{"id":14161,"date":"2018-06-15T19:22:30","date_gmt":"2018-06-15T19:22:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/performing-vector-addition-and-scalar-multiplication\/"},"modified":"2021-10-05T16:32:27","modified_gmt":"2021-10-05T16:32:27","slug":"performing-vector-addition-and-scalar-multiplication","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/performing-vector-addition-and-scalar-multiplication\/","title":{"raw":"Performing Vector Addition and Scalar Multiplication","rendered":"Performing Vector Addition and Scalar Multiplication"},"content":{"raw":"Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector [latex]u[\/latex] [latex]=\\langle x,y\\rangle [\/latex] as an arrow or directed line segment from the origin to the point [latex]\\left(x,y\\right)[\/latex], vectors can be situated anywhere in the plane. The sum of two vectors u<\/em><\/strong> and v<\/em><\/strong>, or vector addition<\/strong>, produces a third vector u<\/em><\/strong>+ v<\/em><\/strong>, the resultant<\/strong> vector.\r\n\r\nTo find u<\/em><\/strong> + v<\/em><\/strong>, we first draw the vector u<\/em><\/strong>, and from the terminal end of u<\/em><\/strong>, we drawn the vector v<\/em><\/strong>. In other words, we have the initial point of v<\/em><\/strong> meet the terminal end of u<\/em><\/strong>. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u<\/em><\/strong> + v<\/em><\/strong> is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u<\/em><\/strong> to the end of v<\/em><\/strong> in a straight path, as shown in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Diagrams Figure 8<\/b>[\/caption]\r\n\r\nVector subtraction is similar to vector addition. To find u<\/em><\/strong> \u2212 v<\/em><\/strong>, view it as u<\/em><\/strong> + (\u2212v<\/em><\/strong>). Adding \u2212v<\/em><\/strong> is reversing direction of v<\/em><\/strong> and adding it to the end of u<\/em><\/strong>. The new vector begins at the start of u<\/em><\/strong> and stops at the end point of \u2212v<\/em><\/strong>. See Figure 9\u00a0for a visual that compares vector addition and vector subtraction using parallelograms<\/strong>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Showing Figure 9<\/b>[\/caption]\r\n\r\n
\r\n

Example 5: Adding and Subtracting Vectors<\/h3>\r\nGiven [latex]u[\/latex] <\/strong> [latex]=\\langle 3,-2\\rangle [\/latex] and [latex]v[\/latex] <\/strong> [latex]=\\langle -1,4\\rangle [\/latex], find two new vectors u<\/em><\/strong> + v<\/em><\/strong>, and u<\/em><\/strong> \u2212 v<\/strong>.\r\n\r\n<\/div>\r\n
\r\n

Solution<\/h3>\r\nTo find the sum of two vectors, we add the components. Thus,\r\n
[latex]\\begin{array}{l}u+v=\\langle 3,-2\\rangle +\\langle -1,4\\rangle \\hfill \\\\ =\\langle 3+\\left(-1\\right),-2+4\\rangle \\hfill \\\\ =\\langle 2,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nSee Figure 10(a).\r\n\r\nTo find the difference of two vectors, add the negative components of [latex]v[\/latex] to [latex]u[\/latex]. Thus,\r\n
[latex]\\begin{array}{l}u+\\left(-v\\right)=\\langle 3,-2\\rangle +\\langle 1,-4\\rangle \\hfill \\\\ =\\langle 3+1,-2+\\left(-4\\right)\\rangle \\hfill \\\\ =\\langle 4,-6\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nSee Figure 10(b).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"Further Figure 10.<\/b> (a) Sum of two vectors (b) Difference of two vectors[\/caption]\r\n\r\n<\/div>\r\n

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

A General Note: 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]v[\/latex] [latex]=\\langle a,b\\rangle [\/latex] by [latex]k[\/latex] , we have\r\n
[latex]kv=\\langle ka,kb\\rangle [\/latex]<\/div>\r\nOnly the magnitude changes, unless [latex]k[\/latex] is negative, and then the vector reverses direction.\r\n\r\n<\/div>\r\n
\r\n

Example 6: Performing Scalar Multiplication<\/h3>\r\nGiven vector [latex]v[\/latex] [latex]=\\langle 3,1\\rangle [\/latex], find 3v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]v[\/latex], <\/strong>and \u2212v<\/strong>.\r\n\r\n<\/div>\r\n
\r\n

Solution<\/h3>\r\nSee Figure 11\u00a0for a geometric interpretation. If [latex]v[\/latex] <\/strong> [latex]=\\langle 3,1\\rangle [\/latex], then\r\n
[latex]\\begin{array}{l}3v=\\langle 3\\cdot 3,3\\cdot 1\\rangle \\hfill \\\\ =\\langle 9,3\\rangle \\hfill \\\\ \\frac{1}{2}v=\\langle \\frac{1}{2}\\cdot 3,\\frac{1}{2}\\cdot 1\\rangle \\hfill \\\\ =\\langle \\frac{3}{2},\\frac{1}{2}\\rangle \\hfill \\\\ -v=\\langle -3,-1\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Showing Figure 11<\/b>[\/caption]\r\n\r\n<\/div>\r\n
\r\n

