{"id":806,"date":"2021-08-27T19:53:41","date_gmt":"2021-08-27T19:53:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=806"},"modified":"2022-10-21T00:03:02","modified_gmt":"2022-10-21T00:03:02","slug":"working-with-vectors-in-three-dimensions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/working-with-vectors-in-three-dimensions\/","title":{"raw":"Working with Vectors in Three Dimensions","rendered":"Working with Vectors in Three Dimensions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Perform vector operations in [latex]\\mathbb{R}^3[\/latex].\u00a0<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1163723968905\">Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.<\/p>\r\n\r\n<div>Three-dimensional vectors can also be represented in component form. The notation [latex]\\mathbf{v} = \\langle x,y,z \\rangle[\/latex] is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, [latex](0, 0, 0)[\/latex], and terminal point [latex](x, y, z)[\/latex]. The zero vector is [latex]\\mathbf{0} = \\langle 0,0,0 \\rangle[\/latex]. So, for example, the three dimensional vector [latex]\\mathbf{v} = \\langle 2,4,1 \\rangle[\/latex] is represented by a directed line segment from point [latex](0, 0, 0)[\/latex] to point [latex](2, 4, 1)[\/latex] (Figure 1).<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div>[caption id=\"attachment_5049\" align=\"aligncenter\" width=\"409\"]<img class=\"size-full wp-image-5049\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26193624\/Figure-2.39.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has a vector drawn. The initial point of the vector is the origin. The terminal point of the vector is (2, 4, 1). The vector is labeled \u201cv = &lt;2, 4, 1&gt;.\u201d\" width=\"409\" height=\"470\" \/> Figure 1. Vector [latex]v=\u27e82,4,1\u27e9[\/latex] is represented by a directed line segment from point [latex](0,0,0)[\/latex] to point [latex](2,4,1)[\/latex].[\/caption]<\/div>\r\n<div><\/div>\r\n<div>\r\n<p id=\"fs-id1163723674155\">Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If [latex]\\mathbf{v} = \\langle x_1,y_1,z_1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle x_2,y_2,z_2 \\rangle[\/latex] are vectors, and [latex]k[\/latex] is a scalar, then<\/p>\r\n\r\n<center>[latex]\\mathbf{v+w} = \\langle x_1 + x_2,y_1 + y_2,z_1 +z_2 \\rangle[\/latex] and [latex]k\\mathbf{v} = \\langle k x_1,k y_1,k z_1 \\rangle.[\/latex]<\/center>\r\n<p id=\"fs-id1163723721671\">If [latex]k=-1[\/latex], then [latex]k\\mathbf{v} = (-1)\\mathbf{v} [\/latex] is written as [latex]-\\mathbf{v} [\/latex], and vector subtraction is defined by\u00a0[latex]\\mathbf{v-w} = \\mathbf{v+ (-w)} = \\mathbf{v} + (-1)\\mathbf{w} [\/latex].<\/p>\r\n<p id=\"fs-id1163723901906\">The standard unit vectors extend easily into three dimensions as well\u2014[latex]\\mathbf{i} = \\langle 1,0,0 \\rangle[\/latex], [latex]\\mathbf{j} = \\langle 0,1,0 \\rangle[\/latex], and [latex]\\mathbf{k} = \\langle 0,0,1 \\rangle[\/latex]\u2014and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in [latex]\\mathbb{R}^3[\/latex] in the following ways:<\/p>\r\n\r\n<center>[latex]\\mathbf{v} = \\langle x,y,z \\rangle = x\\mathbf{i} + y\\mathbf{j} + z\\mathbf{k}[\/latex].<\/center><\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: vector representations<\/h3>\r\nLet [latex]\\overrightarrow{PQ}[\/latex] be the vector with initial point [latex]P=(3, 12, 6)[\/latex] and terminal point [latex]Q=(-4, -3, 2)[\/latex] as shown in\u00a0Figure 2. Express [latex]\\overrightarrow{PQ}[\/latex] in both component form and using standard unit vectors.\r\n\r\n[caption id=\"attachment_5058\" align=\"aligncenter\" width=\"460\"]<img class=\"size-full wp-image-5058\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204541\/Figure-2.40.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two points labeled. The first point is P = (3, 12, 6). The second point is Q = (-4, -3, 2). There is a vector from P to Q.\" width=\"460\" height=\"470\" \/> Figure 2. The vector with initial point [latex]P=(3,12,6)[\/latex] and terminal point [latex]Q=(\u22124,\u22123,2)[\/latex].