Question

Angle between two vectors?

I'm not really sure if I'm overthinking this or not, but how does the change highlighted in the picture occur? I'm looking for the work that happens for it to become radical 40 and 25, essentially. Thanks!

Answer 1

Its the result of using the Pythagorean Theorem, `a^2+b^2=c^2` rewritten to solve for `c`

Explanation:

What is happening in the above figure is that the magnitude of the vector is being calculated.

The `x` and `y` terms are orthagonal to each other (`90^@` between each other), and can be represented as two legs of a triangle.

The Pythagorean Theorem can be used to find the hypotenuse of this triangle, which is the total magnitudinal distance. The only thing that we must be aware of is that the magnitude will always be a positive number, since we are squaring both `x` and `y`

So, if `a^2+b^2=c^2` we can re-write this in terms of the vectors and magnitude:

`x^2+y^2=V^2`

`sqrt(x^2+y^2)=V`

Where `V` is the magnitude of the vector.

Using this formula, we can plug in the two vectors and see that they match up to what is shown in the document:

`sqrt(6^2+2^2)=sqrt(36+4)=color(green)(sqrt(40))`

`sqrt((-4)^2+3^2)=sqrt(16+9)=color(blue)(sqrt(25))`

Answer 2

The radicals come from the computation of the magnitudes.

If `veca = < a_x,a_y>`, then `|veca| = sqrt(a_x^2+a_y^2)`

Explanation:

Given: `vecu = < 6, 2 >`, then `|vecu| = sqrt(6^2+2^2) = sqrt40`

Given: `vecv = < -4, 3 >`, then `|vecu| = sqrt((-4)^2+3^2) = sqrt25`

Starting at the equation:

`cos(theta) = (-24+6)/(sqrt40sqrt25)`

`sqrt40` factors into `sqrt4sqrt10` which becomes `2sqrt10` :

`cos(theta) = (-24+6)/(2sqrt10sqrt25)`

The numerator becomes -18:

`cos(theta) = (-18)/(2sqrt10sqrt25)`

Which divides evenly by 2:

`cos(theta) = (-9)/(sqrt10sqrt25)`

The `sqrt25` is `5`

`cos(theta) = (-9)/(5sqrt10)`

The above is the simplification in slow motion.