{"id":240994,"date":"2025-07-03T14:10:51","date_gmt":"2025-07-03T14:10:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=240994"},"modified":"2025-07-03T14:10:53","modified_gmt":"2025-07-03T14:10:53","slug":"find-the-projection-of-v-onto-u-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/find-the-projection-of-v-onto-u-2\/","title":{"rendered":"Find the projection of v onto u."},"content":{"rendered":"\n<p>Find the projection of v onto u.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-65.png\" alt=\"\" class=\"wp-image-241001\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the correct answer:<\/p>\n\n\n\n<p>proj_u(v) <\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>&#91; 4\/3 ]\n&#91;-2\/3 ]\n&#91;-4\/3 ]<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>To find the projection of a vector&nbsp;<strong>v<\/strong>&nbsp;onto another vector&nbsp;<strong>u<\/strong>, we use the formula:<\/p>\n\n\n\n<p>proj_u(<strong>v<\/strong>) = ( (<strong>v<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>) \/ (<strong>u<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>) ) *&nbsp;<strong>u<\/strong><\/p>\n\n\n\n<p>This formula calculates a scalar value (the fraction part) and then multiplies it by the vector&nbsp;<strong>u<\/strong>. The resulting vector is the component of&nbsp;<strong>v<\/strong>&nbsp;that lies in the direction of&nbsp;<strong>u<\/strong>. The calculation can be broken down into three main steps.<\/p>\n\n\n\n<p><strong>Step 1: Calculate the dot product of v and u (v \u00b7 u).<\/strong><\/p>\n\n\n\n<p>The dot product is found by multiplying the corresponding components of the two vectors and then summing the results.<\/p>\n\n\n\n<p><strong>v<\/strong>&nbsp;= [2, 2, -2]<br><strong>u<\/strong>&nbsp;= [1\/4, -1\/8, -1\/4]<\/p>\n\n\n\n<p><strong>v<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= (2 * 1\/4) + (2 * -1\/8) + (-2 * -1\/4)<br><strong>v<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 2\/4 &#8211; 2\/8 + 2\/4<br><strong>v<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 1\/2 &#8211; 1\/4 + 1\/2<br><strong>v<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 1 &#8211; 1\/4 = 3\/4<\/p>\n\n\n\n<p><strong>Step 2: Calculate the dot product of u with itself (u \u00b7 u).<\/strong><\/p>\n\n\n\n<p>This is equivalent to finding the squared magnitude of&nbsp;<strong>u<\/strong>.<\/p>\n\n\n\n<p><strong>u<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= (1\/4 * 1\/4) + (-1\/8 * -1\/8) + (-1\/4 * -1\/4)<br><strong>u<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 1\/16 + 1\/64 + 1\/16<br>To add these fractions, we find a common denominator, which is 64.<br><strong>u<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 4\/64 + 1\/64 + 4\/64<br><strong>u<\/strong>&nbsp;\u00b7&nbsp;<strong>u<\/strong>&nbsp;= 9\/64<\/p>\n\n\n\n<p><strong>Step 3: Calculate the projection.<\/strong><\/p>\n\n\n\n<p>Now, we substitute the values from the first two steps into the projection formula.<\/p>\n\n\n\n<p>proj_u(<strong>v<\/strong>) = ( (3\/4) \/ (9\/64) ) *&nbsp;<strong>u<\/strong><\/p>\n\n\n\n<p>First, we simplify the scalar fraction. Dividing by a fraction is the same as multiplying by its reciprocal.<\/p>\n\n\n\n<p>Scalar = (3\/4) * (64\/9) = (3 * 64) \/ (4 * 9) = 192 \/ 36<br>Simplifying this fraction gives 16\/3.<\/p>\n\n\n\n<p>Finally, we multiply this scalar by the vector&nbsp;<strong>u<\/strong>.<\/p>\n\n\n\n<p>proj_u(<strong>v<\/strong>) = (16\/3) * [1\/4, -1\/8, -1\/4]<br>proj_u(<strong>v<\/strong>) = [ (16\/3) * (1\/4), (16\/3) * (-1\/8), (16\/3) * (-1\/4) ]<br>proj_u(<strong>v<\/strong>) = [ 16\/12, -16\/24, -16\/12 ]<\/p>\n\n\n\n<p>Simplifying each component gives the final answer:<br>proj_u(<strong>v<\/strong>) = [ 4\/3, -2\/3, -4\/3 ]<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-63.jpeg\" alt=\"\" class=\"wp-image-241009\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the projection of v onto u. The Correct Answer and Explanation is: Here is the correct answer: proj_u(v) Explanation: To find the projection of a vector&nbsp;v&nbsp;onto another vector&nbsp;u, we use the formula: proj_u(v) = ( (v&nbsp;\u00b7&nbsp;u) \/ (u&nbsp;\u00b7&nbsp;u) ) *&nbsp;u This formula calculates a scalar value (the fraction part) and then multiplies it by [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-240994","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240994","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=240994"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240994\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240994"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240994"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240994"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}