[\/caption][reveal-answer q=\"682367238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"682367238\"]\r\n<p id=\"fs-id1163723674687\">In component form,<\/p>\r\n\r\n<center>[latex]\r\n\\begin{align*}\r\n\\overrightarrow{PQ} &amp;= \\langle x_2 - x_1,y_2 - y_1,z_2 - z_1 \\rangle \\\\\r\n&amp;= \\langle -4-3, -3-12,2-6 \\rangle = \\langle -7,-15,-4 \\rangle\\\\\r\n\\end{align*}\r\n[\/latex].<\/center>.\r\n<p id=\"fs-id1163723340586\">In standard unit form,<\/p>\r\n\r\n<center>[latex]\\overrightarrow{PQ} =-7\\mathbf{i} -15\\mathbf{j}-4\\mathbf{k} [\/latex].<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet [latex]S=(3, 8, 2)[\/latex] and [latex]T=(2, -1, 3)[\/latex]. Express [latex]\\overrightarrow{ST}[\/latex] in component form and in standard unit form.\r\n\r\n[reveal-answer q=\"736898611\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736898611\"]\r\n\r\n<center>[latex]\\overrightarrow{ST} = \\langle-1,-9,1 \\rangle = -\\mathbf{i}-9\\mathbf{j}+\\mathbf{k}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\nAs described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure 3).\r\n\r\n[caption id=\"attachment_5059\" align=\"aligncenter\" width=\"263\"]<img class=\"size-full wp-image-5059\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204706\/Figure-2.41.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. It has has three vectors in standard position. The first vector is labeled \u201cA.\u201d The second vector is labeled \u201cB.\u201d The third vector is labeled \u201cA + B.\u201d This vector is in between vectors A and B.\" width=\"263\" height=\"203\" \/> Figure 3. To add vectors in three dimensions, we follow the same procedures we learned for two dimensions.[\/caption]\r\n\r\nWe have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">rule: properties of vectors in space<\/h3>\r\nLet [latex]\\mathbf{v} = \\langle x_1,y_1,z_1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle x_2,y_2,z_2 \\rangle[\/latex] be vectors, and let [latex]k[\/latex] be a scalar.\r\n<strong data-effect=\"bold\">Scalar multiplication:\u00a0<\/strong> [latex]k\\mathbf{v} = \\langle kx_1,ky_1,kz_1 \\rangle[\/latex]\r\n<strong data-effect=\"bold\">Vector addition:<\/strong> [latex]\\mathbf{v+ w} = \\langle x_1,y_1,z_1 \\rangle + \\langle x_2,y_2,z_2 \\rangle = \\langle x_1 +x_2,y_1 +y_2,z_1 +z_2 \\rangle[\/latex]\r\n<strong data-effect=\"bold\">Vector subtraction:<\/strong> [latex]\\mathbf{v- w} = \\langle x_1,y_1,z_1 \\rangle - \\langle x_2,y_2,z_2 \\rangle = \\langle x_1 -x_2,y_1 -y_2,z_1 -z_2 \\rangle[\/latex]\r\n<strong data-effect=\"bold\">Vector magnitude:<\/strong> [latex]\\|\\mathbf{v}\\| = \\sqrt{x_1^2 + y_1 ^2 + z_1 ^2}[\/latex]\r\n<strong data-effect=\"bold\">Unit vector in the direction of v:<\/strong> [latex]\\frac{1}{\\|\\mathbf{v}\\|}\\mathbf{v} = \\frac{1}{\\|\\mathbf{v}\\|} \\langle x_1,y_1,z_1 \\rangle = \\langle \\frac{x_1}{\\|\\mathbf{v}\\|}\\frac{y_1}{\\|\\mathbf{v}\\|}\\frac{z_1}{\\|\\mathbf{v}\\|} \\rangle[\/latex] if\u00a0[latex]\\mathbf{v} \\neq \\mathbf{0}[\/latex].\r\n\r\n<\/div>\r\nWe have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: vector operations in three dimensions<\/h3>\r\n<p id=\"fs-id1163723762800\">Let [latex]\\mathbf{v} = \\langle -2,9,5 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle 1,-1,0 \\rangle[\/latex] (Figure 4). Find the following vectors.<\/p>\r\n\r\n<ol id=\"fs-id1163723287304\" type=\"a\">\r\n \t<li>[latex]3\\mathbf{v} -2\\mathbf{w}[\/latex]<\/li>\r\n \t<li>[latex]5\\|\\mathbf{w}\\|[\/latex]<\/li>\r\n \t<li>[latex]\\|5\\mathbf{w}\\|[\/latex]<\/li>\r\n \t<li>A unit vector in the direction of [latex]\\textbf v[\/latex]<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n[caption id=\"attachment_5060\" align=\"aligncenter\" width=\"472\"]<img class=\"size-full wp-image-5060\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204822\/Figure-2.42.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cv = &lt;-2, 9, 5&gt;.\u201d The second vector is labeled \u201cw = &lt;1, -1, 0&gt;.\u201d\" width=\"472\" height=\"470\" \/> Figure 4. The vectors [latex]{\\bf{v}}=\u27e8\u22122,9,5\u27e9[\/latex] and [latex]{\\bf{w}}=\u27e81,\u22121,0\u27e9[\/latex].