A General Note: Magnitude and Direction of a Vector<\/h3>\n
Given a position vector [latex]v[\/latex] [latex]=\\langle a,b\\rangle [\/latex], the magnitude is found by [latex]|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], as illustrated in Figure 5.<\/p>\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] 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<\/div>\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<\/div>\n First, find the position vector<\/strong>.\n We use the Pythagorean Theorem to find the magnitude.<\/p>\n The direction is given as<\/p>\n However, 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].\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] 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.<\/p>\n<\/div>\n As shown in Figure 7, draw the vector [latex]v[\/latex] starting at initial [latex]\\left(5,-3\\right)[\/latex] and terminal point [latex]\\left(-1,2\\right)[\/latex]. Draw the vector [latex]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.<\/p>\n Next, find and sketch the position vector for v<\/em><\/strong> and u<\/em><\/strong>. We have<\/p>\n Since the position vectors are the same, v<\/em><\/strong> and u<\/em><\/strong> are the same.<\/p>\n An 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.<\/p>\n As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives<\/p>\n However, 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].\n[caption id=\"\" align=\"aligncenter\" width=\"487\"] 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 Given a position vector [latex]v[\/latex] [latex]=\\langle a,b\\rangle[\/latex], the magnitude is found by [latex]|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], as illustrated in Figure 5.<\/p>\n Figure 5<\/b><\/p>\n<\/div>\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<\/div>\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<\/div>\n First, find the position vector<\/strong>.\n<\/p>\n We use the Pythagorean Theorem to find the magnitude.<\/p>\n The direction is given as<\/p>\n However, 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].<\/p>\n Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n 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.<\/p>\n<\/div>\n As shown in Figure 7, draw the vector [latex]v[\/latex] starting at initial [latex]\\left(5,-3\\right)[\/latex] and terminal point [latex]\\left(-1,2\\right)[\/latex]. Draw the vector [latex]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.<\/p>\n Next, find and sketch the position vector for v<\/em><\/strong> and u<\/em><\/strong>. We have<\/p>\n Since the position vectors are the same, v<\/em><\/strong> and u<\/em><\/strong> are the same.<\/p>\n An 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.<\/p>\n As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives<\/p>\n However, 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].<\/p>\n Figure 7<\/b><\/p>\n<\/div>\n<\/div>\n\n\t\t\t ![\"Standard](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3360\/2018\/06\/15192207\/CNX_Precalc_Figure_08_08_017new2.jpg\")
Figure 5<\/b>[\/caption]\nExample 3: Finding the Magnitude and Direction of a Vector<\/h3>\n
Solution<\/h3>\n
![\"Plot](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3360\/2018\/06\/15192210\/CNX_Precalc_Figure_08_08_0182.jpg\")
Figure 6<\/b>[\/caption]\n<\/div>\nExample 4: Showing That Two Vectors Are Equal<\/h3>\n
Solution<\/h3>\n
![\"Plot](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3360\/2018\/06\/15192212\/CNX_Precalc_Figure_08_08_005n2.jpg\")
Figure 7<\/b>[\/caption]\n<\/div>\n\n","rendered":"A General Note: Magnitude and Direction of a Vector<\/h3>\n
<\/p>\nExample 3: Finding the Magnitude and Direction of a Vector<\/h3>\n
Solution<\/h3>\n
<\/p>\nExample 4: Showing That Two Vectors Are Equal<\/h3>\n
Solution<\/h3>\n
<\/p>\nCandela Citations<\/h3>\n\t\t\t\t\t