Angle between Two Vectors?

Question

I have this vectors:
V= (2 5 -4)
W= (1 -2 -3)

V*W= 2*1+5*-2+(-4)*(-3) = 4

cos (V,W) = V*W / IVI * IWI

Can anyone explaind me how can I calculate the below part.

I am sorry, yes i was talking about the denominator of the fraction, English is not my first language.

Answer 1

|V|=sqrt(2^2 +5^2 + (-4)^2) = sqrt(4+25+16) = sqrt(45) = 3sqrt(5)
|W|= sqrt(1+4+9) = sqrt(14)
so |V|*|W| = 3*sqrt(70)

so cos(angle) = 4/(3sqrt(70))
so angle = arccos(4/(3sqrt(70))

Answer 2

V*W is the "Dot" product of the vectors V and W. You find it by simply multiplying the corresponding vector components and adding the results.

V * W = (2*1 + 5*(-2) + (-4)*(-3) ) = (2 -10 + 12) = 4
|V| is the "magnitude" or "frobius norm" of V. You find it by squaring the individual vector components of V, then adding those results together and taking the square root of that.
|V| = sqrt( 2^2 + 5^2 + (-4)^2) = sqrt(4 + 25 + 16) = sqrt (45)

|W| is found the same way as |V|.

Work it all out and you get a number between -1 and 1 that represents to cosine of the angle between vectors V and W.

Answer 3

The 'below' part??? I'm guessing that you mean the denominator of the fraction ☺

|W| and |V| are called the 'norm' of W and V and are just their Euclidean length. In your case:
|W|*|V| = √((2²+5²+(-4)²)*(1+(-2)²+(-3)²))


Doug

Answer 4

the part below is simplY;
|v|=sqrt(2^2+5^2+(-4)^2)=
= sqrt( 4+25+16)
=sqrt( 45)

|w|=sqrt(1+4+9)=sqrt(14)

the upper part is
v.w=2-10+12=4
so cos(v,w)= 4/sqrt(45)sqrt(14)

Answer 5

|V|=sqrt(2^2+5^2+(-4)^2)=6.7

|W|=sqrt(1^2+(-2)^2+(-3)^2)=
3.74

cos (V,W) =4/(6.7*3.74)=.1596

You did the hard part mate.


-v4m

Answer 6

hit upon a.B , the dot product which has value |A.B| = |A||B| cos m m = perspective between vectors in radians So m = arccos (|A.B|/|A||B|) ... discover it. additionally you would be able to apply the bypass product and use arcsin...

Answer 7

lVl is the square root of the dot product V*V

Ana