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<h2 class="hd hd-2 unit-title">Vector Skill Challenge Introduction</h2>
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<h2> Skill Challenge 0: Vectors </h2><p> Vectors are an important mathematical tool for physics, and mastery of vector manipulation is crucial to success in mechanics. The ability to fully utilize vectors is crucial to studying mechanics. There are videos and practice questions as well as a few required problems. </p>
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<h2 class="hd hd-2 unit-title">SC0v1: Vectors vs. Scalars</h2>
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<h3 class="hd hd-2">SC0v1: Vectors vs. Scalars</h3>
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<h2 class="hd hd-2 unit-title">SC0v2: Vector Operators</h2>
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<h3 class="hd hd-2">SC0v2: Vector Operators</h3>
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<h2 class="hd hd-2 unit-title">SC0Q1: Vector Operations Graphically (Practice)</h2>
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SC0Q1-1: Adding Vectors Graphically (Practice)
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<p>
Consider vectors [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] shown on the left figure above. What is the best choice for vector [mathjaxinline]\vec{\textbf{C}}=\vec{\textbf{A}}+\vec{\textbf{B}}[/mathjaxinline]? </p>
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<text> c</text>
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SC0Q1-2: Subtracting Vectors Graphically (Practice)
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Consider vectors [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline]. What is the best choice for vector [mathjaxinline]\vec{\textbf{C}}=\vec{\textbf{A}}-\vec{\textbf{B}}[/mathjaxinline]? </p>
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<h2 class="hd hd-2 unit-title">SC0Q2: Adding Vectors Graphically</h2>
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Adding Vectors Graphically
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<p>Three forces are applied to a rectangular box as shown below. The total force on the box is obtained by adding the three forces: \(\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3\).</p>
<p>Use your mouse or finger to draw the net force \(\vec{F}_{\text{net}}\) on the figure below. <em>You may also drag the vectors \(\vec{F}_1\), \(\vec{F}_2\), and \(\vec{F}_3\), which may be helpful.</em></p>
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Video tutorial on Vectors vs. Scalars </a></li><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-SC0v02" target="_blank">Video tutorial on Graphical Operations</a></li><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-mod1_lp_a" target="_blank">Practice: Adding and Subtracting Vectors Graphically</a></li></ul>
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<h2 class="hd hd-2 unit-title">SC0Q3: Unit Vectors (Practice)</h2>
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SC0Q3: Unit Vectors (Practice)
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Which of the figures represents a unit vector? Check <em>all</em> that apply. </p>
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<h2 class="hd hd-2 unit-title">SC0v4: Vectors - Magnitude and Direction</h2>
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<h2 class="hd hd-2 unit-title">SC0Q4: Magnitude and Direction (Practice)</h2>
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SC0Q4: Magnitude and Direction (Practice)
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<p>
Consider the vectors [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline]. The magnitude of these vectors is approximately: </p>
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<p style="display:inline">Magnitude of A:</p>
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<p style="display:inline">Magnitude of B:</p>
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The direction defined as the angle between the vector and the +x axis measured in degrees is approximately: </p>
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<p style="display:inline">Direction of B:</p>
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<option value="20"> 20</option>
<option value="40"> 40</option>
<option value="110"> 110</option>
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<h2 class="hd hd-2 unit-title">SC0v5: Vector Decomposition into components</h2>
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<h3 class="hd hd-2">SC0v5: Vector Decomposition into components</h3>
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<h2 class="hd hd-2 unit-title">SC0Q5: Vector Decomposition (Practice)</h2>
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SC0Q5-1: Vector Decomposition (Practice)
