Scalar product in vector/coordinate form

Question

As I know, $a*b = |a|*|b|*cos(a,b)$ in vector form

And $a*b = (a_1,a_2)*(b_1,b_2) = a_1*b_1+a_2*b_2$.

1)

$$a*b=?$$

$$a=2i-3j+5k$$

$$b=i+2j+8k$$

SOLUTION: $a*b = (2,-3,5)*(1,2,8) = 2*1+(-3)*2+5*8 = 36.$

I'm unsure about the angle between them? Do I have to do only multiplication of those two?

$$a*b = (2i-3j+5k)*(i+2j+8k) = \dots$$

2)

$$a*b=?$$ $$a=(1,2,3)$$ $$b=(4,5,6)$$

Do I have to do the same?

SOLUTION: $(a*b) = (1,2,3)*(4,5,6) = 1*4+2*5+3*6 = 32$

3) Find the angle between $a$ and $b$?

$$a=2i-j+2k$$ $$b=i+j+k$$

SOLUTION: a*b = (2,-1,2)*(1,1,1) = 3

|a| = 3

|b| = sqrt(3)

cos(a,b)=?

What's the formula for calculating the angle?

Answer 1

Hint: for 1, you are given the components, so use the formula you have after "And" but with three components and you are there. Note that $i,j,k$ are unit vectors, so $i\cdot i=j \cdot j = k \cdot k =1$ For 2, the same. For 3, get the dot product the same way, evaluate $|a|=\sqrt {a \cdot a}$ and $|b|$, then use these to get $\cos (a,b)$ by combining your first two equations.