Question

4part question

Answer 1

okay so:

(-6+-3)=-9
(4+6)=10

=1

Answer 2

17a) is the dot product, so remember the steps:

1. multiply the x-coordinates
2. multiply the y-coordinates
3. sum the results from step 1 + 2

Answer 3

meant to say multiply instead of add

Answer 4

dot product = (-6)(-3) + 4(6) = ?

Answer 5

42

Answer 6

good

now for b) we need to use the angle between two vectors formula

Answer 7

the dot product is in the numerator (which we already calculated in part a)

the denominator is the product of the magnitudes

so you will need to calculate the magnitude of v1 and v2 first then multiply the results

Answer 8

do i leave it in its radical form

Answer 9

yes

Answer 10

magnitude = sqrt(x^2 + y^2)
=sqrt((-6)^2+4^2)
=2sqrt13


magnitude= sqrt(x^2 + y^2)
=sqrt((-3)^2+6^2)
=3sqrt5

2sqrt13 * 3sqrt5 = 6sqrt65

answer: 42/6sqrt65

Answer 11

good but that gives us cos(theta), so to solve for theta we just need to take the arccos of that

so
arccos(42/6sqrt65) = 29.74 degrees (solution for b)

Answer 12

for c) the scalar projection of vector b onto a is given as

(dot product of a and b) / (magnitude of a)

therefore: for our problem, the scalar projection of v1 onto v2 =

(dot product of v1 and v2) / (magnitude of v2) = ?

Answer 13

42/3sqrt5

=14sqrt5/5

Answer 14

awesome

now for d) the scalar projection is

[ (dot product of v1 and v2) / (magnitude of v2)^2 ] * vector v2 = ?

Answer 15

[42/(3sqrt5)^2]* (-3,6)

Answer 16

good, now simplify the [42/(3sqrt5)^2] and then distribute to the (-3,6) part

Answer 17

(-14/5)
(28/5)

Answer 18

good, so the final answer combines those two coordinates to get

(-14/5 , 28/5)

probably going to get something to eat soon

Answer 19

Sure have a good lunch

Answer 20

alright cool i'm back lmk if you have any other problems to work on