Vector Geometry question

[sigma]

Not sure if this is the right place for this question, but the question reads as:

Let u = (2,a,5) and v = (a, 8, 10) be two vectors in \(\displaystyle R^3\).

a. For what values of a will the vectors u and v be parallel? Explain why.

b. For what values of a will the vectors u and v be orthogonal? Explain why.

So this is what I did for part a using the standard basis vectors notation:

u= 2i + aj + 5k, v= ai + 8j + 10k

taking the dot product:

2a + 8a + 50 = 0
10a = -50
a = -5

Thats doing the dot product but there's also the cross product. For this type of quesiton, do I do the dot or cross? I know that two vectors are parallel only if one is a multiple of the other but I don't know if that means taking the dot product or cross product (since both methods give you multiples of one another).

for b, I did the same thing and I just showed that when a= -5, the dot product is 0. I know that to show if 2 vectors are orthogonal, their dot product must be 0 but something doesn't seem right with what I'm doing.

When a= -5, 10(-5)= -50
-50=-50
0=-50+50 = 0

This doesn't seem right though. Help is greatly appreciated. Thanks.

 

[pka]

a) Find a value of a for which this is true: \(\displaystyle \L
\frac{2}{a} = \frac{a}{8} = \frac{5}{{10}}\) .

b) Find a value of a for which this is true:\(\displaystyle \L
\left\langle {2,a,5} \right\rangle \cdot \left\langle {a,8,10} \right\rangle = 0\)

 

[galactus]

Notice, what value of a would make the vectors scalar multiples of each other, therefore, parallel.

Also, the cross product of parallel vectors, u and v, is 0.

 

[sigma]

Ok. I hope figured out b this is what I did. I treated the "a" like a 1. So doing the cross product of u and v:

\(\displaystyle \
\L\
\begin{array}{l}
u \times v = \left[ \begin{array}{l}
2{\rm }1{\rm }5 \\
1{\rm }8{\rm }10 \\
\end{array} \right] \\
u \times v = \left( {\left| \begin{array}{l}
1{\rm 5} \\
8{\rm 10} \\
\end{array} \right|, - \left| \begin{array}{l}
2{\rm 5} \\
{\rm 1 10} \\
\end{array} \right|,\left| \begin{array}{l}
2{\rm 1} \\
{\rm 1 8} \\
\end{array} \right|} \right) \\
= \left( {\left( {10 - 40} \right), - \left( {20 - 5} \right),\left( {16 - 1} \right)} \right) \\
= \left( { - 30, - 15,15} \right) \\
\end{array}
\\)

Now taking the dot product of that and the values for u and v seperate, they come to zero.

\(\displaystyle \
\L\
\begin{array}{l}
\left( { - 30, - 15,15} \right) \bullet \left( {2,1,5} \right) \\
= - 60 - 15 + 75 = 0 \\
\end{array}
\\)

\(\displaystyle \
\L\
\begin{array}{l}
\left( { - 30, - 15,15} \right) \bullet \left( {1,8,10} \right) \\
= - 30 - 120 + 150 = 0 \\
\end{array}
\\)

Is that right? Unfortunately, I'm still having some trouble with part a. How did you get that pka? I would have never figured that out.

 

[pka]

So doing the cross product of u and v. I'm still having some trouble with part a. How did you get that pka? I would have never figured that out.

I would never suggest using ‘cross products’ on either of the two questions.

a) Two vectors a parallel if they are multiples of each other.
Therefore, their direction numbers are proportional: \(\displaystyle a = 4\quad \Rightarrow \quad 2 < 2,a,5 > = < a,8,10 >\) .

b) Two vector are perpendicular if their dot-product is zero.
\(\displaystyle \begin{eqnarray}
< 2,a,5 > \cdot < a,8,10 > & = & 0 \\
2a + 8a + 50 & = & 0 \\
a & = & - 5 \\
\end{array}\)

 

[sigma]

Ok, what I was doing in part a before was what I was suppose to do in part b, so I understand that. And I understand how you got a = 4 in part a but getting that setup I'm just having a hard time grasping that concept of that parallel vectors rule so bare with me for a sec, but I want to understand why there is a 2 in front of that first part and why they are equated to each other and then I can hopefully lay this question to rest.

 

[pka]

\(\displaystyle \L
2 < 2,4,5 > = \left\langle {\left( 2 \right)2,\left( 2 \right)4,\left( 2 \right)5} \right\rangle = < 4,8,10 > .\)

 

[sigma]

Thanks. I understand this now. It was the whole parallel proportion thing that was throwing me off. My text book is very poor and does not show any kind of example like this or even anything close to resembling a problem like this.