{"id":2321,"date":"2025-08-12T21:44:51","date_gmt":"2025-08-12T21:44:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2321"},"modified":"2025-08-13T17:09:24","modified_gmt":"2025-08-13T17:09:24","slug":"2321","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/2321\/","title":{"raw":"Operations with Vectors: Learn It 2","rendered":"Operations with Vectors: Learn It 2"},"content":{"raw":"

Multiplying By a Scalar<\/h2>\r\nWhile adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar<\/strong>, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.\r\n\r\n
\r\n

scalar multiplication<\/h3>\r\nScalar multiplication<\/strong> involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle [\/latex] by [latex]k[\/latex] , we have\r\n

[latex]k\\boldsymbol{v}=\\langle ka,kb\\rangle [\/latex]<\/p>\r\nOnly the magnitude changes, unless [latex]k[\/latex] is negative, and then the vector reverses direction.\r\n\r\n<\/section>

Given vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle [\/latex], find 3v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] v<\/em>, <\/strong>and \u2212v<\/strong><\/em>.[reveal-answer q=\"962766\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962766\"]If [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle [\/latex], then\r\n

[latex]\\begin{align}3\\boldsymbol{v}&=\\langle 3\\cdot 3,3\\cdot 1\\rangle \\\\ &=\\langle 9,3\\rangle \\\\ \\frac{1}{2}\\boldsymbol{v}&=\\langle \\frac{1}{2}\\cdot 3,\\frac{1}{2}\\cdot 1\\rangle \\\\ &=\\langle \\frac{3}{2},\\frac{1}{2}\\rangle \\\\ -\\boldsymbol{v}&=\\langle -3,-1\\rangle \\end{align}[\/latex]<\/p>\r\n\"Showing\r\n

Analysis of the Solution<\/h4>\r\nNotice that the vector 3v<\/em><\/strong> is three times the length of v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]\\boldsymbol{v}[\/latex] is half the length of v<\/em><\/strong>, and \u2013v<\/em><\/strong> is the same length of v<\/em><\/strong>, but in the opposite direction.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Find the scalar multiple<\/strong>\u00a0[latex]3\\boldsymbol{u}[\/latex] given [latex]\\boldsymbol{u}[\/latex] [latex]=\\langle 5,4\\rangle [\/latex].[reveal-answer q=\"453643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453643\"][latex]3\\boldsymbol{u}=\\langle 15,12\\rangle [\/latex][\/hidden-answer]<\/section>
Given [latex]\\boldsymbol{u}=\\langle 3,-2\\rangle [\/latex] and [latex]\\boldsymbol{v}=\\langle -1,4\\rangle [\/latex], find a new vector w<\/em><\/strong> = 3u<\/em><\/strong> + 2v<\/strong>.[reveal-answer q=\"658063\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"658063\"]First, we must multiply each vector by the scalar.\r\n

[latex]\\begin{align}3\\boldsymbol{u}&=3\\langle 3,-2\\rangle \\\\ &=\\langle 9,-6\\rangle \\\\ 2\\boldsymbol{v}&=2\\langle -1,4\\rangle \\\\ &=\\langle -2,8\\rangle \\end{align}[\/latex]<\/p>\r\nThen, add the two together.\r\n

[latex]\\begin{align}\\boldsymbol{w}&=3\\boldsymbol{u}+2\\boldsymbol{v} \\\\ &=\\langle 9,-6\\rangle +\\langle -2,8\\rangle \\\\ &=\\langle 9 - 2,-6+8\\rangle \\\\ &=\\langle 7,2\\rangle \\end{align}[\/latex]<\/p>\r\nSo, [latex]\\boldsymbol{w}=\\langle 7,2\\rangle [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]173921[\/ohm_question]<\/section>
Evaluate [latex]2\\boldsymbol{a} - 3\\boldsymbol{b}[\/latex] where [latex]\\boldsymbol{a} = \\boldsymbol{i} + 5\\boldsymbol{j}[\/latex] and [latex]\\boldsymbol{b} = -2\\boldsymbol{i} + 4\\boldsymbol{j}[\/latex].<\/section><\/section>","rendered":"

Multiplying By a Scalar<\/h2>\n

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar<\/strong>, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.<\/p>\n

\n

scalar multiplication<\/h3>\n

Scalar multiplication<\/strong> involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle a,b\\rangle[\/latex] by [latex]k[\/latex] , we have<\/p>\n

[latex]k\\boldsymbol{v}=\\langle ka,kb\\rangle[\/latex]<\/p>\n

Only the magnitude changes, unless [latex]k[\/latex] is negative, and then the vector reverses direction.<\/p>\n<\/section>\n

Given vector [latex]\\boldsymbol{v}[\/latex] [latex]=\\langle 3,1\\rangle[\/latex], find 3v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] v<\/em>, <\/strong>and \u2212v<\/strong><\/em>.<\/p>\n