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Vectors [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] in the figure can be written in terms of the unit vectors [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline] as [mathjaxinline]\vec{\textbf{A}}= A_ x\hat{\textbf{i}}+A_ y\hat{\textbf{j}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}= B_ x\hat{\textbf{i}}+B_ y\hat{\textbf{j}}[/mathjaxinline]. </p>
<p>
Find the x and y components of vectors [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] and [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]A_ x[/mathjaxinline] = </p>
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<p style="display:inline">[mathjaxinline]A_ y[/mathjaxinline] = </p>
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<p style="display:inline">[mathjaxinline]B_ x[/mathjaxinline] = </p>
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<p style="display:inline">[mathjaxinline]B_ y[/mathjaxinline] = </p>
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SC0Q5-2: Vector Decomposition (Practice)
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Which of the following figures best represents vector [mathjaxinline]\vec{\textbf{A}} = 4\hat{\textbf{i}} - 3\hat{\textbf{j}}[/mathjaxinline]? </p>
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SC0Q5-3: Vector Decomposition (Practice)
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The forces acting on a box sliding down a frictionless incline are the normal force, [mathjaxinline]\vec{\textbf{F}}_ N[/mathjaxinline], perpendicular to the incline surface, and the gravitational force, [mathjaxinline]\vec{\textbf{F}}_ g[/mathjaxinline], pointing down. Consider the (x,y) coordinate system and the unit vectors [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline] shown in the figure. </p>
<p><b class="bfseries">(Part a)</b> If the normal force is expressed as [mathjaxinline]\vec{\textbf{F}}_ N=F_{Nx}\hat{\textbf{i}} + F_{Ny}\hat{\textbf{j}}[/mathjaxinline], find [mathjaxinline]F_{Nx}[/mathjaxinline] and [mathjaxinline]F_{Ny}[/mathjaxinline] in terms of the magnitude of the normal force [mathjaxinline]F_ N=|\vec{\textbf{F}}_ N|[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline]. Use F_N for [mathjaxinline]F_ N[/mathjaxinline] and theta for [mathjaxinline]\theta[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]F_{Nx} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]F_{Ny} =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> If the gravitational force is expressed as [mathjaxinline]\vec{\textbf{F}}_ g=F_{gx}\hat{\textbf{i}} + F_{gy}\hat{\textbf{j}}[/mathjaxinline], find [mathjaxinline]F_{gx}[/mathjaxinline] and [mathjaxinline]F_{gy}[/mathjaxinline] in terms of the magnitude of the gravitational force [mathjaxinline]F_ g=|\vec{\textbf{F}}_ g|[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline]. Use F_g for [mathjaxinline]F_ g[/mathjaxinline] and theta for [mathjaxinline]\theta[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]F_{gx} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]F_{gy} =[/mathjaxinline] </p>
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SC0Q5-4: Vector Decomposition (Practice)
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The forces acting on a box sliding down a frictionless incline are the normal force, [mathjaxinline]\vec{\textbf{F}}_ N[/mathjaxinline], perpendicular to the incline surface, and the gravitational force, [mathjaxinline]\vec{\textbf{F}}_ g[/mathjaxinline], pointing down. Consider the (x,y) coordinate system and the unit vectors [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline] shown in the figure. </p>
<p><b class="bfseries">(Part a)</b> If the normal force is expressed as [mathjaxinline]\vec{\textbf{F}}_ N=F_{Nx}\hat{\textbf{i}} + F_{Ny}\hat{\textbf{j}}[/mathjaxinline], find [mathjaxinline]F_{Nx}[/mathjaxinline] and [mathjaxinline]F_{Ny}[/mathjaxinline] in terms of the magnitude of the normal force [mathjaxinline]F_ N=|\vec{\textbf{F}}_ N|[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline]. Use F_N for [mathjaxinline]F_ N[/mathjaxinline] and theta for [mathjaxinline]\theta[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]F_{Nx} =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> If the gravitational force is expressed as [mathjaxinline]\vec{\textbf{F}}_ g=F_{gx}\hat{\textbf{i}} + F_{gy}\hat{\textbf{j}}[/mathjaxinline], find [mathjaxinline]F_{gx}[/mathjaxinline] and [mathjaxinline]F_{gy}[/mathjaxinline] in terms of the magnitude of the gravitational force [mathjaxinline]F_ g=|\vec{\textbf{F}}_ g|[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline]. Use F_g for [mathjaxinline]F_ g[/mathjaxinline] and theta for [mathjaxinline]\theta[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]F_{gx} =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">SC0Q6: Vector Decomposition and Rotation of Coordinate System - Part I</h2>
