{"id":2328,"date":"2025-08-12T21:46:23","date_gmt":"2025-08-12T21:46:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2328"},"modified":"2025-10-21T18:24:19","modified_gmt":"2025-10-21T18:24:19","slug":"vectors-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/vectors-learn-it-3\/","title":{"raw":"Vectors: Learn It 3","rendered":"Vectors: Learn It 3"},"content":{"raw":"

Finding Component Form<\/h2>\r\nIn some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph that the position vector [latex]\\langle 2,3\\rangle [\/latex] comes from adding the vectors v<\/em><\/strong>1<\/sub> and v<\/em><\/strong>2<\/sub>. We have v<\/em><\/strong>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].\r\n
\r\n

[latex]\\begin{align}{\\boldsymbol{v}}_{1}&=\\langle 2 - 0,0 - 0\\rangle \\\\ &=\\langle 2,0\\rangle \\end{align}[\/latex]<\/p>\r\nWe also have v<\/em><\/strong>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].\r\n

[latex]\\begin{align}{\\boldsymbol{v}}_{2}&=\\langle 0 - 0,3 - 0\\rangle \\\\ &=\\langle 0,3\\rangle \\end{align}[\/latex]<\/div>\r\nTherefore, the position vector is\r\n
[latex]\\begin{align}\\boldsymbol{v}&=\\langle 2+0,3+0\\rangle \\\\ &=\\langle 2,3\\rangle \\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean Theorem, the magnitude of v<\/em><\/strong>1<\/sub> is 2, and the magnitude of v<\/em><\/strong>2<\/sub> is 3. To find the magnitude of v<\/em><\/strong>, use the formula with the position vector.\r\n
[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{|{\\boldsymbol{v}}_{1}{|}^{2}+|{\\boldsymbol{v}}_{2}{|}^{2}} \\\\ &=\\sqrt{{2}^{2}+{3}^{2}} \\\\ &=\\sqrt{13} \\end{align}[\/latex]<\/div>\r\nThe magnitude of v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].\r\n
[latex]\\begin{align}&\\tan \\theta =\\frac{{\\boldsymbol{v}}_{2}}{{\\boldsymbol{v}}_{1}} \\\\ &\\tan \\theta =\\frac{3}{2} \\\\ &\\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\end{align}[\/latex]<\/div>\r\n\"Diagram\r\n\r\nThus, the magnitude of [latex]\\boldsymbol{v}[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.\r\n\r\n
Find the components of the vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].[reveal-answer q=\"171727\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"171727\"]First find the standard position.\r\n

[latex]\\begin{align}\\boldsymbol{v}&=\\langle 7 - 3,4 - 2\\rangle \\\\ &=\\langle 4,2\\rangle \\end{align}[\/latex]<\/p>\r\n\"Diagram\r\n\r\nThe horizontal component is [latex]{\\boldsymbol{v}}_{1}=\\langle 4,0\\rangle [\/latex] and the vertical component is [latex]{\\boldsymbol{v}}_{2}=\\langle 0,2\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n

\u00a0Finding the Unit Vector in the Direction of v<\/h3>\r\nIn addition to finding a vector\u2019s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector<\/strong>. We can then preserve the direction of the original vector while simplifying calculations.\r\n\r\nUnit vectors are defined in terms of components. The horizontal unit vector is written as [latex]\\boldsymbol{i}=\\langle 1,0\\rangle [\/latex] and is directed along the positive horizontal axis. The vertical unit vector is written as [latex]\\boldsymbol{j}=\\langle 0,1\\rangle [\/latex] and is directed along the positive vertical axis.\r\n\r\n \r\n\r\n\"Plot\r\n\r\n
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unit vector<\/h3>\r\nIf [latex]\\boldsymbol{v}[\/latex] is a nonzero vector, then [latex]\\dfrac{\\boldsymbol{v}}{|\\boldsymbol{v}|}[\/latex] is a unit vector in the direction of [latex]\\boldsymbol{v}[\/latex]. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.\r\n\r\n<\/section> \r\n\r\n
Find a unit vector in the same direction as [latex]\\boldsymbol{v}=\\langle -5,12\\rangle [\/latex].[reveal-answer q=\"326943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326943\"]First, we will find the magnitude.\r\n

[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{{\\left(-5\\right)}^{2}+{\\left(12\\right)}^{2}} \\\\ &=\\sqrt{25+144} \\\\ &=\\sqrt{169} \\\\ &=13\\end{align}[\/latex]<\/p>\r\nThen we divide each component by [latex]|v|[\/latex], which gives a unit vector in the same direction as v<\/strong>:\r\n

[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=-\\frac{5}{13}i+\\frac{12}{13}j[\/latex]<\/p>\r\nor, in component form\r\n

