<h2>How to Add Vectors Using Analytical Methods</h2>
<p>To see how to add vectors using perpendicular components, consider the figure below, in which the vectors \(\mathbf{A}\) and \(\mathbf{B}\) are added to produce the resultant \(\mathbf{R}\).</p>
<div class="wp-caption alignnone" style="width: 400px"><img alt="Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown." height="235" src="https://data.ulearngo.com/assets/3058cc44-f8d1-417a-abb3-d4468fdf9604" width="400"/><p class="wp-caption-text">Vectors \(\mathbf{A}\) and \(\mathbf{B}\) are two legs of a walk, and \(\mathbf{R}\) is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of \(\mathbf{R}\).</p></div>
<p>If \(\mathbf{A}\) and \(\mathbf{B}\) represent two legs of a walk (two displacements), then \(\mathbf{R}\) is the total displacement. The person taking the walk ends up at the tip of \(\mathbf{R}.\) There are many ways to arrive at the same point. In particular, the person could have walked first in the <em>x</em>-direction and then in the <em>y</em>-direction. Those paths are the <em>x</em>- and <em>y</em>-components of the resultant, \({\mathbf{R}}_{x}\) and \({\mathbf{R}}_{y}\).</p>
<p>If we know \({\mathbf{\text{R}}}_{x}\) and \({\mathbf{R}}_{y}\), we can find \(R\) and \(\theta \) using the equations \(A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}\) and \(\theta ={\text{tan}}^{–1}\left({A}_{y}/{A}_{x}\right)\). When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.</p>
<h3><em><strong>Step 1.</strong></em><em> Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes</em>.</h3>
<p>Use the equations \({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) and \({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) to find the components. In the figure below, these components are \({A}_{x}\), \({A}_{y}\), \({B}_{x}\), and \({B}_{y}\). The angles that vectors \(\mathbf{A}\) and \(\mathbf{B}\) make with the <em>x</em>-axis are \({\theta }_{\text{A}}\) and \({\theta }_{\text{B}}\), respectively.</p>
<div class="wp-caption alignnone" style="width: 400px"><img alt="Two vectors A and B are shown. The tail of the vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The horizontal and vertical components of the vectors A and B are shown with the help of dotted lines. The vectors labeled as A sub x and A sub y are the components of vector A, and B sub x and B sub y as the components of vector B.." height="231" src="https://data.ulearngo.com/assets/00fc1492-5511-4820-bbc8-b9ae38228c0b" width="400"/><p class="wp-caption-text">To add vectors \(\mathbf{A}\) and \(\mathbf{B}\), first determine the horizontal and vertical components of each vector. These are the dotted vectors \({\mathbf{A}}_{x}\), \({\mathbf{A}}_{y}\), \({\mathbf{B}}_{x}\) and \({\mathbf{\text{B}}}_{y}\) shown in the image.</p></div>
<h3><em><strong>Step 2.</strong></em><em> Find the components of the resultant along each axis by adding the components of the individual vectors along that axis</em>.</h3>
<p>That is, as shown in the figure below,</p>
<div>\({R}_{x}={A}_{x}+{B}_{x}\)</div>
<p>and</p>
<div>\({R}_{y}={A}_{y}+{B}_{y}\text{.}\)</div>
<div class="wp-caption alignnone" style="width: 400px"><img alt="Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The vectors A and B are resolved into the horizontal and vertical components shown as dotted lines parallel to x axis and y axis respectively. The horizontal components of vector A and vector B are labeled as A sub x and B sub x and the horizontal component of the resultant R is labeled at R sub x and is equal to A sub x plus B sub x. The vertical components of vector A and vector B are labeled as A sub y and B sub y and the vertical components of the resultant R is labeled as R sub y is equal to A sub y plus B sub y." height="250" src="https://data.ulearngo.com/assets/f4d70c36-18ac-462c-8c17-546b3f7409a5" width="400"/><p class="wp-caption-text">The magnitude of the vectors \({\mathbf{A}}_{x}\) and \({\mathbf{B}}_{x}\) add to give the magnitude \({R}_{x}\) of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors \({\mathbf{A}}_{y}\) and \({\mathbf{B}}_{y}\) add to give the magnitude \({R}_{y}\) of the resultant vector in the vertical direction.</p></div>
