{"id":3641,"date":"2023-01-28T14:20:26","date_gmt":"2023-01-28T14:20:26","guid":{"rendered":"http:\/\/calculuscoaches.com\/?page_id=3641"},"modified":"2023-08-03T14:22:02","modified_gmt":"2023-08-03T14:22:02","slug":"projection-vector-calculator","status":"publish","type":"page","link":"https:\/\/calculuscoaches.com\/index.php\/projection-vector-calculator\/","title":{"rendered":"Vector Projections in 3D Space: Intuitive Explanations and Examples"},"content":{"rendered":"\n<div style=\"border: 1px solid black; padding: 10px;\">1. <strong>Scalar Projection:<\/strong> The first step is to find the scalar projection of <strong>a<\/strong> onto <strong>b<\/strong>. This is a measure of how much of <strong>a<\/strong> lies in the direction of <strong>b<\/strong>. It&#8217;s calculated as the dot product of <strong>a<\/strong> and <strong>b<\/strong>, divided by the magnitude of <strong>b<\/strong>: \\( c = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{||\\mathbf{b}||} \\) Here, \\( \\mathbf{a} \\cdot \\mathbf{b} \\) is the dot product of <strong>a<\/strong> and <strong>b<\/strong>, and \\( ||\\mathbf{b}|| \\) is the magnitude of <strong>b<\/strong>. The dot product gives a measure of the extent to which <strong>a<\/strong> and <strong>b<\/strong> point in the same direction, and dividing by the magnitude of <strong>b<\/strong> scales this measure by the length of <strong>b<\/strong>. <br \/>2. <strong>Vector Projection:<\/strong> The second step is to multiply this scalar projection by the unit vector in the direction of <strong>b<\/strong>. The unit vector of <strong>b<\/strong> is given by \\( \\mathbf{b} \/ ||\\mathbf{b}|| \\), so the vector projection of <strong>a<\/strong> onto <strong>b<\/strong> is: \\( \\text{proj}_{\\mathbf{b}}(\\mathbf{a}) = c \\times \\left(\\frac{\\mathbf{b}}{||\\mathbf{b}||}\\right) = \\left(\\frac{\\mathbf{a} \\cdot \\mathbf{b}}{||\\mathbf{b}||}\\right) \\times \\left(\\frac{\\mathbf{b}}{||\\mathbf{b}||}\\right) = \\left(\\frac{\\mathbf{a} \\cdot \\mathbf{b}}{||\\mathbf{b}||^2}\\right) \\mathbf{b} \\) This gives a new vector that points in the direction of <strong>b<\/strong> and has length equal to the scalar projection of <strong>a<\/strong> onto <strong>b<\/strong>.<\/div>\n<p><iframe loading=\"lazy\" style=\"border: 1px solid #e4e4e4; border-radius: 4px;\" src=\"https:\/\/www.geogebra.org\/classic\/qgkm5p5p?embed\" width=\"800\" height=\"600\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div style=\"border: 1px solid black; padding: 10px;\">Consider two vectors in 3D:<br \/><strong>a<\/strong> = (3, -2, 5)<br \/><strong>b<\/strong> = (1, 4, -2)<br \/>Step 1: Calculate the dot product of <strong>a<\/strong> and <strong>b<\/strong>.<br \/>The dot product of two vectors <strong>a<\/strong> and <strong>b<\/strong> is given by:<br \/>\\( \\mathbf{a} \\cdot \\mathbf{b} = a_x \\cdot b_x + a_y \\cdot b_y + a_z \\cdot b_z \\)<br \/>Substitute the values:<br \/>\\( \\mathbf{a} \\cdot \\mathbf{b} = (3 \\cdot 1) + (-2 \\cdot 4) + (5 \\cdot -2) = 3 &#8211; 8 &#8211; 10 = -15 \\)<br \/>Step 2: Calculate the magnitude of <strong>b<\/strong>.<br \/>The magnitude of a vector <strong>b<\/strong> is given by:<br \/>\\( ||\\mathbf{b}|| = \\sqrt{b_x^2 + b_y^2 + b_z^2} \\)<br \/>Substitute the values:<br \/>\\( ||\\mathbf{b}|| = \\sqrt{1^2 + 4^2 + (-2)^2} = \\sqrt{1 + 16 + 4} = \\sqrt{21} \\)<br \/>Step 3: Calculate the scalar projection of <strong>a<\/strong> onto <strong>b<\/strong>.<br \/>The scalar projection of <strong>a<\/strong> onto <strong>b<\/strong> is given by:<br \/>\\( c = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{||\\mathbf{b}||} \\)<br \/>Substitute the values:<br \/>\\( c = \\frac{-15}{\\sqrt{21}} \\)<br \/>Step 4: Calculate the vector projection of <strong>a<\/strong> onto <strong>b<\/strong>.