# Are these vectors orthogonal?

1. Feb 2, 2008

### tony873004

1. The problem statement, all variables and given/known data
Determine whether the given vectors are orthogonal, parallel, or neither.

2. Relevant equations
$$\cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}\,\, \Rightarrow \,\,\theta = \cos ^{ - 1} \left( {\frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}} \right)$$

3. The attempt at a solution
$$\begin{array}{l} \overrightarrow {\rm{a}} = 2i + 6j - 4k,\,\,\,\,\overrightarrow {\rm{b}} = - 3{\rm{\hat i}} - {\rm{9\hat j}}\,{\rm{ + }}\,{\rm{6\hat k}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = \left( {2 \cdot - 3} \right) + \left( {6 \cdot - 9} \right) + \left( { - 4 \cdot 6} \right) = - 6 + \left( { - 54} \right) + \left( { - 10} \right) = - 58 \ne 0{\rm{__not_ orthogonal}} \\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} }}{{\left| {\overrightarrow {\rm{a}} } \right|\left| {\overrightarrow {\rm{b}} } \right|}}} \right) = \cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {2^2 + 6^2 + \left( { - 4} \right)^2 } \sqrt {\left( { - 3} \right)^2 + \left( { - 9} \right)^2 + 6^2 } }}} \right) = \\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {4 + 36 + 16} \sqrt {9 + 81 + 36} }}} \right) = \\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos ^{ - 1} \left( {\frac{{ - 58}}{{\sqrt {56} \sqrt {126} }}} \right) \approx 133.67^\circ \ne 0^\circ \,{\rm{not_ parallel,}}\,\, \\ \end{array}$$

But the back of the book says parallel.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 2, 2008

### Dick

-4*6=(-24). -6-54-24=-84. That's the dot product of a and b.

3. Feb 2, 2008

### tony873004

oops. back to 2nd grade for me:rofl:. Thanks, Dick.

4. Feb 2, 2008

### Dick

BTW you can also see that they are parallel by inspection. (-3/2)*a=b.

5. Feb 2, 2008

### foxjwill

\begin{align*} \textbf{a}\cdot\textbf{b} = 0 & &\Rightarrow & &\textbf{a} \perp \textbf{b}\\ \textbf{a}\times\textbf{b} = \textbf{0} & &\Rightarrow & &\textbf{a} \parallel \textbf{b} \end{align*}

6. Feb 2, 2008

### tony873004

Another student did it this way, but I don't see it. Where does -3 and 2 come from?

Cross product is next chapter, but thanks, that gives me a good preview of what's to come!

7. Feb 2, 2008

### foxjwill

Solve each of these for x:

\begin{align*} 2x &= -3\\ 6x &= -9\\ -4x &= 6 \end{align*}

8. Feb 2, 2008

### tony873004

I still don't get it. How does solving them demonstrate that a and b are parallel? Sorry, but the book does not explain this method.

9. Feb 2, 2008

### tony873004

now I get it. The -3 is factored out of b, and the 2 is factored out of a, The remaining parts of a and b are equal, therefore parallel.

10. Feb 2, 2008

### foxjwill

Exactly. In order for two vectors to be parallel, one must be a constant multiple (in this case $$-\frac{3}{2}$$) of the other.

11. Feb 2, 2008

### tony873004

Now I'm glad I made my original dumb mistake of 6*4=10, or I wouldn't have posted and learned this easier method.

12. Feb 3, 2008

### foxjwill

They say the only way to learn is to make mistakes. ;)