[\/caption][reveal-answer q=\"762367374\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"762367374\"]\r\n<ol id=\"fs-id1163723340173\" type=\"a\">\r\n \t<li>First, use scalar multiplication of each vector, then subtract:<span data-type=\"newline\">\r\n<\/span><center>[latex]\r\n\\begin{align*}\r\n3\\mathbf{v} -2\\mathbf{w} &amp;= 3\\langle -2,9,5 \\rangle - 2\\langle 1,-1,0 \\rangle\\\\\r\n&amp;= \\langle -6,27,15 \\rangle - \\langle 2,-2,0 \\rangle\\\\\r\n&amp;= \\langle -6-2,27-(-2),15-0 \\rangle \\\\\r\n&amp;= \\langle -8,29,15 \\rangle \\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>Write the equation for the magnitude of the vector, then use scalar multiplication:<span data-type=\"newline\">\r\n<\/span>\r\n<center>[latex]5\\|\\mathbf{w}\\| = 5\\sqrt{1^2 + (-1)^2 +0^2} = 5\\sqrt{2}[\/latex].<\/center><\/li>\r\n \t<li>First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:<span data-type=\"newline\">\r\n<\/span><center>[latex]\\|5\\mathbf{w}\\| = \\| \\langle 5,-5,0 \\rangle \\|= \\sqrt{5^2 + (-5)^2 +0^2} = \\sqrt{50} = 5\\sqrt{2}[\/latex].<\/center><\/li>\r\n \t<li>Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:<span data-type=\"newline\">\r\n<\/span>\r\n<center>[latex]\r\n\\begin{align*}\r\n\\frac{\\mathbf{v}}{\\|\\mathbf{v}\\|} &amp;= \\frac{1}{\\|\\mathbf{v}\\|}\\langle -2,9,5 \\rangle \\\\\r\n&amp;= \\frac{1}{\\sqrt{(-2)^2 + 9^2 +5^2}}\\langle -2,9,5 \\rangle\\\\\r\n&amp;= \\frac{1}{\\sqrt{110}}\\langle -2,9,5 \\rangle \\\\\r\n&amp;= \\langle \\frac{-2}{\\sqrt{110}},\\frac{9}{\\sqrt{110}},\\frac{5}{\\sqrt{110}} \\rangle \\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet [latex]\\mathbf{v} = \\langle -1,-1,1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle 2,0,1 \\rangle[\/latex]. Find a unit vector in the direction of [latex]5\\mathbf{v}+3\\mathbf{w}[\/latex].\r\n\r\n[reveal-answer q=\"467325473\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467325473\"]\r\n\r\n[latex]\\langle \\frac{1}{3\\sqrt{10}},-\\frac{5}{3\\sqrt{10}},\\frac{8}{3\\sqrt{10}} \\rangle[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7713628&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4P7oGFpfAuk&amp;video_target=tpm-plugin-swx82xkd-4P7oGFpfAuk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.19_transcript.html\">\u201cCP 2.19\u201d here (opens in new window)<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]242490[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: throwing a forward pass<\/h3>\r\nA quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback\u2019s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of [latex]30^\\circ[\/latex] (see the following figure). Write the initial velocity vector of the ball, [latex]\\textbf v[\/latex], in component form.\r\n\r\n[caption id=\"attachment_5067\" align=\"aligncenter\" width=\"717\"]<img class=\"size-full wp-image-5067\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26205807\/Example-2.20.jpg\" alt=\"This figure is an image of two football players with the first one throwing the football to the second one. There is a line segment from each player to the bottom of the image. The distance from the first player to the bottom of the image is 20 yards. The distance from the second player to the same point on the bottom of the image is 15 yards. The two line segments are perpendicular. There is a broken line segment from the first player to the second player. There is a vector from the first player. The angle between the broken line and the vector is 30 degrees.\" width=\"717\" height=\"259\" \/> Figure 5. The receiver is standing [latex]20[\/latex] yd down the field and [latex]15[\/latex] yd to the quarterback\u2019s left. The quarterback throws the ball at a velocity of [latex]60[\/latex] mph toward the receiver at an upward angle of [latex]30\u00b0[\/latex][\/caption][reveal-answer q=\"117362554\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"117362554\"]\r\n\r\nThe first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector [latex]\\textbf w[\/latex] extending from the quarterback\u2019s arm to a point directly above the receiver\u2019s head at an angle of [latex]30^\\circ[\/latex] (see the following figure). This vector would have the same direction as [latex]\\textbf v[\/latex], but it may not have the right magnitude.