<br \/>The vector projection of <strong>a<\/strong> onto <strong>b<\/strong> is given by:<br \/>\\( \\text{proj}_{\\mathbf{b}}(\\mathbf{a}) = c \\times \\left(\\frac{\\mathbf{b}}{||\\mathbf{b}||}\\right) \\)<br \/>Substitute the values and multiply:<br \/>\\( \\text{proj}_{\\mathbf{b}}(\\mathbf{a}) = \\frac{-15}{\\sqrt{21}} \\times \\left(\\frac{1}{\\sqrt{21}}, \\frac{4}{\\sqrt{21}}, \\frac{-2}{\\sqrt{21}}\\right) \\)<br \/>Simplify:<br \/>\\( \\text{proj}_{\\mathbf{b}}(\\mathbf{a}) = \\left(\\frac{-15}{21}, \\frac{-60}{21}, \\frac{30}{21}\\right) \\)<br \/>Step 5: Simplify the result.<br \/>The simplified vector projection is:<br \/>\\( \\text{proj}_{\\mathbf{b}}(\\mathbf{a}) = \\left(\\frac{-5}{7}, \\frac{-20}{7}, \\frac{10}{7}\\right) \\)<br \/>So, the projection of vector <strong>a<\/strong> onto vector <strong>b<\/strong> in 3D is \\( \\left(\\frac{-5}{7}, \\frac{-20}{7}, \\frac{10}{7}\\right) \\). This means that the component of <strong>a<\/strong> in the direction of <strong>b<\/strong> is \\( \\frac{-5}{7} \\) in the x-direction, \\( \\frac{-20}{7} \\) in the y-direction, and \\( \\frac{10}{7} \\) in the z-direction.<\/div>\n<div style=\"border: 1px solid black; padding: 10px;\">Consider two vectors in 3D:<br \/><strong>u<\/strong> = (2, 1, 3)<br \/><strong>v<\/strong> = (3, -2, 4)<br \/>Step 1: Calculate the dot product of <strong>u<\/strong> and <strong>v<\/strong>.<br \/>The dot product of two vectors <strong>u<\/strong> and <strong>v<\/strong> is given by:<br \/>\\( \\mathbf{u} \\cdot \\mathbf{v} = u_x \\cdot v_x + u_y \\cdot v_y + u_z \\cdot v_z \\)<br \/>Substitute the values:<br \/>\\( \\mathbf{u} \\cdot \\mathbf{v} = (2 \\cdot 3) + (1 \\cdot -2) + (3 \\cdot 4) = 6 &#8211; 2 + 12 = 16 \\)<br \/>Step 2: Calculate the magnitude of <strong>v<\/strong>.<br \/>The magnitude of a vector <strong>v<\/strong> is given by:<br \/>\\( ||\\mathbf{v}|| = \\sqrt{v_x^2 + v_y^2 + v_z^2} \\)<br \/>Substitute the values:<br \/>\\( ||\\mathbf{v}|| = \\sqrt{3^2 + (-2)^2 + 4^2} = \\sqrt{9 + 4 + 16} = \\sqrt{29} \\)<br \/>Step 3: Calculate the scalar projection of <strong>u<\/strong> onto <strong>v<\/strong>.<br \/>The scalar projection of <strong>u<\/strong> onto <strong>v<\/strong> is given by:<br \/>\\( c = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{v}||} \\)<br \/>Substitute the values:<br \/>\\( c = \\frac{16}{\\sqrt{29}} \\)<br \/>Step 4: Calculate the vector projection of <strong>u<\/strong> onto <strong>v<\/strong>.<br \/>The vector projection of <strong>u<\/strong> onto <strong>v<\/strong> is given by:<br \/>\\( \\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = c \\times \\left(\\frac{\\mathbf{v}}{||\\mathbf{v}||}\\right) \\)<br \/>Substitute the values and multiply:<br \/>\\( \\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\frac{16}{\\sqrt{29}} \\times \\left(\\frac{3}{\\sqrt{29}}, \\frac{-2}{\\sqrt{29}}, \\frac{4}{\\sqrt{29}}\\right) \\)<br \/>Simplify:<br \/>\\( \\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\left(\\frac{48}{29}, \\frac{-32}{29}, \\frac{64}{29}\\right) \\)<br \/>Step 5: Simplify the result.<br \/>The simplified vector projection is:<br \/>\\( \\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\left(\\frac{48}{29}, \\frac{-32}{29}, \\frac{64}{29}\\right) \\)<br \/>So, the projection of vector <strong>u<\/strong> onto vector <strong>v<\/strong> in 3D is \\( \\left(\\frac{48}{29}, \\frac{-32}{29}, \\frac{64}{29}\\right) \\). This means that the component of <strong>u<\/strong> in the direction of <strong>v<\/strong> is \\( \\frac{48}{29} \\) in the x-direction, \\( \\frac{-32}{29} \\) in the y-direction, and \\( \\frac{64}{29} \\) in the z-direction.