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SC0Q6: Vector Decomposition and Rotation of Coordinate System - Part I
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<p class="problemIntro">A square block is sitting on a ramp that makes an angle \(\theta\) with the horizontal. The magnitude of the force shown in blue is \(F\).</p>
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<h2 class="hd hd-2 unit-title">SC0Q7: Vector Decomposition and Rotation of Coordinate System - Part II</h2>
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SC0Q7: Vector Decomposition and Rotation of Coordinate System - Part II
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<div><p>A force of magnitude \(F\) is being applied to a block at the corner (red dot) sitting on a ramp. The ramp forms an angle of \(\theta\) with respect to the horizontal. The expression for the force vector is</p>
\[\vec{F} = F\cos{(\theta)}\hat{i} - F\sin{(\theta)\hat{j}}\]
<p>where \(\theta\) is the same as the \(\theta\) between the ramp and the horizontal.</p>
<p>In the following two situations with different choices of coordinate system, drag the corresponding force vector into the diagram. It does not matter where in the diagram you put the arrow. Note that for each diagram, the expression above for [mathjaxinline]\vec{F}[/mathjaxinline] should be interpreted as a <em>vector expression in the particular choice of coordinate system in that diagram</em>.</p>
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<h2 class="hd hd-2 unit-title">SC0Q8: Magnitude and Direction or Components (Practice)</h2>
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SC0Q8-1: Find Components given the Magnitude and Direction (Practice)
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At a given instant of time a plane is 100 km from the airport in a direction [mathjaxinline]30^\circ[/mathjaxinline] south of east from the airport. If the [mathjaxinline]+x[/mathjaxinline] axis is pointing east and the [mathjaxinline]+y[/mathjaxinline] axis points north, then the plane's position vector at that instant is [mathjaxinline]\vec{\textbf{r}}=x\hat{\textbf{i}}+y\hat{\textbf{j}}[/mathjaxinline], where </p>
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<span class="trailing_text" id="trailing_text_module_mod1_lp_10_2_1">[mathjaxinline]\mathrm{km}[/mathjaxinline]</span>
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SC0Q8-2: Find Magnitude and Direction given the Components (Practice)
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Given the vector [mathjaxinline]\vec{\textbf{A}} = -3\hat{\textbf{i}}+2\hat{\textbf{j}}[/mathjaxinline] in the xy-plane, find its magnitude, [mathjaxinline]|\vec{\textbf{A}}|[/mathjaxinline], and its direction, [mathjaxinline]\theta[/mathjaxinline], measured with respect to the +x axis measuring in a counter-clockwise direction. </p>
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<p style="display:inline">[mathjaxinline]|\vec{\textbf{A}}| =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">SC0Q9: Magnitude and Direction</h2>
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SC0Q9: Magnitude and Direction
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Calculate the magnitude and the direction measured with respect to the +x axis measured in the counter-clockwise direction of the total force obtained by adding the 3 forces, [mathjaxinline]\vec{\textbf{F}}_{Total}= \vec{\textbf{F}}_1+\vec{\textbf{F}}_2+\vec{\textbf{F}}_3[/mathjaxinline]. </p>
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<p style="display:inline">Magnitude = </p>
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<span class="trailing_text" id="trailing_text_module_mod1_sk_2_3_1">degrees</span>
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<p>If you need help, you might find some of the following content useful:</p><ul><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-SC0v04" target="_blank">Video tutorial on Magnitude and Direction </a></li><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-mod1_lp_c" target="_blank">Practice: Magnitude and Direction</a></li><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-SC0v06" target="_blank">Video tutorial Going Between Representations</a></li><li><a href="/courses/course-v1:MITx+8.01.1x+3T2018/jump_to_id/vert-mod1_lp_e" target="_blank">Practice: Find Magnitude and Direction given the components</a></li></ul>
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<h2 class="hd hd-2 unit-title">SC0Q10: Adding and Subtracting Vectors (Practice)</h2>
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SC0Q10-1: Adding Vectors Analytically (Practice)