\r\n\r\n[caption id=\"attachment_5069\" align=\"aligncenter\" width=\"727\"]<img class=\"size-full wp-image-5069\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26210150\/Example-2.20.2.jpg\" alt=\"This figure is the image of two football players with the first player throwing the football to the second player. The distance between the two players is represented with a broken line segment. There is a vector from the first player. The angle between the vector and the broken line segment is 30 degrees. There is a vertical broken line segment from the second player. Also, there is a right triangle formed from the two broken line segments and the vector from the first player is labeled \u201cw\u201d and is the hypotenuse.\" width=\"727\" height=\"338\" \/> Figure 6. Vector [latex]{\\bf{w}}[\/latex] extending from the quarterback\u2019s arm at an angle of [latex]30\u00b0[\/latex][\/caption]\r\n<p id=\"fs-id1163723143015\">The receiver is 20 yd down the field and 15 yd to the quarterback\u2019s left. Therefore, the straight-line distance from the quarterback to the receiver is<\/p>\r\n\r\n<center>[latex]\\text{Dist from QB to receiver} = \\sqrt{15^2 + 20^2} = \\sqrt{225 + 400} = \\sqrt{625} = 25 [\/latex] yd.<\/center>\r\n<p id=\"fs-id1163723995663\">We have [latex]\\frac{25}{\\|\\mathbf{w}\\|} = \\cos{30^\\circ}[\/latex]. Then the magnitude of [latex]\\textbf w[\/latex] is given by<\/p>\r\n\r\n<center>[latex]\\|\\mathbf{w}\\| = \\frac{25}{\\cos{30^\\circ}} = \\frac{25*2}{\\sqrt{3}} = \\frac{50}{\\sqrt{3}}[\/latex]yd<\/center>\r\n<p id=\"fs-id1163723292681\">and the vertical distance from the receiver to the terminal point of [latex]\\textbf w[\/latex] is<\/p>\r\n\r\n<center>[latex]\\text{Vert dist from receiver to terminal point of } \\textbf w = \\textbf w \\sin{30^\\circ} = \\frac{50}{\\sqrt{3}} * \\frac{1}{2} = \\frac{25}{\\sqrt{3}} [\/latex]yd.<\/center>\r\n<p id=\"fs-id1163723763837\">Then [latex]\\mathbf{w} = \\langle 20,15,\\frac{25}{\\sqrt{3}} \\rangle[\/latex], and has the same direction as [latex]\\textbf v[\/latex].<\/p>\r\n<p id=\"fs-id1163723735762\">Recall, though, that we calculated the magnitude of [latex]\\textbf w[\/latex]to be [latex]\\|\\textbf w\\| = \\frac{50}{\\sqrt{3}}[\/latex], and [latex]\\textbf v[\/latex] has magnitude 60 mph. So, we need to multiply vector [latex]\\textbf w[\/latex] by an appropriate constant, [latex]k[\/latex]. We want to find a value of [latex]k[\/latex] so that [latex]\\|k \\textbf{w}\\| = 60[\/latex] mph. We have<\/p>\r\n\r\n<center>[latex]\\|k \\textbf{w} \\|= k \\|\\textbf{w}\\| = k\\frac{50}{\\sqrt{3}} [\/latex] mph,<\/center>\r\n<p id=\"fs-id1163723763378\">so we want<\/p>\r\n\r\n<center>[latex]\r\n\\begin{align*}\r\nk\\frac{50}{\\sqrt{3}} &amp;= 60 \\\\\r\nk &amp;= \\frac{60\\sqrt{3}}{50}\\\\\r\nk &amp;= \\frac{6\\sqrt{3}}{5}\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>.\r\n<p id=\"fs-id1163723290222\">Then<\/p>\r\n\r\n<center>[latex]\\textbf{v} = k\\textbf{w} = k \\langle 20,15,\\frac{25}{\\sqrt{3}} \\rangle= \\frac{6\\sqrt{3}}{5}\\langle 20,15,\\frac{25}{\\sqrt{3}}\\rangle = \\langle 24\\sqrt{3},18\\sqrt{3},30\\rangle [\/latex]<\/center>.\r\n<p id=\"fs-id1163723934700\">Let\u2019s double-check that [latex]\\|\\textbf{v}\\|=60[\/latex]. We have<\/p>\r\n[latex]\\|\\textbf{v}\\| = \\sqrt{(24\\sqrt{3})^2+(18\\sqrt{3})^2+(30)^2} = \\sqrt{1728+972+900} = \\sqrt{3600} = 60 [\/latex]mph.\r\n<p id=\"fs-id1163723752770\">So, we have found the correct components for [latex]\\textbf v[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nAssume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of 40 mph and an angle of [latex]45^\\circ[\/latex]. Write the initial velocity vector of the ball, [latex]\\textbf v[\/latex], in component form.\r\n\r\n[reveal-answer q=\"934222837\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934222837\"]\r\n\r\n[latex]\\textbf{v} = \\langle16\\sqrt{2},12\\sqrt{2},20\\sqrt{2}\\rangle[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Perform vector operations in [latex]\\mathbb{R}^3[\/latex].\u00a0<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1163723968905\">Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.<\/p>\n<div>Three-dimensional vectors can also be represented in component form. The notation [latex]\\mathbf{v} = \\langle x,y,z \\rangle[\/latex] is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, [latex](0, 0, 0)[\/latex], and terminal point [latex](x, y, z)[\/latex]. The zero vector is [latex]\\mathbf{0} = \\langle 0,0,0 \\rangle[\/latex]. So, for example, the three dimensional vector [latex]\\mathbf{v} = \\langle 2,4,1 \\rangle[\/latex] is represented by a directed line segment from point [latex](0, 0, 0)[\/latex] to point [latex](2, 4, 1)[\/latex] (Figure 1).