Help with linear questions. Got wrong on test and correcting them

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Here are some questions that I received on my test. I got most of these wrong but I got a couple points here and there. Here are the questions I had trouble with. At the bottom on the questions I tried to solve them but I'm not sure if i did them correctly. Please try and help me because I have a final coming up this week and I want to learn how to solve these questions.

1. Consider the three vectors u=(4,-1,-5) , v=(1,-4,1) and w= (1,1,-2)
a) compute the scalar triple product of u,v,w.
my attempt:
u * (v X w)
4 -1 -5
1 -4 1
1 1 -2

=4(7) + 1(-3) -5(5)
= |0|
=0

b) What can you deduce about the vectors u,v,w, supposing they have the same initial point?
my attempt: That the answer would be different because it would be multiplying with different multiples…?

2. Considers the points A(1,0,1) B(1,2,3) C(-1,0,2)
a) find the angle between the vectors AB and AC
my attempt:
cos=AB*AC/||AB|| ||AC||
=2/(8)^1/2 (5)^1/2
= 0.3162
=cos-1(0.3162)
=71.56

b) Find a vector equation of the line through A and B
my attempt:
A=(1,0,1)
B= (1,2,3
?

3. Considers the L1(line 1) with symmetric equation (x-1)/-1 = (y+2)/2 =z/3
and the L2(line 2) parallel to v2=(1,0,-1) and throughout the point P2(0,1,0)

a) Find the direction vector v1 for the line L1 and give a point P1 on L1
my attempt:
p1 (1,1, 1/3)
v1 (1,1,1)

b) Find a parametric equation of the line L2
my attempt:
v2= (1,0,-1)
P2= (0,1,0)
L2 should equal {t, 1, -1}

c) show that the lines L1 and L2 are skew lines
my attempt:
|x1x2 * (V1 X V2)| \ ||V1 X V2||

=(0,1,0) * ( -1, -2, -1) / 6^1/2
= -2/6^1/2 d) Find a unit vector u orthogonal to both v1 and v2
my attempt:
v1 (1,1,1)
v2 (1,0, -1)
v3 ( , , )
v1 X v2
1 1 1
1 0 -1
u=( -1,0,-1)

e) find the orthogonal projection P1P2 on u
(p1p2 * u/ ||u|| ) u
=(-2/2^1/2, 0, -2/2^1/2)

f) deduce the distance d=||Proju P1P || between the lines L1 and L24. If u and v are vector in n-space, Simplify: ( u+v) * (u-v)
my attempt:
=uu –uv + uv - vv
=||u||^2 - ||v||^2

b) use your previous result to show that the parallelogram defined by u and v is a rhombus if and only if its diagonals are perpendicular.
A rhombus has 4 sides that are equal and this would prove it.
 
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olliebear said:
1. Consider the three vectors u=(4,-1,-5) , v=(1,-4,1) and w= (1,1,-2)
a) compute the scalar triple product of u,v,w.
my attempt:
u * (v X w)
4 -1 -5
1 -4 1
1 1 -2

=4(7) + 1(-3) -5(5)
= |0|
=0

b) What can you deduce about the vectors u,v,w, supposing they have the same initial point?
my attempt: That the answer would be different because it would be multiplying with different multiples…?
Geometrically, the scalar triple product is the (signed) volume of the parallelepiped defined by the three given vectors. If the volume of this parallelepiped is zero, then...?
 
it means that the parallelepiped is planar and has no volume. This means that the given three vectors are linearly dependent?
 
olliebear said:
it means that the parallelepiped is planar and has no volume. This means that the given three vectors are linearly dependent?

Yes.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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