Analysis of the Solution<\/h3>\r\nNotice that the vector 3v<\/em><\/strong> is three times the length of v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]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<\/div>\r\n
\r\n

Try It 2<\/h3>\r\nFind the scalar multiple<\/strong> 3 [latex]u[\/latex] given [latex]u[\/latex] [latex]=\\langle 5,4\\rangle [\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
\r\n

Example 7: Using Vector Addition and Scalar Multiplication to Find a New Vector<\/h3>\r\nGiven [latex]u=\\langle 3,-2\\rangle [\/latex] and [latex]v=\\langle -1,4\\rangle [\/latex], find a new vector w<\/em><\/strong> = 3u<\/em><\/strong> + 2v<\/strong>.\r\n\r\n<\/div>\r\n
\r\n

Solution<\/h3>\r\nFirst, we must multiply each vector by the scalar.\r\n
[latex]\\begin{array}{l}3u=3\\langle 3,-2\\rangle \\hfill \\\\ =\\langle 9,-6\\rangle \\hfill \\\\ 2v=2\\langle -1,4\\rangle \\hfill \\\\ =\\langle -2,8\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nThen, add the two together.\r\n
[latex]\\begin{array}{l}w=3u+2v\\hfill \\\\ =\\langle 9,-6\\rangle +\\langle -2,8\\rangle \\hfill \\\\ =\\langle 9 - 2,-6+8\\rangle \\hfill \\\\ =\\langle 7,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nSo, [latex]w=\\langle 7,2\\rangle [\/latex].\r\n\r\n<\/div>\r\n

Finding Component Form<\/h2>\r\nIn some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph in Figure 12\u00a0that the position vector [latex]\\langle 2,3\\rangle [\/latex] comes from adding the vectors v<\/em><\/strong>1<\/sub> and v<\/em><\/strong>2<\/sub>. We have v<\/em><\/strong>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].\r\n
\r\n

[latex]\\begin{array}{l}{v}_{1}=\\langle 2 - 0,0 - 0\\rangle \\hfill \\\\ =\\langle 2,0\\rangle \\hfill \\end{array}[\/latex]<\/p>\r\nWe also have v<\/em><\/strong>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].\r\n

[latex]\\begin{array}{l}{v}_{2}=\\langle 0 - 0,3 - 0\\rangle \\hfill \\\\ =\\langle 0,3\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nTherefore, the position vector is\r\n
[latex]\\begin{array}{l}v=\\langle 2+0,3+0\\rangle \\hfill \\\\ =\\langle 2,3\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nUsing the Pythagorean Theorem, the magnitude of v<\/em><\/strong>1<\/sub> is 2, and the magnitude of v<\/em><\/strong>2<\/sub> is 3. To find the magnitude of v<\/em><\/strong>, use the formula with the position vector.\r\n
[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}|v|=\\sqrt{|{v}_{1}{|}^{2}+|{v}_{2}{|}^{2}}\\hfill \\\\ \\begin{array}{l}=\\sqrt{{2}^{2}+{3}^{2}}\\hfill \\\\ =\\sqrt{13}\\hfill \\end{array}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\nThe magnitude of v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].\r\n
[latex]\\begin{array}{l}\\tan \\theta =\\frac{{v}_{2}}{{v}_{1}}\\hfill \\\\ \\tan \\theta =\\frac{3}{2}\\hfill \\\\ \\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\hfill \\end{array}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Diagram Figure 12<\/b>[\/caption]\r\n\r\nThus, the magnitude of [latex]v[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.\r\n
\r\n

Example 8: Finding the Components of the Vector<\/h3>\r\nFind the components of the vector [latex]v[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].\r\n\r\n<\/div>\r\n
\r\n

Solution<\/h3>\r\nFirst find the standard position.\r\n
[latex]\\begin{array}{l}v=\\langle 7 - 3,4 - 2\\rangle \\hfill \\\\ =\\langle 4,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\r\nSee the illustration in Figure 13.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Diagram Figure 13<\/b>[\/caption]\r\n\r\nThe horizontal component is [latex]{v}_{1}=\\langle 4,0\\rangle [\/latex] and the vertical component is [latex]{v}_{2}=\\langle 0,2\\rangle [\/latex].\r\n\r\n<\/div>\r\n<\/div>","rendered":"

