Question

Given the vectors v=<-3,4> and u=2i-4j Determine a)2v+u
b)vdotu
c)the angle between u and v.

Answer 1

let's us for rewrite the 2nd vector, understanding that v = ai + bj = <a, b>
so
v=<-3,4>
u = <2, -4>

Answer 2

2v is just a scalar multiplication
so 2 * <-3, 4> => < 2(-3), 2(4)>

Answer 3

addition is just <a, b> + <c, d> => <a+c, b+d>

Answer 4

the "dot product"
<a, b> dot <c, d> => ac + bd <--- it's just a number

Answer 5

Ok

Answer 6

thanks so the first one would be for a i

Answer 7

v=<-3,4>
u = <2, -4>

2v
< 2(-3), 2(4)> => <-6, 8>
2v+u
<-6, 8 > + <2, -4> => <-6+2, 8+(-4)> = <-4, 4>

Answer 8

v=<-3,4>
u = <2, -4>

now the "dot product"
\(\large <-3, 4> \bullet <2, -4> \implies -3(2) + 4(-4) = -22\)

Answer 9

now, to find the angle in between both vectors
you'd need to use the formula
$$
cos(\theta)=\frac{v\times u}{||v||\times ||u||}
$$
where the "magnitude" of say "u" -> \(||u||=\sqrt{a^2+b^2}\)

Answer 10

Ohhh ok i see thanks for your help

Answer 11

$$
||v|| = \sqrt{(-3)^2+4^2} \implies \sqrt{25} \implies 5\\
||u||=\sqrt{2^2+(-4)^2} \implies \sqrt{20} \implies 4\sqrt{5}
$$

Answer 12

$$
||v|| = \sqrt{(-3)^2+4^2} \implies \sqrt{25} \implies 5\\
||u||=\sqrt{2^2+(-4)^2} \implies \sqrt{20} \implies 2\sqrt{5}
$$
rather

Answer 13

the product atop from the "dot product" of the numerator will give you a number
the denominator multiplication of the magnitudes will give you another
divide them, that will be the \(\large cos(\theta)\)
to find \(\large \theta\) just get the \(cos^{-1}(\theta)\) to both left and right sides, and that'd be the angle