Hey, everyone. Let's give this problem a try. So here we're asked to find the vector û, the unit vector, in the direction of vector u. Now to solve this problem, we're really just trying to calculate this unit vector. And recall for the unit vector û, it's going to be vector u divided by the magnitude of u. This is always the equation that you use when trying to find the unit vector. Now I can see that we are given vector u that's negative i plus 2j. And you can also write that as negative 1i + 2j, so that way we can clearly see what the x and y components are, and then, that's all going to be divided by the magnitude of vector u. Now the magnitude of u is something we don't have yet, so that's what we're going to have to calculate. Now to find the magnitude of any vector, what you want to do is take the square root of the x component squared plus the y component squared, and this is how you can calculate it. Now what I can see from the situation is that our x component is what's in front of the i negative one. So we're going to have negative one squared plus the y component, which is 2. Now the square root of negative one squared well or I should just say negative one squared comes out to positive one, and then two squared comes out to 4. So I have the square root of 1 plus 4, which is the square root of 5. So that means that the magnitude of u is the square root of 5. Now from here, now that we found the magnitude of u, what I need to do is divide that into the vector on top. So we're going to have negative 1 i plus 2 j, that's going to be divided by the square root of 5. And dividing each of these components by the square root of 5, we're going to have negative one over the square root of 5i, plus 2 over the square root of 5j. So this is what the vector ends up being for our unit vector, but notice that we end up with a square root in the denominator of each of these fractions. Typically, we don't want our final answer to have this radical on the bottom. So what I'm going to do is rationalize the denominator by multiplying the top and bottom of both fractions by the square root of 5. Now what this is going to do is get these square roots to cancel in each of the denominators. So for vector û, I'm going to end up with 1 times the square root of 5, which is just the square root of 5, divided by 5 i, and then this is negative. And that's going to be plus 2 times the square root of 5 divided by 5 j. And this right here is the unit vector and the solution to this problem. So I hope you found this video helpful. Thanks for watching.
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
Unit Vectors and i & j Notation
Video duration:
2mPlay a video:
Related Videos
Related Practice