Question

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

The projection of v1 onto v2.

Answer 1

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

Answer 2

from here it's just plug and play.

Answer 3

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

Answer 4

@Kiritina ?

Answer 5

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

Answer 6

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

scalar projection = \(\sf comp_{\vec v_2}\vec v_1\)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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 \(\vec v_1\)