1
$\begingroup$

I'm self-studying from Vector Analysis and Cartesian Tensors by Kendall because my lecturer is somewhat lacking, and I got conceptually stuck on excersise 1.11 dealing with direction cosines. What does it mean when we say that a line makes an obtuse angle with an axis? Is there a norm on which direction you're supposed to be looking from?

Also, a cosine squared appears in here (which gives 4 candidate solutions), but even if I reject two because it's obtuse or acute, that still leaves me with two potential answers and the solutions only say 135 degrees.

enter image description here

$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

The exercise makes more sense if you imagine that it means "ray" when it says "line". (A ray is half of a line, starting at a particular point and going straight towards infinity in some direction).

These rays then have a natural direction that provides a distinction between acute and obtuse angles. (For the coordinate axes you're considering the rays that start at the origin and grow towards positive coordinates, of course).

"Inclined at an obtuse angle to the $z$-axis" now means that the ray you're looking has its infinite end below the $xy$-plane, where the $z$-coordinates are negative.

This gives you only one solution for the angle -- in space geometry it doesn't make sense to count angles with sign; all angles measures are expected to be between $0$ and $180^\circ$.

$\endgroup$
1
  • $\begingroup$ Okay. I think that makes sense. Also, if I imagine there being cones around the x and y axes, there is only one intersection in the positive x,y,z cube, so that makes sense. So, angle from an axis is measured as if the tip of the positive direction was rotating anticlockwise around origin? (Assuming a horizontal axis growing to the right.) $\endgroup$
    – fazan
    Jul 14, 2018 at 0:56
0
$\begingroup$

The omitted word is "positive." It should say "to the positive $z$-axis." Likewise, "$x$-axis" and "$y$-axis" should (respectively) be replaced by "positive $x$-axis" and "positive $y$-axis$."

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .