Trigonometry Dot Product Question?

Question

I am having a problem figuring out this dot product thing. i am to find the smallest angle between the pair of vectors <-3,2>,<4,5>. ok, so i have cosa=<-3,2><4,5> / |<-3,2>| |<4,5>|. i have a -2 for the top part of the fraction but i have no clue how to do the bottom. the answer says it's (sqrt13)(sqrt14) and then go from there. how do i get the square roots of 13 and 14?

Answer 1

The norm of , denoted ||, is supposed to represent the length or strength of the vector . Because of Pythagorean theorem reasons, this is defined to be sqrt(x^2 + y^2). Another way of expressing it is sqrt(length|^2). Now imagine a right triangle with sides x and y. Then x^2 + y^2 = z^2, where z is the hypotenuse. The hypotenuse is the length we want, so we get sqrt(z^2) = sqrt(x^2 + y^2).

For your problem. |(-3,2)| = sqrt(3^2+2^2) = sqrt(9+4) = sqrt(13). The other term is |<4,5>| = sqrt(4^2+5^2) = sqrt(41). You got the digits turned around the wrong way.

Answer 2

|<-3,2>| = sqrt(3*3 + 2*2) = sqrt(9 + 4) = sqrt(13)
|<4,5>| = sqrt(4*4 + 5*5) = sqrt(16 + 25) = sqrt(41)

Answer 3

The should be the norm value.

The length of vector v is ||v||=sqrt(v1^2+v2^2+...vn^2).

So ||<-3,2>||=sqrt((-3)^2+(2)^2) =sqrt(9+4)=sqrt(13)

and

||<4,5>||=sqrt(4^2+5^2) =sqrt(16+25) =sqrt(41) .... you sure it is s'posed to be sqrt(14)?

Answer 4

The bars on the bottom mean magnitude of a vector. The magnitude of vectro = sqrt(x^2 + y^2)

Answer 5

it relies upon, i wager, on the scope of your subject at this aspect. (a.b) supplies a huge style as its answer. a consistent huge style could have a dot product with c, if it is what your syllabus helps.