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector [latex]u[\/latex] [latex]=\\langle x,y\\rangle[\/latex] as an arrow or directed line segment from the origin to the point [latex]\\left(x,y\\right)[\/latex], vectors can be situated anywhere in the plane. The sum of two vectors u<\/em><\/strong> and v<\/em><\/strong>, or vector addition<\/strong>, produces a third vector u<\/em><\/strong>+ v<\/em><\/strong>, the resultant<\/strong> vector.<\/p>\n

To find u<\/em><\/strong> + v<\/em><\/strong>, we first draw the vector u<\/em><\/strong>, and from the terminal end of u<\/em><\/strong>, we drawn the vector v<\/em><\/strong>. In other words, we have the initial point of v<\/em><\/strong> meet the terminal end of u<\/em><\/strong>. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u<\/em><\/strong> + v<\/em><\/strong> is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u<\/em><\/strong> to the end of v<\/em><\/strong> in a straight path, as shown in Figure 8.<\/p>\n

\"Diagrams<\/p>\n

Figure 8<\/b><\/p>\n<\/div>\n

Vector subtraction is similar to vector addition. To find u<\/em><\/strong> \u2212 v<\/em><\/strong>, view it as u<\/em><\/strong> + (\u2212v<\/em><\/strong>). Adding \u2212v<\/em><\/strong> is reversing direction of v<\/em><\/strong> and adding it to the end of u<\/em><\/strong>. The new vector begins at the start of u<\/em><\/strong> and stops at the end point of \u2212v<\/em><\/strong>. See Figure 9\u00a0for a visual that compares vector addition and vector subtraction using parallelograms<\/strong>.<\/p>\n

\"Showing<\/p>\n

Figure 9<\/b><\/p>\n<\/div>\n

\n

Example 5: Adding and Subtracting Vectors<\/h3>\n

Given [latex]u[\/latex] <\/strong> [latex]=\\langle 3,-2\\rangle[\/latex] and [latex]v[\/latex] <\/strong> [latex]=\\langle -1,4\\rangle[\/latex], find two new vectors u<\/em><\/strong> + v<\/em><\/strong>, and u<\/em><\/strong> \u2212 v<\/strong>.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

To find the sum of two vectors, we add the components. Thus,<\/p>\n

[latex]\\begin{array}{l}u+v=\\langle 3,-2\\rangle +\\langle -1,4\\rangle \\hfill \\\\ =\\langle 3+\\left(-1\\right),-2+4\\rangle \\hfill \\\\ =\\langle 2,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

See Figure 10(a).<\/p>\n

To find the difference of two vectors, add the negative components of [latex]v[\/latex] to [latex]u[\/latex]. Thus,<\/p>\n

[latex]\\begin{array}{l}u+\\left(-v\\right)=\\langle 3,-2\\rangle +\\langle 1,-4\\rangle \\hfill \\\\ =\\langle 3+1,-2+\\left(-4\\right)\\rangle \\hfill \\\\ =\\langle 4,-6\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

See Figure 10(b).<\/p>\n

\"Further<\/p>\n

Figure 10.<\/b> (a) Sum of two vectors (b) Difference of two vectors<\/p>\n<\/div>\n<\/div>\n

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

A General Note: 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]v[\/latex] [latex]=\\langle a,b\\rangle[\/latex] by [latex]k[\/latex] , we have<\/p>\n

[latex]kv=\\langle ka,kb\\rangle[\/latex]<\/div>\n

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

\n

Example 6: Performing Scalar Multiplication<\/h3>\n

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

\n

Solution<\/h3>\n

See Figure 11\u00a0for a geometric interpretation. If [latex]v[\/latex] <\/strong> [latex]=\\langle 3,1\\rangle[\/latex], then<\/p>\n

[latex]\\begin{array}{l}3v=\\langle 3\\cdot 3,3\\cdot 1\\rangle \\hfill \\\\ =\\langle 9,3\\rangle \\hfill \\\\ \\frac{1}{2}v=\\langle \\frac{1}{2}\\cdot 3,\\frac{1}{2}\\cdot 1\\rangle \\hfill \\\\ =\\langle \\frac{3}{2},\\frac{1}{2}\\rangle \\hfill \\\\ -v=\\langle -3,-1\\rangle \\hfill \\end{array}[\/latex]<\/div>\n
\"Showing<\/p>\n

Figure 11<\/b><\/p>\n<\/div>\n<\/div>\n

\n

Analysis of the Solution<\/h3>\n

Notice that the vector 3v<\/em><\/strong> is three times the length of v<\/em><\/strong>, [latex]\\frac{1}{2}[\/latex] [latex]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.<\/p>\n<\/div>\n

\n

Try It 2<\/h3>\n

Find the scalar multiple<\/strong> 3 [latex]u[\/latex] given [latex]u[\/latex] [latex]=\\langle 5,4\\rangle[\/latex].<\/p>\n

