Direction Cosines ...

the_monkey

Hi ,
From p585 Engineering maths by KA Stroud , anyway

At first this seems easy, but looking at the angles as they are in 3D i get confused ...
for example take β
how is Cos β b/r ? at first i was thinking this is a standard right angle triangle but it's not, maybe its a bad diagram ....

amacachi

I could be wrong but I think the Beta angle is from the vector OP to the Y axis, so looking as though you were at y it's "tilted". If you take O and P as two vertices and the point where the corner of the box is on the Y axis as the other vertex then you have a right-angled triangle, because the j component of P is the same as the vector going to that point.
I'm very much open to correction on all this though, 3d stuff really hurts my head. :pac:

MathsManiac

Amacachi is correct. It is indeed a right-angle triangle.

The right-hand face of the dashed rectangular box is a plane perpendicular to the y-axis. So any line drawn on it will be perpendicular to the y-axis.

the_monkey

No, that would mean r would have to be the hypotenuse ??

doesn't work ...

I think the diagram is bad...

ZorbaTehZ

the_monkey wrote: »

No, that would mean r would have to be the hypotenuse ?

Why doesn't that work?
Think of a vector of length r in R^2 you would have r as the "hypotenuse" when calculating the angles too.

sponsoredwalk

Perhaps this will make it a bit clearer?



All the "direction cosines" are doing is telling you how the vector is
orientated with respect to the axes, you have to imagine the angle
between the vector and the x-axis as being two dimensional and
calculating the cosine, then you imagine the vector and the
y-axis as being two dimensional and calculating the cosine and the
same with the z-axis.

A clearer way to see the connection is to think of the norm of a vector.

[latex] \overline{A} \ = \ x \hat{i} \ + \ y \hat{j} \ + \ z \hat{k} [/latex]
[latex] \overline{A} \ = \ || \overline{A} || \ \hat{\lambda } \ = \ A \ \hat{\lambda }[/latex]
[latex] \overline{A} \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} \ \hat{\lambda } [/latex]
[latex] \overline{A} \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} \ (cos(\gamma) \ \hat{i} \ + \ cos(\alpha) \ \hat{j} \ + \ cos(\beta) \ \hat{k}) [/latex]

The [latex] \hat{\lambda} \ = \ (cos(\gamma) \ \hat{i} \ + \ cos(\alpha) \ \hat{j} \ + \ cos(\beta) \ \hat{k})[/latex] term is just a unit vector
pointing in the direction of the larger vector, as seen in black in the
next picture. The [latex] ||\overline{A}|| = A \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} [/latex]
term is just a constant, a scale factor, i.e. it scales/extends a unit vector
of length 1 up to whatever value your actual vector is.
Notice the angles, I labelled them slightly different to the way your book
has but the same idea goes. If you want to make things clearer:



[latex] \overline{A} \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} \ (cos(\alpha) \ \hat{i} \ + \ cos(\beta) \ \hat{j} \ + \ cos(\gamma) \ \hat{k}) \ = \ A \ (cos(\alpha) \ \hat{i} \ + \ cos(\beta) \ \hat{j} \ + \ cos(\gamma) \ \hat{k}) [/latex]

Okay, I'll go back to naming alpha α as the angle with the x-axis (white),
beta β as the angle with the y-axis (blue) and gamma γ as the angle with
the z-axis (yellow) as your book has it.

[latex] \overline{A} \ = \ || \overline{A} || \ \hat{\lambda } \ = \ A \hat{\lambda} \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} \ (cos(\alpha) \ \hat{i} \ + \ cos(\beta) \ \hat{j} \ + \ cos(\gamma) \ \hat{k}) \ = \ \sqrt{x^2 \ + \ y^2 \ + \ z^2} \ (\frac{a}{r} \ \hat{i} \ + \ \frac{b}{r} \ \hat{j} \ + \ \frac{c}{r} \ \hat{k}) [/latex]


Zoomed in version of the unit vector portion of the vector r

[latex] \overline{A} \ = \ || \overline{A} || \ \hat{\lambda } \ = \ A (\frac{a}{r} \hat{i} \ + \ \frac{b}{r} \hat{j} \ + \ \frac{c}{r} \hat{k}) [/latex]

But here r = 1 as we're talking about a unit vector so:

[latex] \overline{A} \ = \ || \overline{A} || \ \hat{\lambda } \ = \ A (a \hat{i} \ + \ b \hat{j} \ + \ c \hat{k}) \ = \ aA \hat{i} \ + \ bA \hat{j} \ + \ cA \hat{k}[/latex]

Remember, a, b & c are all less than one so it makes sense.