<p>Components along the same axis, say the <em>x</em>-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the <em>y</em>-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.)</p>
<p>So resolving vectors into components along common axes makes it easier to add them. Now that the components of \(\mathbf{R}\) are known, its magnitude and direction can be found.</p>
<h3><em><strong>Step 3.</strong></em><em> To get the magnitude \(R\) of the resultant, use the Pythagorean theorem:</em></h3>
<div>\(R=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}\text{.}\)</div>
<h3><em><strong>Step 4.</strong></em><em> To get the direction of the resultant:</em></h3>
<div>\(\theta ={\text{tan}}^{-1}\left({R}_{y}/{R}_{x}\right)\text{.}\)</div>
<p>The following example illustrates this technique for adding vectors using perpendicular components.</p>
<div class="ns-school-emp">
<h3>Example: Adding Vectors Using Analytical Methods</h3>
<div class="ns-school-defn">
<p>Add the vector \(\mathbf{A}\) to the vector \(\mathbf{B}\) shown in the figure below, using perpendicular components along the <em>x</em>- and <em>y</em>-axes. The <em>x</em>- and <em>y</em>-axes are along the east–west and north–south directions, respectively.</p>
<p>Vector \(\mathbf{A}\) represents the first leg of a walk in which a person walks \(\text{53}\text{.}\text{0 m}\) in a direction \(\text{20}\text{.}0\text{º}\) north of east. Vector \(\mathbf{B}\) represents the second leg, a displacement of \(\text{34}\text{.}\text{0 m}\) in a direction \(\text{63}\text{.}0\text{º}\) north of east.</p>
<div class="wp-caption alignnone" style="width: 400px"><img alt="Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B." height="218" src="https://data.ulearngo.com/assets/11338d47-75a8-4da5-9676-248423cc7ddd" width="400"/><p class="wp-caption-text">Vector \(\mathbf{A}\) has magnitude \(\text{53}\text{.}\text{0 m}\) and direction \(\text{20}\text{.}0º\) north of the x-axis. Vector \(\mathbf{B}\) has magnitude \(\text{34}\text{.}\text{0 m}\) and direction \(\text{63}\text{.}0\text{º}\) north of the x-axis. You can use analytical methods to determine the magnitude and direction of \(\mathbf{R}\).</p></div>
<p><strong>Strategy</strong></p>
<p>The components of \(\mathbf{A}\) and \(\mathbf{B}\) along the <em>x</em>- and <em>y</em>-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.</p>
<p><strong>Solution</strong></p>
<p>Following the method outlined above, we first find the components of \(\mathbf{A}\) and \(\mathbf{B}\) along the <em>x</em>- and <em>y</em>-axes. Note that \(A=53.0 m\text{}\), \({\theta }_{\text{A}}=20.0º\), \(B=34.0 m\text{}\), and \({\theta }_{\text{B}}=63.0º\). We find the <em>x</em>-components by using \({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \), which gives</p>
<p>\(A_{x} = A\text{ cos }\theta_{A} = (53.0\text{ m})(\text{cos }20.0^{\circ})\)<br/>
 \(A_{x} = (53.0\text{ m})(0.940) = 49.8\text{ m}\)</p>
<p>and</p>
<p>\(B_{x} = B\text{ cos }\theta_{B} = (34.0\text{ m})(\text{cos }63.0^{\circ})\)<br/>
 \(B_{x} = (34.0\text{ m})(0.454) = 15.4\text{ m}\)</p>
<p>Similarly, the <em>y</em>-components are found using \({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{A}\):</p>
<p>\(A_{y} = A\text{ sin }\theta_{A} = (53.0\text{ m})(\text{sin }20.0^{\circ})\)<br/>
 \(A_{y} = (53.0\text{ m})(0.342) = 18.1\text{ m}\)</p>
<p>and</p>
<p>\(B_{y} = B\text{ sin }\theta_{A} = (34.0\text{ m})(\text{sin }63.0^{\circ})\)<br/>
 \(B_{y} = (34.0\text{ m})(0.891) = 30.3\text{ m}\)</p>
<p>The <em>x</em>- and <em>y</em>-components of the resultant are thus</p>
<p>\({R}_{x}={A}_{x}+{B}_{x}=\text{49}\text{.}8 m\text{}+\text{15}\text{.}4 m\text{}=\text{65}\text{.}2 m\text{}\)</p>
<p>and</p>
<p>\({R}_{y}={A}_{y}+{B}_{y}=\text{18}\text{.}1 m+\text{30}\text{.}3 m=\text{48}\text{.}4 m\text{.}\)</p>
<p>Now we can find the magnitude of the resultant by using the Pythagorean theorem:</p>
