The vector A whose magnitude is 1.72 units makes equal angles with the coordinates axes. Find Ax,Ay,Az

Question

I have just done the problem but at the moment when I apply Pythagoras the magnitude comes out wrong, but it should be right. I did it with polar coordinates and then turned it into Cartesian coordinates.

Now here I just applied Pythagoras, equalling the magnitude and the answer seems to be right, but if I apply that result into the prior it gives me something different.

WHY AREN'T THESE RESULTS EQUAL?

Answer 1

Why did you set $\alpha = \beta = \gamma = 45^\circ$? That is incorrect. The vector does not lie in any of $xy$, $yz$, and $zx$ planes. The correct answer is $A_x = A_y = A_z = 1$.

Answer 2

The results are different because you took the liberty to claim that the angle was $45^\circ$. Whereas the angle must be $\cos \frac{-1}{\sqrt3}\simeq54^\circ$ (approx). Which can be found by the given data.

Since,

1. $A_x=A_y=A_z=0.999 \ $ & $ \ A_x=A.\cos(a)=0.999\simeq1$ (approx)

Also,

2. Let $A=1.732=\sqrt3$

Thus we have,

3. $\cos(a)=\frac{A_x}{A}=\frac1{\sqrt{3}}$ ; $a\simeq54^\circ$