If v1=(-2, 5) and v2=(4, -3), then the angle between the two vectors is ____. Round your answer to two decimal places.
this assumes you know the dot product equivalence
\[|u||v|~\cos(\alpha)=u\cdot v\]
otherwise you use the law of cosine
@amistre64 the law of cosines is a^2=b^2+c^2-2bc*cos(A) right?
yes, so we still have to determine the length of the other side
how do I know what is a and b
b and c are the lengths of the vectors stated, since a is across from A |dw:1430099999615:dw|
a is the length of the vector defined as v1-v2
(-6, 8)
v1=(-2, 5) -v2=(-4, 3) -6,8 is fine for vector a how do we determine lengths?
is that Pythagorean Theorem a^2 +b^2 =c^2
yep square, add, sqrt the parts .... pythagorean style sqrt(4+25) sqrt(16+9) sqrt(36+64)
i spose the sqrting is not required at the moment
29 25 100
100 = 25 + 29 - 2(sqrt(25))(sqrt(29)) cosA yep
what do I plug in for A
A is the angle you are looking for
so do i solve this 25 + 29 - 2(sqrt(25))(sqrt(29)) cosA
yeah, thats what the parts filled in look like equate it to 100 of course
a^2 = b^2 + c^2 -2cb cosA b^2 + c^2 - a^2 -------------- = cosA 2cb
is a=29 b=25 and c=100? is so I got 1223/625 =1.9568 or 1.96
a^2=100 b^2=25 c^2=29
54-100 ---------- 2(5)sqrt(29) 5.4-10 ---------- sqrt(29) -4.6 ---------- sqrt(29) cos^(-1)(-46/sqrt(29))
-0.8541 -0.8541 -0.8541
-8.54068457 or -8.54
i forgot how to add .... regardless http://www.wolframalpha.com/input/?i=cos%5E%28-1%29%28%2825%2B29-100%29%2F%282*sqrt%2825%29*sqrt%2829%29%29
2.59?
do they want radians or degrees?
when I add those three up i get -2.5623
it did not specify all it says is to round your answer to two decimal places
25+29-100 = -46 -46/2(5)(sqrt(29)) = -23/5sqrt(29)
then its either 2.59 radians or 148.67 degrees
im thinking its degrees they want but its just a hunch
oh ok yeah I guess if they say angle 148.67 degrees might be it then
im sorry I looked at the picture they want degrees
thank you :)
youre welcome
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