{"id":14167,"date":"2018-06-15T19:22:38","date_gmt":"2018-06-15T19:22:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/calculating-the-component-form-of-a-vector-direction\/"},"modified":"2021-10-05T16:34:37","modified_gmt":"2021-10-05T16:34:37","slug":"calculating-the-component-form-of-a-vector-direction","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/calculating-the-component-form-of-a-vector-direction\/","title":{"raw":"Calculating the Component Form of a Vector: Direction","rendered":"Calculating the Component Form of a Vector: Direction"},"content":{"raw":"We 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 [latex]|v|[\/latex] <\/strong>replacing [latex]r[\/latex]. <\/strong>\r\n
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A General Note: Vector Components in Terms of Magnitude and Direction<\/h3>\r\nGiven a position vector [latex]v=\\langle x,y\\rangle [\/latex] and a direction angle [latex]\\theta [\/latex],\r\n
[latex]\\begin{array}{lll}\\cos \\theta =\\frac{x}{|v|}\\hfill & \\text{and}\\begin{array}{cc}& \\end{array}\\hfill & \\sin \\theta =\\frac{y}{|v|}\\hfill \\\\ x=|v|\\cos \\theta \\begin{array}{cc}& \\end{array}\\hfill & \\hfill & y=|v|\\sin \\theta \\hfill \\end{array}[\/latex]<\/div>\r\nThus, [latex]v=xi+yj=|v|\\cos \\theta i+|v|\\sin \\theta j[\/latex], and magnitude is expressed as [latex]|v|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].\r\n\r\n<\/div>\r\n
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Example 13: Writing a Vector in Terms of Magnitude and Direction<\/h3>\r\nWrite a vector with length 7 at an angle of 135\u00b0 to the positive\u00a0x<\/em>-axis in terms of magnitude and direction.\r\n\r\n<\/div>\r\n
\r\n

Solution<\/h3>\r\nUsing the conversion formulas [latex]x=|v|\\cos \\theta i[\/latex] and [latex]y=|v|\\sin \\theta j[\/latex], we find that\r\n
[latex]\\begin{array}{l}x=7\\cos \\left(135^\\circ \\right)i\\hfill \\\\ =-\\frac{7\\sqrt{2}}{2}\\hfill \\\\ y=7\\sin \\left(135^\\circ \\right)j\\hfill \\\\ =\\frac{7\\sqrt{2}}{2}\\hfill \\end{array}[\/latex]<\/div>\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=-\\frac{7\\sqrt{2}}{2}i+\\frac{7\\sqrt{2}}{2}j[\/latex]<\/div>\r\n<\/div>\r\n
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Try It 4<\/h3>\r\nA vector travels from the origin to the point [latex]\\left(3,5\\right)[\/latex]. Write the vector in terms of magnitude and direction.\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
<\/div>","rendered":"

We 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.<\/p>\n

Calculating 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 [latex]|v|[\/latex] <\/strong>replacing [latex]r[\/latex]. <\/strong><\/p>\n

\n

A General Note: Vector Components in Terms of Magnitude and Direction<\/h3>\n

Given a position vector [latex]v=\\langle x,y\\rangle[\/latex] and a direction angle [latex]\\theta[\/latex],<\/p>\n

[latex]\\begin{array}{lll}\\cos \\theta =\\frac{x}{|v|}\\hfill & \\text{and}\\begin{array}{cc}& \\end{array}\\hfill & \\sin \\theta =\\frac{y}{|v|}\\hfill \\\\ x=|v|\\cos \\theta \\begin{array}{cc}& \\end{array}\\hfill & \\hfill & y=|v|\\sin \\theta \\hfill \\end{array}[\/latex]<\/div>\n

Thus, [latex]v=xi+yj=|v|\\cos \\theta i+|v|\\sin \\theta j[\/latex], and magnitude is expressed as [latex]|v|=\\sqrt{{x}^{2}+{y}^{2}}[\/latex].<\/p>\n<\/div>\n

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Example 13: Writing a Vector in Terms of Magnitude and Direction<\/h3>\n

Write a vector with length 7 at an angle of 135\u00b0 to the positive\u00a0x<\/em>-axis in terms of magnitude and direction.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

Using the conversion formulas [latex]x=|v|\\cos \\theta i[\/latex] and [latex]y=|v|\\sin \\theta j[\/latex], we find that<\/p>\n

[latex]\\begin{array}{l}x=7\\cos \\left(135^\\circ \\right)i\\hfill \\\\ =-\\frac{7\\sqrt{2}}{2}\\hfill \\\\ y=7\\sin \\left(135^\\circ \\right)j\\hfill \\\\ =\\frac{7\\sqrt{2}}{2}\\hfill \\end{array}[\/latex]<\/div>\n

This 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<\/p>\n

[latex]v=-\\frac{7\\sqrt{2}}{2}i+\\frac{7\\sqrt{2}}{2}j[\/latex]<\/div>\n<\/div>\n
\n

Try It 4<\/h3>\n

A vector travels from the origin to the point [latex]\\left(3,5\\right)[\/latex]. Write the vector in terms of magnitude and direction.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n

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