Question

Question #24b6e

Answer 1

Use the Law of Cosines:

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

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

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

So
`abs(vec(s))^2 = 5^2+2^2-2(5)(2)(-1/2) = 39`

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

The general formula is:

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

`abs(vec(a)+vec(b))^2 = abs(vec(a))^2+abs(vec(b))^2 + 2abs(vec(a))abs(2vec(b))cos theta`

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

Answer 2

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)`)

Here's the calculation:

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

`=25vec(x)\cdot vec(x)+20vec(x)\cdot vec(y)+4vec(y)\cdot vec(y)`

`=25||vec(x)||^{2}+20||vec(x)|| ||vec(y)|| cos(theta)+4||vec(y)||^{2}`

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

`=25+10+4=39`.

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

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