan^{-1}\left(\frac{-3.0}{3.0}\right) = \tan^{-1}(-1) \approx -45^\circ \] Since our vector is in the fourth quadrant, we add 360° to get an angle measured counter-clockwise from the positive x-axis: \( 315^\circ \).
05

Calculate \( \vec{\mathbf{v}}_1 - \vec{\mathbf{v}}_2 + \vec{\mathbf{v}}_3 \)

First find the vector \( \vec{\mathbf{v}}_1 - \vec{\mathbf{v}}_2 \):\[ \vec{\mathbf{v}}_1 - \vec{\mathbf{v}}_2 = (4.0 - 1)\hat{\mathbf{i}} + (-8.0 - 1)\hat{\mathbf{j}} = 3.0\hat{\mathbf{i}} - 9.0\hat{\mathbf{j}} \]Now add \( \vec{\mathbf{v}}_3 \):\[ \vec{\mathbf{v}} = (3.0 - 2.0)\hat{\mathbf{i}} + (-9.0 + 4.0)\hat{\mathbf{j}} = 1.0\hat{\mathbf{i}} - 5.0\hat{\mathbf{j}} \]

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Key Concepts

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

Vector Components
Vectors can be broken down into components, which represent the contributions along each axis of a coordinate system. The component form of a vector is often expressed using unit vectors, such as \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \), which represent the horizontal and vertical axes, respectively. For example, a vector \( \vec{\mathbf{v}} = 4.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} \) indicates that the vector moves 4 units in the direction of the x-axis and 8 units in the opposite direction of the y-axis.
Breaking vectors into components allows for simple vector arithmetic. To add or subtract vectors, you can do so by their respective components. Given vectors \( \vec{\mathbf{v}}_1 = 4.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} \), \( \vec{\mathbf{v}}_2 = \hat{\mathbf{i}} + \hat{\mathbf{j}} \), and \( \vec{\mathbf{v}}_3 = -2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}} \), these vectors can be added to find their resultant by simply adding or subtracting the \( \hat{\mathbf{i}} \) components together and doing the same for the \( \hat{\mathbf{j}} \) components. Through this, the new vector \( \vec{\mathbf{v}}_1 + \vec{\mathbf{v}}_2 + \vec{\mathbf{v}}_3 \) becomes \( 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}} \).
Understanding vector components is crucial as it simplifies complex problems into manageable numbers that can be added or subtracted algebraically, making calculations easier to handle.
Vector Magnitude
The magnitude of a vector is a measure of its length or size. It represents the distance from the vector’s initial point to its terminal point in the coordinate space. To find the magnitude of a vector with components \( (a, b) \), we use the Pythagorean theorem, expressed as:
\[ |\vec{\mathbf{v}}| = \sqrt{a^2 + b^2} \]
In the given problem, the resultant vector from adding the vectors has components \( 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}} \). To determine its magnitude, we calculate:
\[ |\vec{\mathbf{v}}| = \sqrt{(3.0)^2 + (-3.0)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \]
Knowing the magnitude is significant because it tells you about the size or extent of the vector's displacement, important when you're dealing with physical quantities like force, velocity, or acceleration, as they give a sense of how much and in which aspect (size-wise) a vector is involved in its application.
Vector Direction
The direction of a vector is described by the angle it makes with a reference axis, typically the positive x-axis. This angle helps in understanding where the vector is pointing in the coordinate system. To find this angle, we use the inverse tangent function, capturing the ratio of the vector's y-component \( b \) to its x-component \( a \):
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
In our example, the resultant vector from the sum of vectors has components \( 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}} \). Therefore, its direction is given by:
\[ \theta = \tan^{-1}\left(\frac{-3.0}{3.0}\right) = \tan^{-1}(-1) \approx -45^\circ \]
Since standard convention measures angle counter-clockwise from the positive x-axis, and the vector is pointing in the fourth quadrant, we correct this by adding 360°, giving a proper directional angle of 315°.
Figuring out the vector's direction is crucial where specificity in positioning or alignment is required – like determining the angle in a wind's direction or the heading course in navigation.

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Most popular questions from this chapter

When Babe Ruth hit a homer over the 8.0 -m-high rightfield fence \(98 \mathrm{~m}\) from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit \(1.0 \mathrm{~m}\) above the ground and its path initially made a \(36^{\circ}\) angle with the ground.

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