<\/div>\n<div style=\"border: 1px solid black; padding: 10px;\">\n<p>\u00a0<\/p>\n<\/div>\n<div style=\"border: 1px solid black; padding: 10px; line-height: 1.6;\">\n<p>Imagine a tiny person walking along vectors in 3D space to understand the concept of vector projection.<\/p>\n<p>Imagine the tiny person starting at the origin (0, 0, 0) and standing on vector <strong>u<\/strong> = (2, 1, 3). This vector represents the direction the person is facing and the distance they move in each step.<\/p>\n<p>Now, let&#8217;s introduce another vector <strong>v<\/strong> = (3, -2, 4). This vector represents the direction of a path or a direction the person wants to walk towards.<\/p>\n<p>Step 1: The dot product of <strong>u<\/strong> and <strong>v<\/strong> (16) tells us how much the tiny person is aligned with the direction of vector <strong>v<\/strong>. If the person is walking exactly in the direction of <strong>v<\/strong>, the dot product will be the magnitude of <strong>v<\/strong> (which is \\( \\sqrt{29} \\)).<\/p>\n<p>Step 2: The magnitude of <strong>v<\/strong> (\\( \\sqrt{29} \\)) gives the length of the path the tiny person wants to walk on. It represents how far the person wants to walk along the direction of <strong>v<\/strong>.<\/p>\n<p>Step 3: The scalar projection (\\( c = \\frac{16}{\\sqrt{29}} \\)) tells us how far the tiny person has actually moved in the direction of <strong>v<\/strong>. If the scalar projection is positive, the person has moved in the same direction as <strong>v<\/strong>. If it&#8217;s negative, the person has moved in the opposite direction of <strong>v<\/strong>.<\/p>\n<p>Step 4: The vector projection of <strong>u<\/strong> onto <strong>v<\/strong> (\\( \\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\left(\\frac{48}{29}, \\frac{-32}{29}, \\frac{64}{29}\\right) \\)) gives the actual displacement of the tiny person along the path represented by <strong>v<\/strong>. It tells us how far the person has moved along <strong>v<\/strong> from the starting point (0, 0, 0).<\/p>\n<p>So, if the tiny person is walking along the vector <strong>u<\/strong> = (2, 1, 3) and wants to move along the direction represented by <strong>v<\/strong> = (3, -2, 4), the vector projection <strong>\\(\\left(\\frac{48}{29}, \\frac{-32}{29}, \\frac{64}{29}\\right)\\)<\/strong> tells us the displacement of the person along the path <strong>v<\/strong>.<\/p>\n<p>In simple terms, the vector projection shows the tiny person&#8217;s progress along the path or direction represented by vector <strong>v<\/strong> while starting from the origin and walking along vector <strong>u<\/strong>. It allows us to understand how much the person is moving in the desired direction and how far they have gone along that direction.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>1. Scalar Projection: The first step is to find the scalar projection of a onto b. This is a measure of how much of a lies in the direction of b. It&#8217;s calculated as the dot product of a and b, divided by the magnitude of b: \\( c = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{||\\mathbf{b}||} \\) Here, &hellip; <\/p>\n<p class=\"link-more\"><a href=\"https:\/\/calculuscoaches.com\/index.php\/projection-vector-calculator\/\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\"> &#8220;Vector Projections in 3D Space: Intuitive Explanations and Examples&#8221;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":155,"comment_status":"open","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-3641","page","type-page","status-publish","hentry"],"aioseo_notices":[],"aioseo_head":"\n\t\t<!-- All in One SEO 5.0.4 - aioseo.com -->\n\t<meta name=\"description\" content=\"1. Scalar Projection: The first step is to find the scalar projection of a onto b. 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