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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Algebraic Operations on Vectors Example 1

Nick

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Let's see if we can solve this example. In this example, we're told if vector a equals 31, and vector b equals negative 49, and vector c is equal to 5 times vector a minus 2 times vector b, we're asked to calculate the magnitude of vector c. Now what I can see here is that we're trying to ultimately find the magnitude of this vector, but this vector depends on the other two vectors we have in this problem. So we definitely have our work cut out for us here. But whenever solving these problems where we have multiple vector operations, I personally like to start small with the problem and then work my way to solving the whole thing. So let's actually do that.

Now the first thing I'm going to do is figure out what 5 times vector a is, because I'm really going to be focusing on this vector c here. To find 5 times a, well 5 times vector a is just going to be 5 times this vector. So we're going to have 5 times 31. Now when multiplying this scalar into each of the components, we'll have 5 times 3, which is 15, and then we'll have 5 times 1, which is 5. So this is the vector 5a.

Next, I'm going to figure out what vector 2 b is. So vector 2b is going to be 2 times vector b. I can see here that vector b is negative 49, so that's going to be our vector, and then what I need to do is distribute the scalar into this vector as well. So we're going to have 2 times negative 4, which is negative 8, and we're going to have 2 times 9, which is 18. So this is vector 2b.

And what I need to do from here is subtract these 2 vectors. To find vector c, it's going to be 5a minus 2b. We can see here that 5a is (15, 5), and 2b is (-8, 18). By subtracting these 2 vectors, we're going to have 15 minus negative 8, which is the same thing as 15 plus 8, which is 23. And then we'll have 5 minus 18, which comes out to negative 13. So this right here is vector c.

Now that I've found vector c, our last step is to just find the magnitude of c. We can use the Pythagorean theorem, and for these vectors, it's going to be the square root of the x-component of c squared plus the y-component of c squared. The x-component is 23, so we're going to have 23 squared, plus the y-component, which is negative 13, squared. So this is what we get: 232 + 132 Now, 23 squared equals 529, and negative 13 squared comes out to a positive 169. So we're going to have 529 plus 169, which comes out to 698. This value, the square root of 698, is not a radical that you can simplify further. So the magnitude of c is the square root of 698, and that is the solution to this problem.

So this is how you can handle multiple vector operations, as well as finding the magnitude of a vector once you've already done operations on that vector. I hope you found this video helpful. Thanks for watching.