# How to determine the cosine of the angle between the vectors u and v?u=3i-4j v=2i+j

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### 2 Answers

We have the vectors u = 3i - 4j and v = 2i + j.

Now the cosine of the angle between two vectors A = a1*i + b1*j and B = a2*i + b2*j is given by

cos theta=(a1*b1+a2*b2)/[sqrt(a1^2+b1^2)*sqrt(a2^2+b2^2)

=> cos theta = (3*2 - 4*1)/ sqrt (9 + 16)* sqrt (4 +1)

=> cos theta = (6 - 4)/ 5*sqrt 5

=> cos theta = 2 / 5 sqrt 5

=> cos theta = 2*sqrt 5 / 25

Therefore the required cosine of the angle between u = 3i - 4j and v = 2i + j is** 2*sqrt 5 / 25**

To determine the cosine of the angle of u and v, we'll ahve to apply the dot product formula;

u*v = |u|*|v|*cos (u,v)

cos (u,v) = u*v/|u|*|v|

We'll multiply the vectors u and v:

(3i-4j)(2i+j) = 6i^2 + 3ij - 8ij - 4j^2

Since i^2 = 1 and j^2 = 1, also i*j = 0, we'll get:

(3i-4j)(2i+j) = 6 - 4

(3i-4j)(2i+j) = 2

We'll calculate the modulus of the vectors u and v:

u = sqrt[3^2 + (-4)^2]

u = sqrt (9 + 16)

u = sqrt 25

uÂ =5

v = sqrt(2^2 + 1^2)

v = sqrt(4 + 1)

v = sqrt 5

cos (u,v) = 2/5*sqrt5

**cos (u,v) = 2sqrt5/25**