hunt3rshadow said:
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 d
r/dt.
\displaystyle\frac{d\textbf{r}}{dt}=\ \left<2,2,\sqrt{2}\ \right>
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:
\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|}
Do similarly for the other two axes.
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