{"id":861,"date":"2025-10-14T14:56:08","date_gmt":"2025-10-14T14:56:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=861"},"modified":"2025-10-20T14:44:16","modified_gmt":"2025-10-20T14:44:16","slug":"vector-basics-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-basics-learn-it-3\/","title":{"raw":"Vector Basics: Learn It 3","rendered":"Vector Basics: Learn It 3"},"content":{"raw":"
Vector Components<\/h2>\r\n
Working with vectors becomes much more straightforward when you use a coordinate system. Instead of relying purely on geometric methods, you can use the coordinates of initial and terminal points to perform precise calculations.<\/p>\r\nSince vectors can be positioned anywhere in the plane, calculations become easier when we place the vector's initial point at the origin. A vector positioned this way is called a standard-position vector<\/strong>.\r\n
For any standard-position vector, the initial point is [latex](0,0)[\/latex], so you only need the terminal point coordinates to describe the vector completely. If vector [latex]\\mathbf{v}[\/latex] has terminal point [latex](x,y)[\/latex], we write:<\/p>\r\n
The scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components<\/strong> of [latex]\\mathbf{v}[\/latex].<\/p>\r\n\r\n\r\n
component form<\/h3>\r\nA vector [latex]\\mathbf{v}[\/latex] with initial point at the origin [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex] can be written in component form as:\r\n\r\n
[latex]\\bf{v}[\/latex] [latex]= \\langle x,y \\rangle[\/latex].<\/center>\r\nThe scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components of [latex]\\bf{v}[\/latex].\r\n\r\n<\/section>Use angle brackets [latex]\\langle x,y \\rangle[\/latex] for vectors in component form, not parentheses [latex](x,y)[\/latex]. Parentheses describe points in the plane, while angle brackets describe vectors. The vector [latex]\\langle x,y \\rangle[\/latex] starts at [latex](0,0)[\/latex] and ends at [latex](x,y)[\/latex].<\/section>\r\n
Converting Any Vector to Component Form<\/h3>\r\n
When you have a vector that's not already in standard position, you can find its component form using two approaches:<\/p>\r\n
Geometric Approach<\/strong>: Sketch the vector in the coordinate plane, then draw an equivalent standard-position vector.<\/p>\r\n
Algebraic Approach<\/strong>: Use the coordinates directly. If a vector has initial point [latex](x_1, y_1)[\/latex] and terminal point [latex](x_2, y_2)[\/latex], then:<\/p>\r\n\r\n
The vector in component form is [latex]\\langle x_2 - x_1, y_2 - y_1 \\rangle[\/latex].<\/p>\r\n\r\n\r\n
converting to component form<\/h3>\r\nLet [latex]\\bf{v}[\/latex] be a vector with initial point [latex](x_i,y_i)[\/latex] and terminal point [latex](x_t,y_t)[\/latex]. Then we can express [latex]\\bf{v}[\/latex] in component form as\r\n\r\n
[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex]<\/center><\/section>Express vector [latex]\\bf{v}[\/latex] with initial point [latex](\u22123,4)[\/latex] and terminal point [latex](1,2)[\/latex] in component form.\r\n
\r\n1. Sketch the vector in the coordinate plane (Figure 13).<\/ol>\r\n2. The terminal point is [latex]4[\/latex] units to the right and [latex]2[\/latex] units down from the initial point.<\/ol>\r\n3. Find the point that is [latex]4[\/latex] units to the right and [latex]2[\/latex] units down from the origin.<\/ol>\r\n4. In standard position, this vector has initial point [latex](0,0)[\/latex] and terminal point [latex](4,\u22122)[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
\r\nIn the first solution, we used a sketch of the vector to see that the terminal point lies [latex]4[\/latex] units to the right. We can accomplish this algebraically by finding the difference of the [latex]x[\/latex]-coordinates:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
To find the magnitude of a vector, you calculate the distance between its initial point and terminal point. For a vector [latex]\\mathbf{v} = \\langle x,y \\rangle[\/latex], the magnitude is:<\/p>\r\n\r\n
[latex]\\bf{||v||}[\/latex] [latex]=\\sqrt{x^2 + y^2}[\/latex]<\/center>This formula comes from the distance formula. Since any vector in component form is equivalent to a standard-position vector with initial point [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex], we calculate:\r\n\r\n
\r\n
