Question #24b6e

2 Answers
Apr 27, 2015

Use the Law of Cosines:

Let vec(s)=5vec(x)+2vec(y)s=5x+2y

abs(vec(s))^2 = abs(5vec(x))^2+abs(2vec(y))^2-2abs(5vec(x))abs(2vec(y))cos(120^@)s2=5x2+2y225x2ycos(120)

(Draw the parallelogram for the addition, The large interior angle opposite vec(s)s is 120^@120.)

So
abs(vec(s))^2 = 5^2+2^2-2(5)(2)(-1/2) = 39s2=52+222(5)(2)(12)=39

abs(vec(s)) = sqrt 39s=39

The general formula is:

For vectors vec(a)a and vec(b)b and the angle between them thetaθ

abs(vec(a)+vec(b))^2 = abs(vec(a))^2+abs(vec(b))^2 + 2abs(vec(a))abs(2vec(b))cos thetaa+b2=a2+b2+2a2bcosθ

(The angle opposite the sum is 180^@ - theta180θ, so its cosine is -cos thetacosθ)

Apr 28, 2015

It's also an interesting exercise to do this with algebraic properties of dot products and their relationship to vector lengths (in particular, ||vec(v)||^{2}=vec(v)\cdot vec(v)v2=vv)

Here's the calculation:

||5vec(x)+2vec(y)||^{2}=(5vec(x)+2vec(y))\cdot (5vec(x)+2vec(y))5x+2y2=(5x+2y)(5x+2y)

=25vec(x)\cdot vec(x)+20vec(x)\cdot vec(y)+4vec(y)\cdot vec(y)=25xx+20xy+4yy

=25||vec(x)||^{2}+20||vec(x)|| ||vec(y)|| cos(theta)+4||vec(y)||^{2}=25x2+20xycos(θ)+4y2

=25\cdot 1^{2}+20\cdot 1\cdot 1\cdot cos(60^{\circ})+4\cdot 1^{2}=2512+2011cos(60)+412

=25+10+4=39=25+10+4=39.

Therefore, ||5vec(x)+2vec(y)||=\sqrt{39}5x+2y=39.

What this really illustrates is that the Law of Cosines and the geometric interpretation of dot products are equivalent.