<\/div>\n<div><\/div>\n<div><\/div>\n<div>\n<div id=\"attachment_5049\" style=\"width: 419px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5049\" class=\"size-full wp-image-5049\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26193624\/Figure-2.39.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has a vector drawn. The initial point of the vector is the origin. The terminal point of the vector is (2, 4, 1). The vector is labeled \u201cv = &lt;2, 4, 1&gt;.\u201d\" width=\"409\" height=\"470\" \/><\/p>\n<p id=\"caption-attachment-5049\" class=\"wp-caption-text\">Figure 1. Vector [latex]v=\u27e82,4,1\u27e9[\/latex] is represented by a directed line segment from point [latex](0,0,0)[\/latex] to point [latex](2,4,1)[\/latex].<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div>\n<p id=\"fs-id1163723674155\">Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If [latex]\\mathbf{v} = \\langle x_1,y_1,z_1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle x_2,y_2,z_2 \\rangle[\/latex] are vectors, and [latex]k[\/latex] is a scalar, then<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{v+w} = \\langle x_1 + x_2,y_1 + y_2,z_1 +z_2 \\rangle[\/latex] and [latex]k\\mathbf{v} = \\langle k x_1,k y_1,k z_1 \\rangle.[\/latex]<\/div>\n<p id=\"fs-id1163723721671\">If [latex]k=-1[\/latex], then [latex]k\\mathbf{v} = (-1)\\mathbf{v}[\/latex] is written as [latex]-\\mathbf{v}[\/latex], and vector subtraction is defined by\u00a0[latex]\\mathbf{v-w} = \\mathbf{v+ (-w)} = \\mathbf{v} + (-1)\\mathbf{w}[\/latex].<\/p>\n<p id=\"fs-id1163723901906\">The standard unit vectors extend easily into three dimensions as well\u2014[latex]\\mathbf{i} = \\langle 1,0,0 \\rangle[\/latex], [latex]\\mathbf{j} = \\langle 0,1,0 \\rangle[\/latex], and [latex]\\mathbf{k} = \\langle 0,0,1 \\rangle[\/latex]\u2014and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in [latex]\\mathbb{R}^3[\/latex] in the following ways:<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{v} = \\langle x,y,z \\rangle = x\\mathbf{i} + y\\mathbf{j} + z\\mathbf{k}[\/latex].<\/div>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: vector representations<\/h3>\n<p>Let [latex]\\overrightarrow{PQ}[\/latex] be the vector with initial point [latex]P=(3, 12, 6)[\/latex] and terminal point [latex]Q=(-4, -3, 2)[\/latex] as shown in\u00a0Figure 2. Express [latex]\\overrightarrow{PQ}[\/latex] in both component form and using standard unit vectors.<\/p>\n<div id=\"attachment_5058\" style=\"width: 470px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5058\" class=\"size-full wp-image-5058\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204541\/Figure-2.40.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two points labeled. The first point is P = (3, 12, 6). The second point is Q = (-4, -3, 2). There is a vector from P to Q.\" width=\"460\" height=\"470\" \/><\/p>\n<p id=\"caption-attachment-5058\" class=\"wp-caption-text\">Figure 2. The vector with initial point [latex]P=(3,12,6)[\/latex] and terminal point [latex]Q=(\u22124,\u22123,2)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q682367238\">Show Solution<\/span><\/p>\n<div id=\"q682367238\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723674687\">In component form,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*}  \\overrightarrow{PQ} &= \\langle x_2 - x_1,y_2 - y_1,z_2 - z_1 \\rangle \\\\  &= \\langle -4-3, -3-12,2-6 \\rangle = \\langle -7,-15,-4 \\rangle\\\\  \\end{align*}[\/latex].<\/div>\n<p>.<\/p>\n<p id=\"fs-id1163723340586\">In standard unit form,<\/p>\n<div style=\"text-align: center;\">[latex]\\overrightarrow{PQ} =-7\\mathbf{i} -15\\mathbf{j}-4\\mathbf{k}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Let [latex]S=(3, 8, 2)[\/latex] and [latex]T=(2, -1, 3)[\/latex]. Express [latex]\\overrightarrow{ST}[\/latex] in component form and in standard unit form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736898611\">Show Solution<\/span><\/p>\n<div id=\"q736898611\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\overrightarrow{ST} = \\langle-1,-9,1 \\rangle = -\\mathbf{i}-9\\mathbf{j}+\\mathbf{k}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space (Figure 3).