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Vector [mathjaxinline]\vec{\textbf{C}}[/mathjaxinline] in the figure is the result of the addition of two vectors, [mathjaxinline]\vec{\textbf{C}} = \vec{\textbf{A}}+\vec{\textbf{B}}[/mathjaxinline]. Take each side of a square in the grid to have a length of one. </p>
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Vector [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] is shown in the figure and vector [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] is unknown. Find the x and the y components of [mathjaxinline]\vec{\textbf{B}} = B_ x\hat{\textbf{i}}+B_ y\hat{\textbf{j}}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]B_ x =[/mathjaxinline] </p>
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SC0Q10-2: Vector Addition and Subtraction (Practice)
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<p><b class="bfseries">(Part a)</b> [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] is 2 units long and [mathjaxinline]60^\circ[/mathjaxinline] above the [mathjaxinline]x[/mathjaxinline]-axis in the first quadrant. [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] is 2 units long and [mathjaxinline]60^\circ[/mathjaxinline] below the [mathjaxinline]x[/mathjaxinline]-axis in the fourth quadrant. Find the vector sum [mathjaxinline]\vec{\textbf{A}} + \vec{\textbf{B}}[/mathjaxinline]. </p>
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Express you answer in terms of some or all of the following: hati for [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline], and hatj for [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\vec{\textbf{A}} + \vec{\textbf{B}} =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> A vector [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline] of magnitude 10 units makes an angle of [mathjaxinline]30^\circ[/mathjaxinline] with a vector [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] of length 6 units. Find the magnitude of the vector difference [mathjaxinline]\vec{\textbf{A}} - \vec{\textbf{B}}[/mathjaxinline], and the magnitude of the angle it makes with the vector [mathjaxinline]\vec{\textbf{A}}[/mathjaxinline]. </p>
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<p style="display:inline">Magnitude = </p>
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<h2 class="hd hd-2 unit-title">SC0Q11: Vector Decomposition</h2>
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SC0Q11-1: Vector Decomposition
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Consider the [mathjaxinline](x,y)[/mathjaxinline] Cartesian coordinate system shown in the figure. [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] is the unit vector parallel to the x-axis, and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline] is the unit vector parallel to the y-axis. </p>
<p><b class="bfseries">(Part a)</b> Write the vector [mathjaxinline]\vec{\textbf{F}}_1[/mathjaxinline] in terms of the unit vectors [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline]. Drag the appropriate terms to construct the correct expression. Pay attention to the difference between [mathjaxinline]\alpha[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline] in the trigonometric functions. </p>
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SC0Q11-2: Vector Decomposition Part 2
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<p><b class="bfseries">(Part b)</b> Write the vector sum [mathjaxinline]\vec{\textbf{F}}_1+\vec{\textbf{F}}_2+\vec{\textbf{F}}_3[/mathjaxinline] in terms of the unit vectors [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline] and [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline]. </p>
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Use [mathjaxinline]F_1=|\vec{\textbf{F}}_1|[/mathjaxinline], [mathjaxinline]F_2=|\vec{\textbf{F}}_2|[/mathjaxinline], and [mathjaxinline]F_3=|\vec{\textbf{F}}_3|[/mathjaxinline] to be the magnitude of the vectors [mathjaxinline]\vec{\textbf{F}}_1[/mathjaxinline], [mathjaxinline]\vec{\textbf{F}}_2[/mathjaxinline], and [mathjaxinline]\vec{\textbf{F}}_3[/mathjaxinline], respectively. </p>
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Drag the appropriate terms to construct the correct expression. Pay attention to the difference between [mathjaxinline]\alpha[/mathjaxinline] and [mathjaxinline]\theta[/mathjaxinline] in the trigonometric functions. </p>
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<h2 class="hd hd-2 unit-title">SC0Q12: Tetrahedron Model of Methane (Challenge Problem - Practice)</h2>
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SC0Q12: Tetrahedron Model of Methane (Challenge Problem - Practice)
