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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.
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.

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Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \sf comp_{\vec v_2} \vec v_1 = \frac{\vec v_1 \cdot \vec v_2}{\vec v_2}\]\[\large \sf proj_{\vec v_2}\vec v_1=\frac{\vec v_1 \cdot \vec v_2}{\vec v_2^2}\cdot \vec v_2 \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1from here it's just plug and play.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\sf \vec v_1 \cdot \vec v_2 = ((6)(3)) + ((4)(6))=~?\]\[\sf \vec v_2 = \sqrt{(3)^2+(6)^2}=~?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This would be the regular projection, not the scalar projection, right? Magnitude of v2=3√5

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1the "regular projection" = \(\sf proj_{\vec v_2}\vec v_1\) scalar projection = \(\sf comp_{\vec v_2}\vec v_1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dot product=42 so the result would be 42/(3√5)^2 *v2. Should magnitude be plugged in for v2 here?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1magnitude = \(\sf \vec v_2\) vector = \(\sf \vec v_2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you multiply \[\frac{ 42 }{ (3√5)^2 }*v2\] What would you plug in for v2? Sorry for all the questions :\

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1As 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
 one year ago
Best ResponseYou've already chosen the best response.1Therefore \(\sf proj_{\vec v_2} \vec v_1 \) creates another vector, what you can see as the "shadow" of \(\sf \vec v_1\) Does that make more sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\sf proj_{\vec v_2}\vec v_1 =\frac{\vec v_1 \cdot \vec v_2}{\vec v_2^2}\cdot \vec v_2~~ =~\frac{42}{(3\sqrt{5})^2} \cdot \langle 3~,~6\rangle \]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\sf proj_{\vec v_2}\vec v_1 = \frac{42}{45}\langle 3~,~6\rangle~ = ~~\langle \frac{14}{15}~,~ \frac{28}{5}\rangle =~~ \frac{1}{5}\langle \frac{14}{3} ~,~ 28\rangle \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh, 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
 one year ago
Best ResponseYou've already chosen the best response.0So would the scalar projection be just plain 42/3√5?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Thats why I was trying to explain to you what a projection is compared to what a component is.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1finding 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
 one year ago
Best ResponseYou've already chosen the best response.1and yes, you are right about the scalar projection

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, I think I got it! The scalar projection would be 14/√5, then. Thank you so much!
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