<\/p>\n<div id=\"attachment_5059\" style=\"width: 273px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5059\" class=\"size-full wp-image-5059\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204706\/Figure-2.41.jpg\" alt=\"This figure is the first octant of the 3-dimensional coordinate system. It has has three vectors in standard position. The first vector is labeled \u201cA.\u201d The second vector is labeled \u201cB.\u201d The third vector is labeled \u201cA + B.\u201d This vector is in between vectors A and B.\" width=\"263\" height=\"203\" \/><\/p>\n<p id=\"caption-attachment-5059\" class=\"wp-caption-text\">Figure 3. To add vectors in three dimensions, we follow the same procedures we learned for two dimensions.<\/p>\n<\/div>\n<p>We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">rule: properties of vectors in space<\/h3>\n<p>Let [latex]\\mathbf{v} = \\langle x_1,y_1,z_1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle x_2,y_2,z_2 \\rangle[\/latex] be vectors, and let [latex]k[\/latex] be a scalar.<br \/>\n<strong data-effect=\"bold\">Scalar multiplication:\u00a0<\/strong> [latex]k\\mathbf{v} = \\langle kx_1,ky_1,kz_1 \\rangle[\/latex]<br \/>\n<strong data-effect=\"bold\">Vector addition:<\/strong> [latex]\\mathbf{v+ w} = \\langle x_1,y_1,z_1 \\rangle + \\langle x_2,y_2,z_2 \\rangle = \\langle x_1 +x_2,y_1 +y_2,z_1 +z_2 \\rangle[\/latex]<br \/>\n<strong data-effect=\"bold\">Vector subtraction:<\/strong> [latex]\\mathbf{v- w} = \\langle x_1,y_1,z_1 \\rangle - \\langle x_2,y_2,z_2 \\rangle = \\langle x_1 -x_2,y_1 -y_2,z_1 -z_2 \\rangle[\/latex]<br \/>\n<strong data-effect=\"bold\">Vector magnitude:<\/strong> [latex]\\|\\mathbf{v}\\| = \\sqrt{x_1^2 + y_1 ^2 + z_1 ^2}[\/latex]<br \/>\n<strong data-effect=\"bold\">Unit vector in the direction of v:<\/strong> [latex]\\frac{1}{\\|\\mathbf{v}\\|}\\mathbf{v} = \\frac{1}{\\|\\mathbf{v}\\|} \\langle x_1,y_1,z_1 \\rangle = \\langle \\frac{x_1}{\\|\\mathbf{v}\\|}\\frac{y_1}{\\|\\mathbf{v}\\|}\\frac{z_1}{\\|\\mathbf{v}\\|} \\rangle[\/latex] if\u00a0[latex]\\mathbf{v} \\neq \\mathbf{0}[\/latex].<\/p>\n<\/div>\n<p>We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: vector operations in three dimensions<\/h3>\n<p id=\"fs-id1163723762800\">Let [latex]\\mathbf{v} = \\langle -2,9,5 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle 1,-1,0 \\rangle[\/latex] (Figure 4). Find the following vectors.<\/p>\n<ol id=\"fs-id1163723287304\" type=\"a\">\n<li>[latex]3\\mathbf{v} -2\\mathbf{w}[\/latex]<\/li>\n<li>[latex]5\\|\\mathbf{w}\\|[\/latex]<\/li>\n<li>[latex]\\|5\\mathbf{w}\\|[\/latex]<\/li>\n<li>A unit vector in the direction of [latex]\\textbf v[\/latex]<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<div id=\"attachment_5060\" style=\"width: 482px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5060\" class=\"size-full wp-image-5060\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26204822\/Figure-2.42.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cv = &lt;-2, 9, 5&gt;.\u201d The second vector is labeled \u201cw = &lt;1, -1, 0&gt;.\u201d\" width=\"472\" height=\"470\" \/><\/p>\n<p id=\"caption-attachment-5060\" class=\"wp-caption-text\">Figure 4. The vectors [latex]{\\bf{v}}=\u27e8\u22122,9,5\u27e9[\/latex] and [latex]{\\bf{w}}=\u27e81,\u22121,0\u27e9[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q762367374\">Show Solution<\/span><\/p>\n<div id=\"q762367374\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723340173\" type=\"a\">\n<li>First, use scalar multiplication of each vector, then subtract:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*}  3\\mathbf{v} -2\\mathbf{w} &= 3\\langle -2,9,5 \\rangle - 2\\langle 1,-1,0 \\rangle\\\\  &= \\langle -6,27,15 \\rangle - \\langle 2,-2,0 \\rangle\\\\  &= \\langle -6-2,27-(-2),15-0 \\rangle \\\\  &= \\langle -8,29,15 \\rangle \\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Write the equation for the magnitude of the vector, then use scalar multiplication:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]5\\|\\mathbf{w}\\| = 5\\sqrt{1^2 + (-1)^2 +0^2} = 5\\sqrt{2}[\/latex].