[latex]||\\mathbf{v}|| = \\sqrt{(x-0)^2 + (y-0)^2} = \\sqrt{x^2 + y^2}[\/latex]<\/p>\r\n\r\n<\/center>Note that [latex]||\\mathbf{v}|| \\geq 0[\/latex] for any vector [latex]\\mathbf{v}[\/latex], and [latex]||\\mathbf{v}|| = 0[\/latex] if and only if [latex]\\mathbf{v} = \\mathbf{0}[\/latex].\r\n\r\nYou can also think of this using the Pythagorean theorem\u2014the vector's components form the legs of a right triangle, and the magnitude is the hypotenuse.\r\n\r\n[caption id=\"attachment_694\" align=\"aligncenter\" width=\"282\"] Figure 5. If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle\u2019s hypotenuse.[\/caption]\r\n
Operations with Component Form<\/h3>\r\n
Expressing vectors in component form makes scalar multiplication and vector addition much simpler\u2014you can perform these operations algebraically instead of geometrically.<\/p>\r\n\r\n\r\n
vector operations in component form<\/h3>\r\n
Let [latex]\\mathbf{v} = \\langle x_1, y_1 \\rangle[\/latex] and [latex]\\mathbf{w} = \\langle x_2, y_2 \\rangle[\/latex] be vectors, and let [latex]k[\/latex] be a scalar.<\/p>\r\n\r\n
Vector addition<\/strong>: [latex]\\mathbf{v} + \\mathbf{w} = \\langle x_1 + x_2, y_1 + y_2 \\rangle[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>Let [latex]\\bf{v}[\/latex] be the vector with initial point [latex](2,5)[\/latex] and terminal point [latex](8,13)[\/latex], and let [latex]\\bf{w}[\/latex][latex] = \\langle\u22122,4\\rangle[\/latex].\r\na. Express [latex]\\bf{v}[\/latex] in component form and find [latex]\\bf{||v||}[\/latex]. Then, using algebra, find<\/ol>\r\nb. [latex]\\bf{v + w}[\/latex],<\/ol>\r\nc. [latex]3[\/latex][latex]\\bf{v}[\/latex], and<\/ol>\r\nd. [latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex].<\/ol>\r\n
\r\na. To place the initial point of [latex]\\bf{v}[\/latex] at the origin, we must translate the vector [latex]2[\/latex] units to the left and [latex]5[\/latex] units down (Figure 2.15). Using the algebraic method, we can express [latex]\\bf{v}[\/latex] as [latex]\\bf{v}[\/latex] [latex] = \\langle 8 \u2212 2,13 \u2212 5\\rangle = \\langle6,8\\rangle[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
\r\nb. To find [latex]\\bf{v + w}[\/latex], add the [latex]x[\/latex]-components and the [latex]y[\/latex]-components separately:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
\r\nc. To find [latex]3[\/latex][latex]\\bf{v}[\/latex], multiply [latex]\\bf{v}[\/latex] by the scalar [latex]k=3[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
\r\nd. To find [latex]\\bf{v}[\/latex][latex]\u22122[\/latex][latex]\\bf{w}[\/latex], find [latex]\u22122[\/latex][latex]\\bf{w}[\/latex] and add it to [latex]\\bf{v}[\/latex]:<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
Vector operations follow familiar algebraic properties. These properties make vector calculations predictable and allow you to manipulate vector expressions just like algebraic expressions.<\/p>\r\n\r\n\r\n
properties of vector operations<\/h3>\r\n
Let [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] be vectors, and let [latex]r[\/latex] and [latex]s[\/latex] be scalars.<\/p>\r\n
[latex]1\\mathbf{u} = \\mathbf{u}[\/latex] and [latex]0\\mathbf{u} = \\mathbf{0}[\/latex] (Identity and Zero)<\/li>\r\n<\/ul>\r\n<\/section>\r\n
Proof<\/strong><\/p>\r\n\r\n\r\n\r\n
Proof of Cummutative Property<\/strong><\/p>\r\nLet [latex]{\\bf{u}}=\\langle x_1, y_1\\rangle[\/latex] and [latex]{\\bf{v}}=\\langle x_2, y_2\\rangle[\/latex]. Apply the commutative property for real numbers:\r\n\r\n
Finding Components from Magnitude and Direction<\/h3>\r\n
Sometimes you know a vector's magnitude and direction angle, but not specific coordinate points. You can use trigonometry to find the components.<\/p>\r\n
Consider a vector [latex]\\mathbf{v}[\/latex] that makes angle [latex]\\theta[\/latex] with the positive [latex]x[\/latex]-axis. The vector's components form the legs of a right triangle, with the vector itself as the hypotenuse.<\/p>\r\n\r\n\r\n[caption id=\"attachment_697\" align=\"aligncenter\" width=\"262\"] Figure 7. The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.[\/caption]\r\n
Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.<\/p>\r\n