<p>\(R=\sqrt{{R}_{x}^{2}+{R}_{y}^{2}}=\sqrt{(65.2)^{2}+(48.4)^{2}\text{ m}}\)</p>
<p>so that</p>
<p>\(R=81.2\text{ m.}\)</p>
<p>Finally, we find the direction of the resultant:</p>
<p>\(\theta ={\text{tan}}^{-1}\left({R}_{y}/{R}_{x}\right)\text{=+}{\text{tan}}^{-1}\left(\text{48}\text{.}4/\text{65}\text{.}2\right)\text{.}\)</p>
<p>Thus,</p>
<p>\(\theta ={\text{tan}}^{-1}\left(0\text{.}\text{742}\right)=\text{36}\text{.}6º\text{.}\)</p>
<div class="wp-caption alignnone" style="width: 800px"><img alt="The addition of two vectors A and B is shown. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. Vector B is of magnitude thirty four units and is inclined at angle sixty three degrees to the horizontal. The components of vector A are shown as dotted vectors A X is equal to forty nine point eight meter along x axis and A Y is equal to eighteen point one meter along Y axis. The components of vector B are also shown as dotted vectors B X is equal to fifteen point four meter and B Y is equal to thirty point three meter. The horizontal component of the resultant R X is equal to A X plus B X is equal to sixty five point two meter. The vertical component of the resultant R Y is equal to A Y plus B Y is equal to forty eight point four meter. The magnitude of the resultant of two vectors is eighty one point two meters. The direction of the resultant R is in thirty six point six degree from the vector A in anticlockwise direction." height="460" src="https://data.ulearngo.com/assets/d922f3e2-5a8c-46a0-b0b3-fa99797153a2" width="800"/><p class="wp-caption-text">Using analytical methods, we see that the magnitude of \(\mathbf{R}\) is \(\text{81}\text{.}\text{2 m}\) and its direction is \(\text{36}\text{.}6º\) north of east.</p></div>
<p><strong>Discussion</strong></p>
<p>This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.</p>
<p>Subtraction of vectors is accomplished by the addition of a negative vector. That is, \(\mathbf{A}-\mathbf{B}\equiv \mathbf{A}+\left(\mathbf{–B}\right)\). Thus, <em>the method for the subtraction of vectors using perpendicular components is identical to that for addition</em>. The components of \(\mathbf{\text{–B}}\) are the negatives of the components of \(\mathbf{B}\). The <em>x</em>- and <em>y</em>-components of the resultant \(\mathbf{A}-\text{B = R}\) are thus</p>
<p>\({R}_{x}={A}_{x}+\left(–{B}_{x}\right)\)</p>
<p>and</p>
<p>\({R}_{y}={A}_{y}+\left(–{B}_{y}\right)\)</p>
<p>and the rest of the method outlined above is identical to that for addition. (See the figure below.)</p>
</div>
</div>
<p>Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next topic, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.</p>
<div class="wp-caption alignnone" style="width: 400px"><img alt="In this figure, the subtraction of two vectors A and B is shown. A red colored vector A is inclined at an angle theta A to the positive of x axis. From the head of vector A a blue vector negative B is drawn. Vector B is in west of south direction. The resultant of the vector A and vector negative B is shown as a black vector R from the tail of vector A to the head of vector negative B. The resultant R is inclined to x axis at an angle theta below the x axis. The components of the vectors are also shown along the coordinate axes as dotted lines of their respective colors." height="258" src="https://data.ulearngo.com/assets/162b494e-3d4c-481d-999d-2b7dd95b6643" width="400"/><p class="wp-caption-text">The subtraction of the two vectors shown earlier. The components of \(\mathbf{\text{–B}}\) are the negatives of the components of \(\mathbf{B}\). The method of subtraction is the same as that for addition.</p></div>

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How to Add Vectors Using Analytical Methods

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Explore the principles of two-dimensional kinematics, including mastering vector addition and subtraction, resolving a vector into components, and analyzing projectile motion and relative velocities.


How to Add Vectors Using Analytical Methods

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