Thinking of things this way it is very easy to see how we're going to
find the direction cosines as the example in your book has asked.

[latex] \overline{A} \ = \ || \overline{A} || \ \hat{\lambda } [/latex]


[latex] \hat{\lambda } \ = \ \frac{ \overline{A}}{||\overline{A}||} \ = \ \frac{x \hat{i} \ + \ y \hat{j} \ + \ z \hat{k}}{||\overline{A}||} \ = \ \frac{x}{||\overline{A}||} \hat{i} \ + \ \frac{y}{||\overline{A}||} \hat{j} \ + \ \frac{z}{||\overline{A}||} \hat{k} [/latex]

This corresponds exactly to what your book has done.

If it's not clear I'll just make sure it is now:

r = 3i - 2j + 6k
||r|| = r = [latex] \sqrt{3^2 + (-2)^2 + 6^2} \ = \ \sqrt{49} \ = \ 7 [/latex]
[latex] \overline{r} \ = \ ||\overline{r}|| \ \hat{\lambda} \ = \ r \ \hat{\lambda} \ = \ 7 \ \hat{\lambda}[/latex]


[latex] \hat{\lambda} \ = \ \frac{\overline{r}}{||\overline{r}||} \ = \ \frac{3 \hat{i} \ - \ 2 \hat{j} \ + \ 6 \hat{k}}{7} [/latex]


[latex] cos(\alpha) \ \hat{i} \ + \ cos(\beta) \ \hat{j} \ + \ cos(\gamma) \ \hat{k} \ = \ \frac{3}{7} \hat{i} \ - \ \frac{2}{7} \hat{j} \ + \ \frac{6}{7} \hat{k} [/latex]

bnt

the_monkey wrote: »

No, that would mean r would have to be the hypotenuse ??

doesn't work ...

I think the diagram is bad...

It is the "hypotenuse", except that the concept of "hypotenuse" is complicated by the fact that P is a vector in three dimensions, rather than the usual two. Go back to the diagram, and imagine looking at the box from the X axis, so that it looks like a square. If you do that, you'll see only two of the three dimensions so it will make more sense. Cos β will be = b/r. Then you can do the same from other sides.

the_monkey

OK i see it , but with Alpha ??? How is cos alpha = a/r ???

Alpha is a big angle .... can't be in a L angle triangle..

sponsoredwalk

Look at my picture that has a yellow background, it shows a on the x-axis
how far the vector has gone in the x-direction. You have to think of
a plane going straight through the vector and the x-axis, like a
2-dimensional sheet tilted so that it contains the vector and the
x-axis line. Then you can use the idea of [latex] cos \theta \ = \ \frac{adjacent}{hypoteneuse}[/latex]
where the vector r is the hypoteneuse. The angle α in your book shows α
to be big but that's not a problem, why would it be?

the_monkey

sponsoredwalk wrote: »

where the vector r is the hypoteneuse. The angle α in your book shows α
to be big but that's not a problem, why would it be?

because then it could not possibly be a right angle triangle ?

and therefore the cos a = adj/hyp would not apply ?

sponsoredwalk

Do you see here? You see the green, this is a representation of the
two-dimensional plane connecting the vector r to the x-axis.
You have to split it up in your mind, when you take the angle of the
vector relative to the y-axis you imagine a two-dimensional plane
connecting the vector to the y-axis and find the angle between the
y-axis and the vector.
You do the same for the vector and the x-axis as shown above,



This is another image of the very same thing, when we're dealing with
the direction cosine in the x-direction we imagine a new entire x-y
plane and the red portion as just being the angle between the vector and
the x-axis. It's a 2-dimensional plane tilted.



Look here, the vector [latex]A_i[/latex], the picture doesn't show it but
use your mind to imagine an angle between the vector [latex]A_i[/latex]
and the [latex]y_3[/latex] axis, i.e. what we would think of as the
z-axis. You can see a 2-dimensional plane coming through the vector
and the axis and you have to imagine the same picture for each
seperate axis, this picture has a plane tilted and an angle α to the
x-axis.

https://www.youtube.com/watch?v=AJfEeKIdmxI

the_monkey

Thanks a lot sponsoredwalk ... that clears it up !