Find Angles of Vector A with Coordinate Axes

[MaxManus]

Homework Statement

Find the angles which the vector A = 3i -6j +2k makes with the coordinate axes

The Attempt at a Solution

Let a, b, c be the angles which A makes with the positive x,y,z axes.
A• i = (A)(i)cos(a) = 7*cos(a)


The Solution says:
A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3

And I do not understand how they get 3 as the answer.

 

[Doc Al]

The Attempt at a Solution

Let a, b, c be the angles which A makes with the positive x,y,z axes.
A• i = (A)(i)cos(a) = 7*cos(a)

You need to take the product of the vector A with i, not the magnitude of A:

\vec{A}\cdot \hat i = A \cos\theta

What do these equal?
i • i = ?
j • i = ?
k • i = ?

 

[MaxManus]

I was supposed to write A• i = (A)(1)cos(a) = sqrt(3**2 + (-6)**2 + 2**2) = 7*cos(a)

second question:

had to check the book and it says
i• i = 1
j• i = j• k = 0

ah, thanks

So
A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3 - 0 + 0 = 3 ?

 

[Doc Al]

I was supposed to write A• i = (A)(1)cos(a) = sqrt(3**2 + (-6)**2 + 2**2) = 7*cos(a)

OK.

So
A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3 - 0 + 0 = 3 ?

Exactly. Now use this result to solve for the angle a in your first equation.

 

[MaxManus]

Thanks

 

[Mark44]

Homework Statement

Find the angles which the vector A = 3i -6j +2k makes with the coordinate axes

The Attempt at a Solution

Let a, b, c be the angles which A makes with the positive x,y,z axes.
A• i = (A)(i)cos(a) = 7*cos(a)

There's nothing wrong with the above, as far as it goes. In addition to the coordinate definition of the dot product, there is the definition that involves the magnitudes of the vecctors and the angle between them.

In this case cos(a) = (length of the projection of A onto the x-axis)/(magnitude of A) = 3/7.

So A \cdot i = 7 * cos(a) = 7 * 3/7 = 3

The Solution says:
A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3

And I do not understand how they get 3 as the answer.

 

[Doc Al]

There's nothing wrong with the above, as far as it goes.

That's true. (I could have explained things better in my first response.)