[latex]\\frac{\\boldsymbol{v}}{|\\boldsymbol{v}|}=\\langle -\\frac{5}{13},\\frac{12}{13}\\rangle [\/latex]<\/p>\r\n\"Plot\r\n\r\nVerify that the magnitude of the unit vector equals 1. The magnitude of [latex]-\\frac{5}{13}\\boldsymbol{i}+\\frac{12}{13}\\boldsymbol{j}[\/latex] is given as\r\n

[latex]\\begin{align}\\sqrt{{\\left(-\\frac{5}{13}\\right)}^{2}+{\\left(\\frac{12}{13}\\right)}^{2}}&=\\sqrt{\\frac{25}{169}+\\frac{144}{169}} \\\\ &=\\sqrt{\\frac{169}{169}}=1 \\end{align}[\/latex]<\/p>\r\nThe vector\u00a0[latex]u=\\frac{-5}{13}[\/latex] i<\/em><\/strong> [latex]+\\frac{12}{13}[\/latex] j<\/em><\/strong> is the unit vector in the same direction as v<\/em><\/strong> [latex]=\\langle -5,12\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

\r\n\r\nGiven a vector [latex]v[\/latex] <\/strong>with initial point [latex]P=\\left({x}_{1},{y}_{1}\\right)[\/latex] and terminal point [latex]Q=\\left({x}_{2},{y}_{2}\\right)[\/latex], v<\/em><\/strong> is written as\r\n

[latex]\\boldsymbol{v}=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{1}-{y}_{2}\\right)\\boldsymbol{j}[\/latex]<\/p>\r\nThe position vector from [latex]\\left(0,0\\right)[\/latex] to [latex]\\left(a,b\\right)[\/latex], where [latex]\\left({x}_{2}-{x}_{1}\\right)=a[\/latex] and [latex]\\left({y}_{2}-{y}_{1}\\right)=b[\/latex], is written as v<\/em><\/strong> = ai<\/strong><\/em> + bj<\/strong><\/em>. This vector sum is called a linear combination of the vectors i<\/em><\/strong> and j<\/strong>.\r\n\r\nThe magnitude of v<\/em><\/strong> = ai<\/strong><\/em> + bj<\/strong><\/em> is given as [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex].\r\n\r\n\"Plot\r\n\r\n<\/div>\r\n

Given a vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]P=\\left(2,-6\\right)[\/latex] and terminal point [latex]Q=\\left(-6,6\\right)[\/latex], write the vector in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].<\/strong>[reveal-answer q=\"997243\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"997243\"]Begin by writing the general form of the vector. Then replace the coordinates with the given values.\r\n

[latex]\\begin{align}\\boldsymbol{v}&=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &=\\left(-6 - 2\\right)\\boldsymbol{i}+\\left(6-\\left(-6\\right)\\right)\\boldsymbol{j} \\\\ &=-8\\boldsymbol{i}+12\\boldsymbol{j} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Given initial point [latex]{P}_{1}=\\left(-1,3\\right)[\/latex] and terminal point [latex]{P}_{2}=\\left(2,7\\right)[\/latex], write the vector [latex]\\boldsymbol{v}[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].[reveal-answer q=\"301789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"301789\"]Begin by writing the general form of the vector. Then replace the coordinates with the given values.\r\n

[latex]\\begin{align}\\boldsymbol{v}&=\\left({x}_{2}-{x}_{1}\\right)\\boldsymbol{i}+\\left({y}_{2}-{y}_{1}\\right)\\boldsymbol{j} \\\\ &=\\left(2-\\left(-1\\right)\\right)\\boldsymbol{i}+\\left(7 - 3\\right)\\boldsymbol{j} \\\\ &=3\\boldsymbol{i}+4\\boldsymbol{j} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Write the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]P=\\left(-1,6\\right)[\/latex] and terminal point [latex]Q=\\left(7,-5\\right)[\/latex] in terms of [latex]\\boldsymbol{i}[\/latex] and [latex]\\boldsymbol{j}[\/latex].[reveal-answer q=\"445026\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"445026\"][latex]\\boldsymbol{u}=8\\boldsymbol{i} - 11\\boldsymbol{j}[\/latex][\/hidden-answer]<\/section>\r\n

Calculating the Component Form of a Vector: Direction<\/h3>\r\nWe have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using [latex]i\\text{and}j[\/latex]. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.\r\n\r\nCalculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with\u00a0|v<\/em>|\u00a0<\/strong>replacing r<\/em><\/strong>.\r\n\r\n
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vector components<\/h3>\r\nGiven a position vector [latex]\\boldsymbol{v}=\\langle x,y\\rangle [\/latex] and a direction angle [latex]\\theta [\/latex],\r\n

