anonymous
  • anonymous
Let v1 = (-6, 4) and v2 = (-3, 6). Compute the following. The scalar projection of v1 onto v2. The projection of v1 onto v2.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
\[\large \sf comp_{\vec v_2} \vec v_1 = \frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|}\]\[\large \sf proj_{\vec v_2}\vec v_1=\frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|^2}\cdot \vec v_2 \]
Jhannybean
  • Jhannybean
from here it's just plug and play.
Jhannybean
  • Jhannybean
\[\sf \vec v_1 \cdot \vec v_2 = ((-6)(-3)) + ((4)(6))=~?\]\[\sf |\vec v_2| = \sqrt{(-3)^2+(6)^2}=~?\]

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Jhannybean
  • Jhannybean
@Kiritina ?
anonymous
  • anonymous
This would be the regular projection, not the scalar projection, right? Magnitude of v2=3√5
Jhannybean
  • Jhannybean
the "regular projection" = \(\sf proj_{\vec v_2}\vec v_1\) scalar projection = \(\sf comp_{\vec v_2}\vec v_1\)
anonymous
  • anonymous
Dot product=42 so the result would be 42/(3√5)^2 *v2. Should magnitude be plugged in for v2 here?
Jhannybean
  • Jhannybean
magnitude = \(\sf |\vec v_2|\) vector = \(\sf \vec v_2\)
anonymous
  • anonymous
How would you multiply \[\frac{ 42 }{ (3√5)^2 }*v2\] What would you plug in for v2? Sorry for all the questions :\
Jhannybean
  • Jhannybean
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|
Jhannybean
  • Jhannybean
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?
Jhannybean
  • Jhannybean
@Kiritina
anonymous
  • anonymous
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.
Jhannybean
  • Jhannybean
Alright.
Jhannybean
  • Jhannybean
\[\sf proj_{\vec v_2}\vec v_1 =\frac{\vec v_1 \cdot \vec v_2}{|\vec v_2|^2}\cdot \vec v_2~~ =~\frac{42}{(3\sqrt{5})^2} \cdot \langle -3~,~6\rangle \]
Jhannybean
  • Jhannybean
\[\sf proj_{\vec v_2}\vec v_1 = \frac{42}{45}\langle -3~,~6\rangle~ = ~~\langle -\frac{14}{15}~,~ \frac{28}{5}\rangle =~~ \frac{1}{5}\langle -\frac{14}{3} ~,~ 28\rangle \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
So would the scalar projection be just plain 42/3√5?
Jhannybean
  • Jhannybean
Thats why I was trying to explain to you what a projection is compared to what a component is.
Jhannybean
  • Jhannybean
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.
Jhannybean
  • Jhannybean
and yes, you are right about the scalar projection
anonymous
  • anonymous
Alright, I think I got it! The scalar projection would be 14/√5, then. Thank you so much!
Jhannybean
  • Jhannybean
Mmhmm.

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