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Consider a regular tetrahedron whose vertices [mathjaxinline]v_1[/mathjaxinline] through [mathjaxinline]v_4[/mathjaxinline] are given by the position vectors [mathjaxinline]\vec{\mathbf{r}}_1 =\vec{\mathbf{0}}[/mathjaxinline], [mathjaxinline]\vec{\mathbf{r}}_2 =a(\mathbf{\hat{i}}+\mathbf{\hat{j}})[/mathjaxinline], [mathjaxinline]\vec{\mathbf{r}}_3 =a(\mathbf{\hat{i}}+\mathbf{\hat{k}})[/mathjaxinline], and [mathjaxinline]\vec{\mathbf{r}}_4 =a(\mathbf{\hat{j}}+\mathbf{\hat{k}})[/mathjaxinline] respectively, where [mathjaxinline]a[/mathjaxinline] is the length of the side of the cube formed by the vertices of the tetrahedron. </p>
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<p><b class="bfseries">(Part a)</b> Verify that each face of a tetrahedron is an equilateral triangle by showing that the distances between any pair of vertices are the same. Express your answer in terms of [mathjaxinline]a[/mathjaxinline]. You may also need to use sqrt() to indicate the square root function. </p>
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<p style="display:inline">[mathjaxinline]r_{12}=r_{13}=r_{14}=r_{23}=r_{24}=r_{34}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> We define the unit vector pointing from point A to point B by the expression </p>
<table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\mathbf{\hat{r}}_{AB} =\frac{\mathbf{\vec{r}}_{AB}}{\left| \mathbf{\vec{r}}_{AB} \right|} =\frac{\mathbf{\vec{r}}_ B -\mathbf{\vec{r}}_ A}{\left| \mathbf{\vec{r}}_{AB}\right|} \, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
Find the unit vectors [mathjaxinline]\mathbf{\hat{r}}_{21}[/mathjaxinline] which points from [mathjaxinline]v_2[/mathjaxinline] to [mathjaxinline]v_1[/mathjaxinline], [mathjaxinline]\mathbf{\hat{r}}_{31}[/mathjaxinline] which points from [mathjaxinline]v_3[/mathjaxinline] to [mathjaxinline]v_1[/mathjaxinline] and [mathjaxinline]\mathbf{\hat{r}}_{41}[/mathjaxinline] which points from [mathjaxinline]v_4[/mathjaxinline] to [mathjaxinline]v_1[/mathjaxinline]. Express your answers interms of some or all of the following: [mathjaxinline]a[/mathjaxinline], hati for [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline], hatj for [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline], and hatk for [mathjaxinline]\hat{\textbf{k}}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathbf{\hat{r}}_{21}=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\mathbf{\hat{r}}_{31}=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\mathbf{\hat{r}}_{41}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> In the methane molecule, [mathjaxinline]\mathrm{CH}_4[/mathjaxinline], each hydrogen atom is at the vertex of a tetrahedron with the carbon atom at the center. In the coordinate system for the four vertices given in <b class="bfseries">(Part a)</b>, what is the position vector of the carbon atom? Express your answer in terms of some or all of the following: [mathjaxinline]a[/mathjaxinline], hati for [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline], hatj for [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline], and hatk for [mathjaxinline]\hat{\textbf{k}}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\vec{\mathbf{r_ C}} =[/mathjaxinline] </p>
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<span id="solution_pset_pset1_6_solution_3"/>
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<p><b class="bfseries">(Part d)</b> Find the vector [mathjaxinline]\mathbf{\vec{r}}_{C1}[/mathjaxinline] from the carbon atom to the hydrogen atom located at [mathjaxinline]v_1[/mathjaxinline], and the vector [mathjaxinline]\mathbf{\vec{r}}_{C2}[/mathjaxinline] from the carbon atom to the hydrogen atom located at [mathjaxinline]v_2[/mathjaxinline]. Express your answers in terms of some or all of the following: [mathjaxinline]a[/mathjaxinline], hati for [mathjaxinline]\hat{\textbf{i}}[/mathjaxinline], hatj for [mathjaxinline]\hat{\textbf{j}}[/mathjaxinline], and hatk for [mathjaxinline]\hat{\textbf{k}}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{\mathbf{r}}_{C1} =[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\vec{\mathbf{r}}_{C2} =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> The angle [mathjaxinline]\theta[/mathjaxinline] between the bonds between the carbon atom and hydrogen atoms located at vertices [mathjaxinline]v_1[/mathjaxinline] and [mathjaxinline]v_2[/mathjaxinline], is given by the expression </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\theta = \cos ^{-1} \left( \frac{\vec{\textbf{r}}_{C1} \cdot \vec{\textbf{r}}_{C2}}{\left|\vec{\textbf{r}}_{C1}\right| \left| \vec{\textbf{r}}_{C2} \right|} \right)\, .[/mathjax]</td>
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What is the angle between these two bonds? Enter your answer in degrees. </p>
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<p style="display:inline">[mathjaxinline]\theta =[/mathjaxinline] </p>
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