The discussion revolves around finding the angles that a vector A = 3i - 6j + 2k makes with the coordinate axes. Participants are exploring the application of the dot product and the geometric interpretation of vector projections in relation to the coordinate system.
Some participants have provided clarifications regarding the dot product and its components, while others express confusion about the derivation of the value 3. The conversation is ongoing, with attempts to reconcile different interpretations of the calculations involved.
There are references to the definitions of the dot product and the need to consider both the geometric and algebraic perspectives. Participants also mention checking external resources for definitions and values related to the dot product.
You need to take the product of the vector A with i, not the magnitude of A:MaxManus said: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)
OK.MaxManus said:I was supposed to write A• i = (A)(1)cos(a) = sqrt(3**2 + (-6)**2 + 2**2) = 7*cos(a)
Exactly. Now use this result to solve for the angle a in your first equation.So
A• i = (3i - 6j + 2k)• i = 3i• i -6j• i + 2k•i = 3 - 0 + 0 = 3 ?
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.MaxManus said: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)
MaxManus said: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.
That's true. (I could have explained things better in my first response.)Mark44 said:There's nothing wrong with the above, as far as it goes.