A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Let v1 = (-6, 4) and v2 = (-3, 6). Compute the following. The scalar projection of v1 onto v2. The projection of v1 onto v2.

  • This Question is Closed
  1. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \sf comp_{\vec v_2} \vec v_1 = \frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|}\]\[\large \sf proj_{\vec v_2}\vec v_1=\frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|^2}\cdot \vec v_2 \]

  2. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    from here it's just plug and play.

  3. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sf \vec v_1 \cdot \vec v_2 = ((-6)(-3)) + ((4)(6))=~?\]\[\sf |\vec v_2| = \sqrt{(-3)^2+(6)^2}=~?\]

  4. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Kiritina ?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This would be the regular projection, not the scalar projection, right? Magnitude of v2=3√5

  6. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the "regular projection" = \(\sf proj_{\vec v_2}\vec v_1\) scalar projection = \(\sf comp_{\vec v_2}\vec v_1\)

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Dot product=42 so the result would be 42/(3√5)^2 *v2. Should magnitude be plugged in for v2 here?

  8. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    magnitude = \(\sf |\vec v_2|\) vector = \(\sf \vec v_2\)

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How would you multiply \[\frac{ 42 }{ (3√5)^2 }*v2\] What would you plug in for v2? Sorry for all the questions :\

  10. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    As I've written, \(\vec v_2\) is the vector. therefore you would plug in the vector 2. The projection of \(\vec v_2\) onto \(\vec v_1\) |dw:1442208512755:dw|

  11. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Therefore \(\sf proj_{\vec v_2} \vec v_1 \) creates another vector, what you can see as the "shadow" of \(\sf \vec v_1\) Does that make more sense?

  12. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Kiritina

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It makes sense in a...like technical term, I can understand what the question is asking but I'm confused about how to solve it mathematically.

  14. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Alright.

  15. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sf proj_{\vec v_2}\vec v_1 =\frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|^2}\cdot \vec v_2~~ =~\frac{42}{(3\sqrt{5})^2} \cdot \langle -3~,~6\rangle \]

  16. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sf proj_{\vec v_2}\vec v_1 = \frac{42}{45}\langle -3~,~6\rangle~ = ~~\langle -\frac{14}{15}~,~ \frac{28}{5}\rangle =~~ \frac{1}{5}\langle -\frac{14}{3} ~,~ 28\rangle \]

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahhh, you multiply them by each part of the vector coordinate individually! I was confusing the end result to be a whole number, not a vector.

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So would the scalar projection be just plain 42/3√5?

  19. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thats why I was trying to explain to you what a projection is compared to what a component is.

  20. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    finding the component means you're looking for "how long" the "shadow" is. finding the projection means you're finding the actual vector that is the shadow created by the first vector.

  21. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and yes, you are right about the scalar projection

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright, I think I got it! The scalar projection would be 14/√5, then. Thank you so much!

  23. Jhannybean
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Mmhmm.

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.