<\/div>\n<\/li>\n<li>First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b.:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]\\|5\\mathbf{w}\\| = \\| \\langle 5,-5,0 \\rangle \\|= \\sqrt{5^2 + (-5)^2 +0^2} = \\sqrt{50} = 5\\sqrt{2}[\/latex].<\/div>\n<\/li>\n<li>Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*}  \\frac{\\mathbf{v}}{\\|\\mathbf{v}\\|} &= \\frac{1}{\\|\\mathbf{v}\\|}\\langle -2,9,5 \\rangle \\\\  &= \\frac{1}{\\sqrt{(-2)^2 + 9^2 +5^2}}\\langle -2,9,5 \\rangle\\\\  &= \\frac{1}{\\sqrt{110}}\\langle -2,9,5 \\rangle \\\\  &= \\langle \\frac{-2}{\\sqrt{110}},\\frac{9}{\\sqrt{110}},\\frac{5}{\\sqrt{110}} \\rangle \\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Let [latex]\\mathbf{v} = \\langle -1,-1,1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle 2,0,1 \\rangle[\/latex]. Find a unit vector in the direction of [latex]5\\mathbf{v}+3\\mathbf{w}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467325473\">Show Solution<\/span><\/p>\n<div id=\"q467325473\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\langle \\frac{1}{3\\sqrt{10}},-\\frac{5}{3\\sqrt{10}},\\frac{8}{3\\sqrt{10}} \\rangle[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7713628&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4P7oGFpfAuk&amp;video_target=tpm-plugin-swx82xkd-4P7oGFpfAuk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.19_transcript.html\">\u201cCP 2.19\u201d here (opens in new window)<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm242490\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=242490&theme=oea&iframe_resize_id=ohm242490&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: throwing a forward pass<\/h3>\n<p>A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback\u2019s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of [latex]30^\\circ[\/latex] (see the following figure). Write the initial velocity vector of the ball, [latex]\\textbf v[\/latex], in component form.<\/p>\n<div id=\"attachment_5067\" style=\"width: 727px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5067\" class=\"size-full wp-image-5067\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26205807\/Example-2.20.jpg\" alt=\"This figure is an image of two football players with the first one throwing the football to the second one. There is a line segment from each player to the bottom of the image. The distance from the first player to the bottom of the image is 20 yards. The distance from the second player to the same point on the bottom of the image is 15 yards. The two line segments are perpendicular. There is a broken line segment from the first player to the second player. There is a vector from the first player. The angle between the broken line and the vector is 30 degrees.\" width=\"717\" height=\"259\" \/><\/p>\n<p id=\"caption-attachment-5067\" class=\"wp-caption-text\">Figure 5. The receiver is standing [latex]20[\/latex] yd down the field and [latex]15[\/latex] yd to the quarterback\u2019s left. The quarterback throws the ball at a velocity of [latex]60[\/latex] mph toward the receiver at an upward angle of [latex]30\u00b0[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q117362554\">Show Solution<\/span><\/p>\n<div id=\"q117362554\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector [latex]\\textbf w[\/latex] extending from the quarterback\u2019s arm to a point directly above the receiver\u2019s head at an angle of [latex]30^\\circ[\/latex] (see the following figure). This vector would have the same direction as [latex]\\textbf v[\/latex], but it may not have the right magnitude.<\/p>\n<div id=\"attachment_5069\" style=\"width: 737px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5069\" class=\"size-full wp-image-5069\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/26210150\/Example-2.20.2.jpg\" alt=\"This figure is the image of two football players with the first player throwing the football to the second player. The distance between the two players is represented with a broken line segment. There is a vector from the first player. The angle between the vector and the broken line segment is 30 degrees. There is a vertical broken line segment from the second player. Also, there is a right triangle formed from the two broken line segments and the vector from the first player is labeled \u201cw\u201d and is the hypotenuse.