Finding Cosine of an Angle in 3D space

[hunt3rshadow]

Homework Statement

a) Find the cosine of the angles that the line r = [-3,2,5] + t [2,2,√2 ] makes with the coordinate axes.

b) If a,b and c are the angles that the line makes with the x,y and z axis respectively, find the value of cos^2a + cos^2b + cos^2c.

c) What is the magnitude of the vector [cosa,cosb,cosc]

d) What can you say about the direction of the vector [cosa,cosb,cosc] and the direction of the vector line?

Homework Equations

I believe we have to use CosTheta = r . t / |r||t| but it doesn't seem right because this is a vector equation involving direction.

 

[gneill]

Homework Statement

a) Find the cosine of the angles that the line r = [-3,2,5] + t [2,2,√2 ] makes with the coordinate axes.

b) If a,b and c are the angles that the line makes with the x,y and z axis respectively, find the value of cos^2a + cos^2b + cos^2c.

c) What is the magnitude of the vector [cosa,cosb,cosc]

d) What can you say about the direction of the vector [cosa,cosb,cosc] and the direction of the vector line?

Homework Equations

I believe we have to use CosTheta = r . t / |r||t| but it doesn't seem right because this is a vector equation involving direction.

Actually, you'll want to use the dot product of a vector that is parallel to the line with unit vectors on your coordinate axes. So if you construct vector [itex] \vec{r} [/itex] to lie parallel to your line, then
[tex] cosa = \frac{\vec{r} \cdot \hat{i}}{|\vec{r}||\hat{i}|} [/tex]
and so on for the unit vectors [itex] \hat{i},\hat{j},\hat{k} [/itex] to yield cosa, cosb, cosc.

. . .


Some final notes: When it is asking for cosines of the angles, are they asking for more then one? And also it asks for each coordinate axes, does that mean we have to find it for x , y and z? Finally, I'm not sure if this belongs here because the course is called Calculus and Vectors, so this type of question is more math based rather then physics, but I'm hoping you guys can still help! Thanks

Yes, there will be three angle cosines to find, one for each axes of the coordinate system.

Your first step should be to find a vector r that is parallel to your line.

 

[hunt3rshadow]

How do I find a the line parallel to R?

 

[gneill]

How do I find a the line parallel to R?

You have a line given in parametric form: [-3,2,5] + t [2,2,√2 ]. The [-3,2,5] bit is just an offset. You could shift the line to pass through the origin by subtracting this offset, and the result would obviously be a line that is parallel to the original. Thus the line r' = t [2,2,√2 ] is parallel to the original line. Plug in any value of t that you like (1 is a good choice!) to find a vector that lies in the line.

 

[SammyS]

Homework Statement

a) Find the cosine of the angles that the line r = [-3,2,5] + t [2,2,√2 ] makes with the coordinate axes.

b) If a,b and c are the angles that the line makes with the x,y and z axis respectively, find the value of cos^2a + cos^2b + cos^2c.

c) What is the magnitude of the vector [cosa,cosb,cosc]

d) What can you say about the direction of the vector [cosa,cosb,cosc] and the direction of the vector line?

Homework Equations

I believe we have to use CosTheta = r . t / |r||t| but it doesn't seem right because this is a vector equation involving direction.

The Attempt at a Solution

Essentially that above

a) (-3)(2)+(2)(2)+(5)(√2) /(√9+4+25)(√4+4+2)

Cos-1 of the number is 74 degrees.

b) c) and d) I am completely lost because I do not understand where I'm suppose to find those values. That most likely means I did a) incorrect.

Some final notes: When it is asking for cosines of the angles, are they asking for more then one? And also it asks for each coordinate axes, does that mean we have to find it for x , y and z? Finally, I'm not sure if this belongs here because the course is called Calculus and Vectors, so this type of question is more math based rather then physics, but I'm hoping you guys can still help! Thanks

Hello hunt3rshadow. Welcome to PF !

Usually you should post no more than two questions when opening a thread. These all look fairly closely related, so I suppose it's OK.

I'll try to get you headed in the right direction, but it's up to you to actually solve the problem(s).

For (a):
The vector, r = [-3,2,5] + t [2,2,√2 ], is a position vector having its tail at the origin, and its head it the point (x, y, z) = ( -3 + 2t, 2 + 2t, 5 + (√2)t ), i.e.

x = -3 + 2t,
y = 2 + 2t,
z = 5 + (√2)t .​

As t goes from -∞ to +∞, the head of vector r(t) traces a straight line. The vector [-3,2,5], has nothing to do with the direction of this line. It only specifies the location of a point, (-3,2,5), on the line that corresponds to t = 0. The direction is given by the vector that t multiplies, namely, [2,2,(√2) ]. Note that this is also equal to dr/dt.

[itex]\displaystyle\frac{d\textbf{r}}{dt}=\ \left<2,2,\sqrt{2}\ \right>[/itex]​

You have to find the cosine of the angle this vector makes with respect to (w.r.t) the x-axis, ... then the cosine of the angle this vector makes w.r.t. the y-axis,... then the cosine of the angle this vector makes w.r.t. the z-axis.

The x-axis is in the direction of [1, 0, 0].

This is where you use that equation you have:

[itex]\displaystyle\cos(\theta)=\frac{ \left\langle 2,\,2,\,\sqrt{2} \right\rangle\cdot \left\langle 1,\,0,\,0 \right\rangle}{\left| \left\langle 2,\,2,\,\sqrt{2} \right\rangle \right|\ \left| \left\langle 1,\,0,\,0 \right\rangle \right|}[/itex]
​

Do similarly for the other two axes.

 

[hunt3rshadow]

Oh my gosh guys I really owe you! You two made it seem so easy

 

[SammyS]

Oh my gosh guys I really owe you! You two made it seem so easy

And sorry about that, I didn't know you could only post one question per thread, they were kinda connected so I thought it was okay! I'll keep that in mind next time.
I'm going out to eat but I understand the gist of the question now. I'll do it when I get home and post the answers to see if I did it correctly. Thanks again!

EDIT: I just tried connecting gneill and SammyS's explanations and there is one contrast I'm slightly confused about. gneill you said to use a parallel line while sammyS chose to use a coordinate which lies on the x,y,z axis (i.e 1,0,0). Maybe I misread the explanation but can somebody tell me the difference between the two or will the answers be the same?

A vector parallel to the line is the vector < 2, 2, (√2) > . That's basically the dr/dt vector I discussed.

 

[hunt3rshadow]

Ahh I see, thanks.

 

[gneill]

Ahh I see, thanks. I got a) :D

Now I'm abit confused on b). I have the angles that the lines make with the x,y,z coordinates. How do I insert it into the eqtn cos^2a + cos^2b + cos^2c? Am I just squaring the angles and adding them? Sorry for so many questions, I'm not good with these concepts.

Just do as the formula states and add the sum of the squares of the cosines of the angles; you don't need to find the angles themselves, just use the cosines. The result should be particularly satisfying .

 

[hunt3rshadow]

so much more simple then I imagined! Thanks