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Hi

I don't understand the description in my book of direction cosines of vectors. There is a picture below.

Why, for example, is a/r = cos(alpha)? Is it using a right angled triangle that I can't see or some trig identity that I don't know or am I just missing the point entirely?

thanks

cosines.gif

I don't understand the description in my book of direction cosines of vectors. There is a picture below.

Why, for example, is a/r = cos(alpha)? Is it using a right angled triangle that I can't see or some trig identity that I don't know or am I just missing the point entirely?

thanks

cosines.gif

sorry i uploaded the wrong picture :(

cosines.jpg

cosines.jpg

"If you have a vector A, whose magnitude is A, and whose coordinates are (ax, ay, az), then consider the the right triangle formed by A with the x-axis (i.e., drop a perpendicular line from the head of A to the x-axis at the coordinate (ax, 0, 0). The angle alpha is then determined by the formula:

cos(alpha) = ax/A"

How do i know that the line from the head of A to the coordinate (ax, 0, 0) will be perpendicular to the x axis?

The diagram on the web page you linked to is clearer than the one i posted so shall we use that one? I've included it below

Image517.gif

A line from (ax,ay,az) to (ax,0,0) is perpendicular to the axis you say. (I'm not saying i don't believe you - just can't see it yet)

ozo said: The yz plane is perpendicular to the x axis, so any line in the yz plane will be perpendicular to the x axiz

How do a know the line from (ax,ay,az) to (ax,0,0) is in the yz plane? Should i be able to tell that from looking at the coordinates? If so, how. When i look at the diagram the line i am drawing in my head doesn't look perpendicular

I've only just learned about dot products today and only in 2d but if you think it would help and were willing to show me a worked example i'd work through it

Image517.jpg

My line E is in the yx plane. I think i can visualise how A = D+E but i still don't see how a line from (ax,0,0) to (ax,ay,az) is perpendicular to the x axis

Do i need dot products for this as I'm already confused

What do you mean project? What do you mean by dark line? Why is it dark? Is it one of the lines already on the diagram?

Ignoring any one of the 3 coordinates leaves you with a 2d situation, and they are still perpendicular.

Also, you can consider the 2d plane that goes through (ax,ay,az), (ax,0,0) and (0,0,0)

I can 'see' that the lines D and E from previous post are perpendicular using dot products but I can't see it in my head. I mean i can't visualise it i guess.

I have drawn some lines in the xy plane on a diagram. Are you saying they are all perpendicular to the x axis because this is what is confusing me. I just can't articulate it or draw it!

235-unit-vector.jpg

Let's start with your original diagram, with a vector u. You can piece u together as a sum of components in each of the directions x, y, and z.

u = ax + by + cz (Pretend I'm using whatever vector notation you are comfortable using, like u = ai + bj + ck or whatever.)

On your diagram, you can actually see the components of u on the x and y directions (but for some reason, they didn't draw it in for the z direction).

Your hidden right-angle triangle consists of the origin, the endpoint of the vector u, and the endpoint of the vector ax, on the x axis (in one of your diagrams you used a purple line to draw the last edge in). It's a little hard to see that it's a right angle, because it's rotated upwards a bit, but perhaps you can imagine it being moved back down onto the x-y plane.

The directional cosine is just the cosine of the angle at the origin for that hidden right-angled triangle, which is a / r. (r being the length of the vector u).

We can make directional cosines for the other two directions by drawing in the other two triangles.

Hope this helps!

Bend a wire at right angles.

Let one leg of that wire be the x axis.

Hold that axis fixed, and spin the wire around that axis.

See how the other leg sweeps out a plane.

The wire stays at a right angle, and anything in that plane is perpendicular to the fixed leg of the wire

glue a pencil at the origin making a 45 degree angle

drop a string from the tip of the pencil to the x-axis - make it perpendicular to the x-axis

look carefully at the string - isn't it parallel to the y-z cardboard plane?

cosines.jpg

Let

Given a vector

The vectors, A

Using the solid rectangular box in the video formed by

If you review the first parts of Linear Algebra, this may become clearer.

I do agree that it is at right angles from the dot product but it just doesn't look it. Was this comment correct: ID: 33497210?

I will watch that video on youtube

-- You drew a line from the tip of A to the x-y plane parallel to the z-axis

-- call intersection point B

-- draw a line from the tip of A to the x-z plane parallel to the y-axis

-- call intersection point C

From geometry (or intuition), do you see that these two line AB and AC are perpendicular to the x-y and x-z planes, respectively? After all, the y-axis is perpendicular to the x-z plane and AC is parallel to the y-axis.

-- You drew a line from B to the x-axis (call intersection D), and this line is parallel to the y-axis also. CABD form a rectangle - a plane that is parallel to the y-z plane.

I would not have realised that the a line in the yz plane was perpendicular to the x axis as you rotate it all the way around. It only looks perpendicular to me at the points of a compass

v1.jpg

v2.jpg

You drew a line from the tip of A to the x-y plane parallel to the z-axis

-- call intersection point B

as I didn't think the line i drew (which is teh orange pipe cleaner above) is parallel to the z axis?

>> -- You drew a line from the tip of A to the x-y plane parallel to the z-axis

In http:#33496429 are only two vertical lines - one is the z-axis and the other is the line that I was referring to. (This line is not in your pipe cleaner construction.)

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