\" width=\"727\" height=\"338\" \/><\/p>\n<p id=\"caption-attachment-5069\" class=\"wp-caption-text\">Figure 6. Vector [latex]{\\bf{w}}[\/latex] extending from the quarterback\u2019s arm at an angle of [latex]30\u00b0[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1163723143015\">The receiver is 20 yd down the field and 15 yd to the quarterback\u2019s left. Therefore, the straight-line distance from the quarterback to the receiver is<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Dist from QB to receiver} = \\sqrt{15^2 + 20^2} = \\sqrt{225 + 400} = \\sqrt{625} = 25[\/latex] yd.<\/div>\n<p id=\"fs-id1163723995663\">We have [latex]\\frac{25}{\\|\\mathbf{w}\\|} = \\cos{30^\\circ}[\/latex]. Then the magnitude of [latex]\\textbf w[\/latex] is given by<\/p>\n<div style=\"text-align: center;\">[latex]\\|\\mathbf{w}\\| = \\frac{25}{\\cos{30^\\circ}} = \\frac{25*2}{\\sqrt{3}} = \\frac{50}{\\sqrt{3}}[\/latex]yd<\/div>\n<p id=\"fs-id1163723292681\">and the vertical distance from the receiver to the terminal point of [latex]\\textbf w[\/latex] is<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Vert dist from receiver to terminal point of } \\textbf w = \\textbf w \\sin{30^\\circ} = \\frac{50}{\\sqrt{3}} * \\frac{1}{2} = \\frac{25}{\\sqrt{3}}[\/latex]yd.<\/div>\n<p id=\"fs-id1163723763837\">Then [latex]\\mathbf{w} = \\langle 20,15,\\frac{25}{\\sqrt{3}} \\rangle[\/latex], and has the same direction as [latex]\\textbf v[\/latex].<\/p>\n<p id=\"fs-id1163723735762\">Recall, though, that we calculated the magnitude of [latex]\\textbf w[\/latex]to be [latex]\\|\\textbf w\\| = \\frac{50}{\\sqrt{3}}[\/latex], and [latex]\\textbf v[\/latex] has magnitude 60 mph. So, we need to multiply vector [latex]\\textbf w[\/latex] by an appropriate constant, [latex]k[\/latex]. We want to find a value of [latex]k[\/latex] so that [latex]\\|k \\textbf{w}\\| = 60[\/latex] mph. We have<\/p>\n<div style=\"text-align: center;\">[latex]\\|k \\textbf{w} \\|= k \\|\\textbf{w}\\| = k\\frac{50}{\\sqrt{3}}[\/latex] mph,<\/div>\n<p id=\"fs-id1163723763378\">so we want<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*}  k\\frac{50}{\\sqrt{3}} &= 60 \\\\  k &= \\frac{60\\sqrt{3}}{50}\\\\  k &= \\frac{6\\sqrt{3}}{5}\\\\  \\end{align*}[\/latex]<\/div>\n<p>.<\/p>\n<p id=\"fs-id1163723290222\">Then<\/p>\n<div style=\"text-align: center;\">[latex]\\textbf{v} = k\\textbf{w} = k \\langle 20,15,\\frac{25}{\\sqrt{3}} \\rangle= \\frac{6\\sqrt{3}}{5}\\langle 20,15,\\frac{25}{\\sqrt{3}}\\rangle = \\langle 24\\sqrt{3},18\\sqrt{3},30\\rangle[\/latex]<\/div>\n<p>.<\/p>\n<p id=\"fs-id1163723934700\">Let\u2019s double-check that [latex]\\|\\textbf{v}\\|=60[\/latex]. We have<\/p>\n<p>[latex]\\|\\textbf{v}\\| = \\sqrt{(24\\sqrt{3})^2+(18\\sqrt{3})^2+(30)^2} = \\sqrt{1728+972+900} = \\sqrt{3600} = 60[\/latex]mph.<\/p>\n<p id=\"fs-id1163723752770\">So, we have found the correct components for [latex]\\textbf v[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of 40 mph and an angle of [latex]45^\\circ[\/latex]. Write the initial velocity vector of the ball, [latex]\\textbf v[\/latex], in component form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934222837\">Show Solution<\/span><\/p>\n<div id=\"q934222837\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\textbf{v} = \\langle16\\sqrt{2},12\\sqrt{2},20\\sqrt{2}\\rangle[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-806\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 2.19. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 2.19\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-806","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/806","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":55,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/806\/revisions"}],"predecessor-version":[{"id":6433,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/806\/revisions\/6433"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/806\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=806"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=806"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=806"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}