Q on Vector in 3D space.
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When it says that the wire is directed on a bearing of 15o:
what angle (counter-clockwise) does it make with x-axis?
what angle (counter-clockwise) does it make with y-axis?
can you write the unit vector along the wire?
Dr.Peterson
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SK gave you a place to begin. I'll suggest a direction to go. (Sounds like the vector equation of a line ...)
Think about the form they are writing equations in, [MATH](\mathbf{r}-\mathbf{a})\times \mathbf{b}=0[/MATH]. Do you see that a will be the position vector of a point on the line, and b is a vector in the direction of the line? Do you see why?
It will be very helpful for this discussion if you can tell us your understanding of the equation (and of the question as a whole), so we can see where you need help. I have to say that this is an unusual way to write the equation of a line, but it appears that they have previously introduced it to you.
Thank you for the help so far. I am just confused how a wire might be horizontal and have 15 degrees bearing. Thats pretty vertical to me. I am sure that meaning of bearing is what I know it as, that its the angle clockwise from north.
Dr.Peterson
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Place a map flat on the ground. It's horizontal, right?
Mark a bearing of 015 on it. It's between North and East. But it is still horizontal (meaning, parallel to the ground).
Now lift the map up in the air. The line you drew represents the wire. It is still horizontal, and still at the same bearing.
My guess is that you are not accustomed to three-dimensional thinking, and are confusing north being "up" on a map with "up" being "up" in the real world ...
I was thinking about it before sleeping. I realised that often with 3 d vectors the x y z are not as one might image, the y tends to horizontal and the z is vertical. but I wanted to sleep over it. I will check now. 3D vectors are nice questions.
I have solved the question now. l=cos 75, m cos 15 and n =0. That give you the direction vector. The a is (0,0,6), so you get the answer by using (r-a).b =0. The other answer is just solving a simultaneous equation and if the values of mu and lamda works for the third equation then the two lines intersect.
Dr.Peterson
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Correct. Of course, the work for solving (b) using that system of equations will be a little complex, which might be a reason a different approach would be better.
It sounds like you skipped over the first paragraph in the problem, which told you explicitly that k is vertical. This is a reason for rereading a problem several times, so that after you raise a question (such as, how can this be horizontal), the next thorough reading will reveal that they have answered your question.
It sounds like you skipped over the first paragraph in the problem, which told you explicitly that k is vertical. This is a reason for rereading a problem several times, so that after you raise a question (such as, how can this be horizontal), the next thorough reading will reveal that they have answered your question.
HallsofIvy
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I am puzzled as to why you say "with 3 d vectors the x y z are not as one might imagine". Although the choice for which directions the axes are is arbitrary, the usual is "x east-west, y north-south, z up-down".
Yes but x is horizontal and y is vertical until you start doing 3D vectors. am in right? Its z that is vertical by convention. It makes life easier if you understand conventions such as there is no reason why l,m, n should be cosine angles for in x, y and z but knowing this makes life a lot easier. They could use u,v and w and different letters (not alphabets!) the following time. I just didnt think in 3D when they started talking about horizontal and vertical. my bad.
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