{"id":2326,"date":"2025-08-12T21:46:02","date_gmt":"2025-08-12T21:46:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2326"},"modified":"2025-12-02T22:46:51","modified_gmt":"2025-12-02T22:46:51","slug":"vectors-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/vectors-learn-it-2\/","title":{"raw":"Vectors: Learn It 2","rendered":"Vectors: Learn It 2"},"content":{"raw":"

Finding Magnitude and Direction<\/h2>\r\nTo work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.\r\n\r\n
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vector magnitude and direction<\/h3>\r\nGiven a position vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle [\/latex], the magnitude is found by [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]. The direction is equal to the angle formed with the x<\/em>-axis, or with the y<\/em>-axis, depending on the application. For a position vector, the direction is found by [latex]\\tan \\theta =\\left(\\frac{b}{a}\\right)\\Rightarrow \\theta ={\\tan }^{-1}\\left(\\frac{b}{a}\\right)[\/latex].\r\n\r\n\"Standard\r\n\r\n \r\n\r\nTwo vectors v<\/em><\/strong> and u<\/em><\/strong> are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.\r\n\r\n<\/section>
Find the magnitude and direction of the vector with initial point [latex]P\\left(-8,1\\right)[\/latex] and terminal point [latex]Q\\left(-2,-5\\right)[\/latex]. Draw the vector.[reveal-answer q=\"241600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"241600\"]First, find the position vector<\/strong>.\r\n

[latex]\\begin{align}\\boldsymbol{u}&=\\langle -2,-\\left(-8\\right),-5 - 1\\rangle \\\\ &=\\langle 6,-6\\rangle \\end{align}[\/latex]<\/p>\r\nWe use the Pythagorean Theorem to find the magnitude.\r\n

[latex]\\begin{align}|\\boldsymbol{u}|&=\\sqrt{{\\left(6\\right)}^{2}+{\\left(-6\\right)}^{2}} \\\\ &=\\sqrt{72} \\\\ &=6\\sqrt{2} \\end{align}[\/latex]<\/p>\r\nThe direction is given as\r\n

[latex]\\begin{align}&\\tan \\theta =\\frac{-6}{6}=-1\\\\ &\\theta ={\\tan }^{-1}\\left(-1\\right) =-45^\\circ \\end{align}[\/latex]<\/p>\r\nHowever, the angle terminates in the fourth quadrant, so we add 360\u00b0 to obtain a positive angle. Thus, [latex]-45^\\circ +360^\\circ =315^\\circ [\/latex].\r\n\r\n\"Plot\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Show that vector v<\/em><\/strong> with initial point<\/strong> at [latex]\\left(5,-3\\right)[\/latex] and terminal point<\/strong> at [latex]\\left(-1,2\\right)[\/latex] is equal to vector u<\/em><\/strong> with initial point at [latex]\\left(-1,-3\\right)[\/latex] and terminal point at [latex]\\left(-7,2\\right)[\/latex]. Draw the position vector on the same grid as v<\/em><\/strong> and u<\/em><\/strong>. Next, find the magnitude and direction of each vector.[reveal-answer q=\"46978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"46978\"]Draw the vector [latex]\\boldsymbol{v}[\/latex] starting at initial [latex]\\left(5,-3\\right)[\/latex] and terminal point [latex]\\left(-1,2\\right)[\/latex]. Draw the vector [latex]\\boldsymbol{u}[\/latex] with initial point [latex]\\left(-1,-3\\right)[\/latex] and terminal point [latex]\\left(-7,2\\right)[\/latex]. Find the standard position for each.\r\n\r\nNext, find and sketch the position vector for v<\/em><\/strong> and u<\/em><\/strong>. We have\r\n

[latex]\\begin{align}\\boldsymbol{v}&=\\langle -1 - 5,2-\\left(-3\\right)\\rangle \\\\ &=\\langle -6,5\\rangle \\\\ \\text{ } \\\\ \\boldsymbol{u}&=\\langle -7-\\left(-1\\right),2-\\left(-3\\right)\\rangle \\\\ &=\\langle -6,5\\rangle \\end{align}[\/latex]<\/p>\r\nSince the position vectors are the same, v<\/em><\/strong> and u<\/em><\/strong> are the same.\r\n\r\nAn alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.\r\n

[latex]\\begin{align}|\\boldsymbol{v}|&=\\sqrt{{\\left(-1 - 5\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &=\\sqrt{36+25} \\\\ &=\\sqrt{61} \\\\ |\\boldsymbol{u}|&=\\sqrt{{\\left(-7-\\left(-1\\right)\\right)}^{2}+{\\left(2-\\left(-3\\right)\\right)}^{2}} \\\\ &=\\sqrt{{\\left(-6\\right)}^{2}+{\\left(5\\right)}^{2}} \\\\ &=\\sqrt{36+25} \\\\ &=\\sqrt{61} \\end{align}[\/latex]<\/p>\r\nAs the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives\r\n

[latex]\\begin{align}&\\tan \\theta =-\\frac{5}{6}\\\\ &\\theta ={\\tan }^{-1}\\left(-\\frac{5}{6}\\right) =-39.8^\\circ \\end{align}[\/latex]<\/p>\r\nHowever, we can see that the position vector terminates in the second quadrant, so we add [latex]180^\\circ [\/latex]. Thus, the direction is [latex]-39.8^\\circ +180^\\circ =140.2^\\circ [\/latex].\r\n\r\n\"Plot\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Finding Magnitude and Direction<\/h2>\n

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.<\/p>\n

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vector magnitude and direction<\/h3>\n

Given a position vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle[\/latex], the magnitude is found by [latex]|\\boldsymbol{v}|=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]. The direction is equal to the angle formed with the x<\/em>-axis, or with the y<\/em>-axis, depending on the application. For a position vector, the direction is found by [latex]\\tan \\theta =\\left(\\frac{b}{a}\\right)\\Rightarrow \\theta ={\\tan }^{-1}\\left(\\frac{b}{a}\\right)[\/latex].<\/p>\n

\"Standard<\/p>\n

 <\/p>\n

Two vectors v<\/em><\/strong> and u<\/em><\/strong> are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.<\/p>\n<\/section>\n

Find the magnitude and direction of the vector with initial point [latex]P\\left(-8,1\\right)[\/latex] and terminal point [latex]Q\\left(-2,-5\\right)[\/latex]. Draw the vector.<\/p>\n