Table of contents

0. Fundamental Concepts of Algebra3h 29m
1. Equations and Inequalities3h 27m
2. Graphs1h 43m
3. Functions & Graphs2h 17m
4. Polynomial Functions1h 54m
5. Rational Functions1h 23m
6. Exponential and Logarithmic Functions2h 28m
7. Measuring Angles39m
8. Trigonometric Functions on Right Triangles2h 5m
9. Unit Circle1h 19m
10. Graphing Trigonometric Functions1h 19m
11. Inverse Trigonometric Functions and Basic Trig Equations1h 41m
12. Trigonometric Identities 2h 34m
13. Non-Right Triangles1h 38m
14. Vectors2h 25m
15. Polar Equations2h 5m
16. Parametric Equations1h 6m
17. Graphing Complex Numbers1h 7m
18. Systems of Equations and Matrices1h 6m
19. Conic Sections2h 36m
20. Sequences, Series & Induction1h 15m
21. Combinatorics and Probability1h 45m
22. Limits & Continuity1h 49m
23. Intro to Derivatives & Area Under the Curve2h 9m

14. Vectors

Vectors in Component Form

Precalculus

14. Vectors

Vectors in Component Form

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Finding Magnitude of a Vector Example 1

Nick

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Let's give this problem a try. So in this problem, we're told if vector v has initial point 1,2 and terminal point 4,4, sketch v as a position vector and calculate its magnitude. Now, to solve this problem, the first thing I'm going to do is figure out what vector v is in component form. This is always a good first step when you're dealing with these types of problems where you're given the initial and terminal point of a vector. To find this component form, what I can do is take the final x and subtract off the initial x, and then take the final y and subtract off the initial y, and that's going to give us our vector.

First off, we can see that for x₂, this is going to be the x value for the terminal point, which is 4. So what I can first do is plug in 4 to this equation. Next, I'm going to subtract off the initial x, and the initial x is 1. So we're going to have 4 minus 1 for the difference in the x's. As for the y's, I can see that we have y₂, and y₂ is going to be this y value for the terminal point, which is 4, and then this is going to be minus the initial y, and the initial y is 2. So we're going to have 4 minus 1, which is 3, and 4 minus 2, which is 2. So this right here is the vector.

Now that we have our vector in component form, what we can do is sketch the position vector. The position vector just means that our vector is going to start at the origin. So if I go here at the origin of our graph, I can see that our vector is 3. So that means on the x-axis, we're going to go over here 1, 2, 3, then I can see that our y value is 2, so that means we're going to go up 2. This is what our vector is going to look like, and that's vector v sketched as a position vector.

We've now found the component form, and we've sketched our position vector. Our last step is going to be to calculate the magnitude of this vector. To calculate the magnitude, we can use this equation: m a g n i t u d e = vx + vy where v_x is the x-component and v_y is the y-component. We have these values that we calculated already, so we're going to have v_x which is 3, so we have $3^2$ plus $2^2$. $3^2$ is 9, so we're going to have the square root of 9 plus 4, which is 13. So the magnitude of our vector 'v' is equal to the square root of 13. And this is the magnitude of our vector, as well as our vector sketched as a position vector, and that is the solution to this problem. I hope you found this video helpful. Thanks for watching.