Direction of cartesian vector

[werson tan]

Homework Statement

if the alpha is larger than 90 degree ( which means the resultant F is lean towards -x axis , then the angle between the Ax and the line from Ax to A will be less than 90 , am i right ?

Homework Equations

The Attempt at a Solution

 

[BvU]

No. It will be exactly 90$^\circ$

(the 90$^\circ$ as indicated in figure 2-26 (a) )

 

[werson tan]

No. It will be exactly 90$^\circ$

(the 90$^\circ$ as indicated in figure 2-26 (a) )

why , i can't visualize it??

 

[werson tan]

No. It will be exactly 90$^\circ$

(the 90$^\circ$ as indicated in figure 2-26 (a) )

If the force is lean towards negative x-axis , it means the angle between the force and positive x -axis is more than 90 degree, right ? then how is it possible for it to remain 90 degree?

 

[BvU]

You were talking about the line from Ax to A, not about the line from the origin to A (i.e. $\vec A$ itself).

 

[werson tan]

You were talking about the line from Ax to A, not about the line from the origin to A (i.e. $\vec A$ itself).

ya , i mean the line from Ax to A, not about the line from the origin to A (i.e. A⃗ itself).
if the force is lean towards the negative x-axis then the , we would have three line , right , namely the line joining origin to Ax , A(which is lean towards negative x-axis ) , and a line which join A to negative x-axis , right ? if it is so , then the i an undersatnd why the angle is 90 degree. if it so , the triangle formed is at opposite of the current one .

 

[BvU]

I think you've got it all right and correct, just that the wording in post #1 was rather unfortunate.
With nowadays visualization possibilities it's almost a stone-age approach to put things in words, but let's do it anyway:

let's look in the blue plane of figure 2-26 (a) for the case $\alpha \in [\pi/2, \pi]$ where all the angles in our story are

[edit] I mean to say that all the angles are in that plane. Sorry for the ambiguity.​

so the vertical line A_yz is in the yz plane and that plane is perpendicular to the x axis.

and as you can see the line from A_x to A (the green line) is parallel to that plane and therefore also perpendicular to the x axis. Hence my 90 $^\circ$.

You of course in your post #1 meant to refer to the angle between $\vec A$ and $\vec A_x$ which as you can see is $\pi-\alpha$ and thereby in the range 0 to 90$^\circ$.

Note that I forgot to draw the arrow above $\vec A_x$ and $\vec A_{yz}$. Nobody is perfect, be we keep trying.
To be specific: in $\vec A_x = A_x\, \hat\imath\$, $\ A_x$ is a number (negative for the alpha in the picture: $A_x = |\vec A | \cos\alpha$​