Solution<\/a><\/p>\n<\/div>\n

\n

Example 7: Using Vector Addition and Scalar Multiplication to Find a New Vector<\/h3>\n

Given [latex]u=\\langle 3,-2\\rangle[\/latex] and [latex]v=\\langle -1,4\\rangle[\/latex], find a new vector w<\/em><\/strong> = 3u<\/em><\/strong> + 2v<\/strong>.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

First, we must multiply each vector by the scalar.<\/p>\n

[latex]\\begin{array}{l}3u=3\\langle 3,-2\\rangle \\hfill \\\\ =\\langle 9,-6\\rangle \\hfill \\\\ 2v=2\\langle -1,4\\rangle \\hfill \\\\ =\\langle -2,8\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

Then, add the two together.<\/p>\n

[latex]\\begin{array}{l}w=3u+2v\\hfill \\\\ =\\langle 9,-6\\rangle +\\langle -2,8\\rangle \\hfill \\\\ =\\langle 9 - 2,-6+8\\rangle \\hfill \\\\ =\\langle 7,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

So, [latex]w=\\langle 7,2\\rangle[\/latex].<\/p>\n<\/div>\n

Finding Component Form<\/h2>\n

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the [latex]x[\/latex] direction, and the vertical component is the [latex]y[\/latex] direction. For example, we can see in the graph in Figure 12\u00a0that the position vector [latex]\\langle 2,3\\rangle[\/latex] comes from adding the vectors v<\/em><\/strong>1<\/sub> and v<\/em><\/strong>2<\/sub>. We have v<\/em><\/strong>1<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(2,0\\right)[\/latex].<\/p>\n

\n

[latex]\\begin{array}{l}{v}_{1}=\\langle 2 - 0,0 - 0\\rangle \\hfill \\\\ =\\langle 2,0\\rangle \\hfill \\end{array}[\/latex]<\/p>\n

We also have v<\/em><\/strong>2<\/sub> with initial point [latex]\\left(0,0\\right)[\/latex] and terminal point [latex]\\left(0,3\\right)[\/latex].<\/p>\n

[latex]\\begin{array}{l}{v}_{2}=\\langle 0 - 0,3 - 0\\rangle \\hfill \\\\ =\\langle 0,3\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

Therefore, the position vector is<\/p>\n

[latex]\\begin{array}{l}v=\\langle 2+0,3+0\\rangle \\hfill \\\\ =\\langle 2,3\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

Using the Pythagorean Theorem, the magnitude of v<\/em><\/strong>1<\/sub> is 2, and the magnitude of v<\/em><\/strong>2<\/sub> is 3. To find the magnitude of v<\/em><\/strong>, use the formula with the position vector.<\/p>\n

[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}|v|=\\sqrt{|{v}_{1}{|}^{2}+|{v}_{2}{|}^{2}}\\hfill \\\\ \\begin{array}{l}=\\sqrt{{2}^{2}+{3}^{2}}\\hfill \\\\ =\\sqrt{13}\\hfill \\end{array}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n

The magnitude of v<\/em><\/strong> is [latex]\\sqrt{13}[\/latex]. To find the direction, we use the tangent function [latex]\\tan \\theta =\\frac{y}{x}[\/latex].<\/p>\n

[latex]\\begin{array}{l}\\tan \\theta =\\frac{{v}_{2}}{{v}_{1}}\\hfill \\\\ \\tan \\theta =\\frac{3}{2}\\hfill \\\\ \\theta ={\\tan }^{-1}\\left(\\frac{3}{2}\\right)=56.3^\\circ \\hfill \\end{array}[\/latex]<\/div>\n
\"Diagram<\/p>\n

Figure 12<\/b><\/p>\n<\/div>\n

Thus, the magnitude of [latex]v[\/latex] is [latex]\\sqrt{13}[\/latex] and the direction is [latex]{56.3}^{\\circ }[\/latex] off the horizontal.<\/p>\n

\n

Example 8: Finding the Components of the Vector<\/h3>\n

Find the components of the vector [latex]v[\/latex] with initial point [latex]\\left(3,2\\right)[\/latex] and terminal point [latex]\\left(7,4\\right)[\/latex].<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

First find the standard position.<\/p>\n

[latex]\\begin{array}{l}v=\\langle 7 - 3,4 - 2\\rangle \\hfill \\\\ =\\langle 4,2\\rangle \\hfill \\end{array}[\/latex]<\/div>\n

See the illustration in Figure 13.<\/p>\n

\"Diagram<\/p>\n

Figure 13<\/b><\/p>\n<\/div>\n

The horizontal component is [latex]{v}_{1}=\\langle 4,0\\rangle[\/latex] and the vertical component is [latex]{v}_{2}=\\langle 0,2\\rangle[\/latex].<\/p>\n<\/div>\n<\/div>\n\n\t\t\t

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