Vector problem

[Dome]

I have this problem and cant think of the solution. I hope someone can solve it.

Consider a vector A = {2; -3; 4}. Find the projection of vector A in vector B, which forms equal acute angles with the coordinate axes.

So I thought that to find the projection one should use the dot product:

A*B = |A||B| cos(t) , where |A| and |B| are the lengths of vectors and t is the angle between those vectors.

|A|*cos(t) = projection of A in B (I'l denote this as A(b)).

So:
A*B = A(b)|B|.
Since B = |B|B0 (where B0 is the unit vector of B), I thought that we can write the dot product as
A*B0|B| = A(b)|B|
The term |B| cancels out and we're left with

A(b) = A*B0

So then I thought that if vector B forms equal acute angles, they should all be equal to 45 degrees. And so the coordinates of vector B0 = {cos(45), cos(45), cos(45)}.
Then the dot product is:
A(b) = A*B0 = 2*cos(45) - 3*cos(45) + 4*cos(45) = 3*cos(45) = 3sqrt(2)/2.

BUT the textbook I'm using says the answer is sqrt(3).
So, can anyone help me please?

 

[daon2]

Your problem is a little poorly worded. The assumption you're making about a 45 degree angle is also not correct.

If a vector is to make equal acute angles with all three axes in three-space (say in quadrant one) then it is the diagonal of a cube (as opposed to a square in two dimensions). For the purpose of obtaining the angle, you can find the angle between vectors (1,0,0) and (1,1,1)

 

[Deleted member 4993]

I have this problem and cant think of the solution. I hope someone can solve it.

Consider a vector A = {2; -3; 4}. Find the projection of vector A in vector B, which forms equal acute angles with the coordinate axes.

So I thought that to find the projection one should use the dot product:

A*B = |A||B| cos(t) , where |A| and |B| are the lengths of vectors and t is the angle between those vectors.

|A|*cos(t) = projection of A in B (I'l denote this as A(b)).

So:
A*B = A(b)|B|.
Since B = |B|B0 (where B0 is the unit vector of B), I thought that we can write the dot product as
A*B0|B| = A(b)|B|
The term |B| cancels out and we're left with

A(b) = A*B0

So then I thought that if vector B forms equal acute angles, they should all be equal to 45 degrees. And so the coordinates of vector B0 = {cos(45), cos(45), cos(45)}.
Then the dot product is:
A(b) = A*B0 = 2*cos(45) - 3*cos(45) + 4*cos(45) = 3*cos(45) = 3sqrt(2)/2.

BUT the textbook I'm using says the answer is sqrt(3).
So, can anyone help me please?

The unit vector along B is conventionally called e_B (instead of Bo) would be:

\(\displaystyle e_B \ = \ \dfrac{i \ + \ j \ + \ k}{\sqrt{1^2 \ + 1^2 \ + \ 1^2}}\)

This is a 3-dimensional case. The angles between the axes and vector B (related to direction cosines) can be called α, ß and Θ and

\(\displaystyle \alpha \ = \ \beta = \ \theta \ =\ \cos^{-1}\left (\dfrac{1}{\sqrt{3}}\right )\) = 54°44'08"

But finding the angle is really unnecessary for this problem.

If I were to do this problem, I would write those as cartesian vectors and:

A = 2 i + (-3) j + 4 (k)

e_B = i/√3 + j/√3 + k/√3

A . e_B = 2/√3 - 3/√3 + 4/√3 ............ continue

 

[Dome]

Thanks for response. I now get the things with angles and unit vectors. But do you think the equation:

A . e_B = A_B

is correct?

 

[HallsofIvy]

Thanks for response. I now get the things with angles and unit vectors. But do you think the equation:

A . e_B = A_B

is correct?

I presume that "\(\displaystyle e_B\)" is a basis vector and "\(\displaystyle A_B\)" is the component of A parallel to \(\displaystyle e_B\).

If B is an "orthonormal" basis- that is, that all basis vectors are orthogonal to one another and unit length, yes, that is true.

 

[Dome]

Alright thanks, got it