[latex]\\mathbf{v} = \\langle ||\\mathbf{v}|| \\cos \\theta, ||\\mathbf{v}|| \\sin \\theta \\rangle[\/latex]<\/p>\r\n\r\nFind the component form of a vector with magnitude [latex]4[\/latex] that forms an angle of [latex]\u221245\u00b0[\/latex] with the [latex]x[\/latex]-axis.\r\n
[reveal-answer q=\"fs-id1167794955754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794955754\"]\r\nLet [latex]x[\/latex] and [latex]y[\/latex] represent the components of the vector (Figure 2.16). Then [latex]x = 4\\cos({\u221245\u00b0}) = 2\\sqrt{2}[\/latex] and [latex]y = 4\\sin({\u221245\u00b0}) = \u22122\\sqrt{2}[\/latex]. The component form of the vector is [latex]\\langle 2\\sqrt{2},\u22122\\sqrt{2} \\rangle[\/latex].[caption id=\"attachment_698\" align=\"aligncenter\" width=\"186\"] Figure 8. Use trigonometric ratios, [latex]x=||{\\bf{v}}||\\cos{\\theta}[\/latex] and[latex]y=||{\\bf{v}}||\\sin{\\theta}[\/latex], to identify the components of the vector.[\/caption]<\/div>\r\n
Working with vectors becomes much more straightforward when you use a coordinate system. Instead of relying purely on geometric methods, you can use the coordinates of initial and terminal points to perform precise calculations.<\/p>\n
Since vectors can be positioned anywhere in the plane, calculations become easier when we place the vector’s initial point at the origin. A vector positioned this way is called a standard-position vector<\/strong>.<\/p>\n
For any standard-position vector, the initial point is [latex](0,0)[\/latex], so you only need the terminal point coordinates to describe the vector completely. If vector [latex]\\mathbf{v}[\/latex] has terminal point [latex](x,y)[\/latex], we write:<\/p>\n
The scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components<\/strong> of [latex]\\mathbf{v}[\/latex].<\/p>\n\n
component form<\/h3>\n
A vector [latex]\\mathbf{v}[\/latex] with initial point at the origin [latex](0,0)[\/latex] and terminal point [latex](x,y)[\/latex] can be written in component form as:<\/p>\n
The scalars [latex]x[\/latex] and [latex]y[\/latex] are called the components of [latex]\\bf{v}[\/latex].<\/p>\n<\/section>\nUse angle brackets [latex]\\langle x,y \\rangle[\/latex] for vectors in component form, not parentheses [latex](x,y)[\/latex]. Parentheses describe points in the plane, while angle brackets describe vectors. The vector [latex]\\langle x,y \\rangle[\/latex] starts at [latex](0,0)[\/latex] and ends at [latex](x,y)[\/latex].<\/section>\n
Converting Any Vector to Component Form<\/h3>\n
When you have a vector that’s not already in standard position, you can find its component form using two approaches:<\/p>\n
Geometric Approach<\/strong>: Sketch the vector in the coordinate plane, then draw an equivalent standard-position vector.<\/p>\n
Algebraic Approach<\/strong>: Use the coordinates directly. If a vector has initial point [latex](x_1, y_1)[\/latex] and terminal point [latex](x_2, y_2)[\/latex], then:<\/p>\n
The vector in component form is [latex]\\langle x_2 - x_1, y_2 - y_1 \\rangle[\/latex].<\/p>\n\n
converting to component form<\/h3>\n
Let [latex]\\bf{v}[\/latex] be a vector with initial point [latex](x_i,y_i)[\/latex] and terminal point [latex](x_t,y_t)[\/latex]. Then we can express [latex]\\bf{v}[\/latex] in component form as<\/p>\n
[latex]\\bf{v}[\/latex] [latex]=\\langle x_t \u2212 x_i,y_t \u2212 y_i\\rangle[\/latex]<\/div>\n<\/section>\nExpress vector [latex]\\bf{v}[\/latex] with initial point [latex](\u22123,4)[\/latex] and terminal point [latex](1,2)[\/latex] in component form.<\/p>\n
![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190801\/2-1-13.jpeg\")
Figure 4.\u00a0These vectors are equivalent.[\/caption]\r\n\r\n \r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17190913\/2-1-14.jpeg\")
Figure 5. If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle\u2019s hypotenuse.[\/caption]\r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191050\/2-1-15.jpeg\")
Figure 6. In component form, [latex]{\\bf{v}}=\\langle{6,8}\\rangle[\/latex].[\/caption]<\/center>\r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191244\/2-1-16.jpeg\")
Figure 7. The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.[\/caption]\r\n![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/17191328\/2-1-17.jpeg\")
Figure 8. Use trigonometric ratios, [latex]x=||{\\bf{v}}||\\cos{\\theta}[\/latex] and[latex]y=||{\\bf{v}}||\\sin{\\theta}[\/latex], to identify the components of the vector.[\/caption]<\/div>\r\n