[latex]\\begin{align}&\\cos \\theta =\\frac{x}{|\\boldsymbol{v}|}&& \\text{and}&& \\sin \\theta =\\frac{y}{|\\boldsymbol{v}|} \\\\ &x=|\\boldsymbol{v}|\\cos \\theta &&&& y=|\\boldsymbol{v}|\\sin \\theta \\end{align}[\/latex]<\/p>\r\nThus, [latex]\\boldsymbol{v}=x\\boldsymbol{i}+y\\boldsymbol{j}=|\\boldsymbol{v}|\\cos \\theta \\boldsymbol{i}+|v|\\sin \\theta \\boldsymbol{j}[\/latex], and magnitude is expressed as [latex]|\\boldsymbol{v}|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].\r\n\r\n<\/section>

Write a vector with length 7 at an angle of 135\u00b0 to the positive\u00a0x<\/em>-axis in terms of magnitude and direction.[reveal-answer q=\"775938\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"775938\"]Using the conversion formulas [latex]x=|\\boldsymbol{v}|\\cos \\theta i[\/latex] and [latex]y=|\\boldsymbol{v}|\\sin \\theta j[\/latex], we find that\r\n

[latex]\\begin{align}x&=7\\cos \\left(135^\\circ \\right)\\boldsymbol{i} \\\\ &=-\\frac{7\\sqrt{2}}{2} \\\\ y&=7\\sin \\left(135^\\circ \\right)\\boldsymbol{j} \\\\ &=\\frac{7\\sqrt{2}}{2} \\end{align}[\/latex]<\/p>\r\nThis vector can be written as [latex]v=7\\cos \\left(135^\\circ \\right)i+7\\sin \\left(135^\\circ \\right)j[\/latex] or simplified as\r\n

[latex]v=-\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{i}+\\dfrac{7\\sqrt{2}}{2}\\boldsymbol{j}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

A vector travels from the origin to the point [latex]\\left(3,5\\right)[\/latex]. Write the vector in terms of magnitude and direction.[reveal-answer q=\"365862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"365862\"][latex]\\boldsymbol{v}=\\sqrt{34}\\cos \\left(59^\\circ \\right)\\boldsymbol{i}+\\sqrt{34}\\sin \\left(59^\\circ \\right)\\boldsymbol{j}[\/latex]\r\nMagnitude = [latex]\\sqrt{34}[\/latex]\r\n[latex]\\theta ={\\tan }^{-1}\\left(\\frac{5}{3}\\right)=59.04^\\circ [\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]149550[\/ohm_question]<\/section>","rendered":"

Finding Component Form<\/h2>\n

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph that the position vector [latex]\\langle 2,3\\rangle[\/latex] comes from adding the vectors v<\/em><\/strong>1<\/sub> and v<\/em><\/strong>2<\/sub>. We have v<\/em><\/strong>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].<\/p>\n

\n

[latex]\\begin{align}{\\boldsymbol{v}}_{1}&=\\langle 2 - 0,0 - 0\\rangle \\\\ &=\\langle 2,0\\rangle \\end{align}[\/latex]<\/p>\n

We also have v<\/em><\/strong>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].<\/p>\n

[latex]\\begin{align}{\\boldsymbol{v}}_{2}&=\\langle 0 - 0,3 - 0\\rangle \\\\ &=\\langle 0,3\\rangle \\end{align}[\/latex]<\/div>\n

Therefore, the position vector is<\/p>\n

[latex]\\begin{align}\\boldsymbol{v}&=\\langle 2+0,3+0\\rangle \\\\ &=\\langle 2,3\\rangle \\end{align}[\/latex]<\/div>\n

Using the Pythagorean Theorem, the magnitude of v<\/em><\/strong>1<\/sub> is 2, and the magnitude of v<\/em><\/strong>2<\/sub> is 3. To find the magnitude of v<\/em><\/strong>, use the formula with the position vector.<\/p>\n

[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{|{\\boldsymbol{v}}_{1}{|}^{2}+|{\\boldsymbol{v}}_{2}{|}^{2}} \\\\ &=\\sqrt{{2}^{2}+{3}^{2}} \\\\ &=\\sqrt{13} \\end{align}[\/latex]<\/div>\n

The magnitude of v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].<\/p>\n

[latex]\\begin{align}&\\tan \\theta =\\frac{{\\boldsymbol{v}}_{2}}{{\\boldsymbol{v}}_{1}} \\\\ &\\tan \\theta =\\frac{3}{2} \\\\ &\\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\end{align}[\/latex]<\/div>\n

\"Diagram<\/p>\n

Thus, the magnitude of [latex]\\boldsymbol{v}[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.<\/p>\n

Find the components of the vector [latex]\\boldsymbol{v}[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].<\/p>\n