1. ## 3d math questions

I have a book with questions at the end of each chapter, but no answers(3d Math Primer for Graphics and Game Development). I was wondering if anyone here could check my answers
Code:
```(dfdp)=dot for dot product
(cfcp)=cross for cross product
norm(x)=normalize vector x

I.
a)  -[3 7]  =  [-3  -7]
b)  ||[-12  5]||  =  13
c)  ||[8  -3  1/2]||  =  8.56(about)
d)  3[4  -7  0]  =  [12  -21  0]
e)  [4  5]/2  =  [2  2.5]

II.
a)  norm([12  5])  =  [12/13  5/13]
b)  norm([8  -3  1/2])  =  [100/107  -75/214  25/428](about)

III.
a)  [3  10  7] - [8  -7  4]  =  [-5  17  3]
b)  3[a  b  c] -  4[2  10  -6]  =  [3a-8  3b-40  3b-24]

IV.  Compute the distance between the pairs of points
a)  [3  10  7],[8  -7  4]  =  17.972(about)
b)  [10  6],[-14  30]  =  33.941(about)

V.
a)  I don't get this one.  [2  6] (dfdp) -38???
b)  3[-2  0  4] (dfdp) ([8  -2  3/2] + [0  9  7])  =  78  (Could someone explain the order of precedence on this one?)

VI.
a)  compute angle betweent the vectors [1  2]  and [-6  3]  =  90

VII.
Given the vectors
v=[4  3  -1], n=[sqrt(2)/2  sqrt(2)/2  0]
separate v into components that are perpendicular and parallel to n.(n is a unit vector)  I don't know how to do this one

VIII.
[3  10  7] (cfcp) [8  -7  4] =  [89  44  -101]```
thanks for any help.

2. III. b. should be :

[3a-8 3b-40 3c+24].

V. a. Doesn't make sense to me.

V. b. First take care of the left hand side.
Distribute the 3 through the vector. Then on the
right hand side, add the vectors together. Now take the dot product of the two vectors, which should be:

(-6, 0, 12) DOT (8, 7, 8.5) = (-6*8 + 0*7 + 12*8.5) = 54

I believe that is correct.

VII.
This just has to do with some simple projections. I'll go through it briefly but I'm sure you can read about it later.

I'll take it straight from my linear algebra book because they can explain it better than I can.

If u and a are vectors in 2-space or 3-space and if a != 0, then

proj a u = ((u DOT a) / (||a||^2)) * a

That yields the vector component of u along a (parallel).

u - proj a u = u - ((u DOT a) / (||a||^2)) * a

That yields the vector component of u orthogonal (perpendicular) to a.

Note: If your vector a is normalized, like in your example you gave, you don't need to divide out by the magnitude squared. That is the general case when you don't have a unit vector to work with.