How find direction given $i,j,k$? [closed]

A proton initially has $\vec v = 8.0\vec i − 3.0\vec j + 5.0\vec k$ and then $4.0 s$ later has $\vec v = −5.0\vec i − 3.0\vec j + 4.0\vec k$ (in meters per second).

For that $4$ sec, what's the direction of average acceleration?

Isn't angle determined by tan inverse $\ast \frac{y}{x}$? And $y = k$ and $x = i$?

closed as unclear what you're asking by Jack's wasted life, Watson, user223391, user91500, Claude LeiboviciSep 27 '16 at 11:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Too small to read it comfortably... – DonAntonio Sep 25 '16 at 12:54
• It's fixed now. – user366783 Sep 25 '16 at 13:02
• I would prefer seeing a typed problem, instead of the image of some webassign homework. – iamvegan Sep 25 '16 at 13:08
• I edited the text in before you replied. – user366783 Sep 25 '16 at 13:39

The situation is that you correctly calculated a vector $-3.25i + 0j - 0.25k$, you correctly calculated the magnitude of that vector, but the online assignment software said you incorrectly calculated the direction of the vector.

The direction of the vector was requested as the angle from the positive $x$ axis in the $x,z$ plane. Conveniently, the vector already is in the $x,z$ plane. And indeed, one step in finding the angle from the positive $x$ axis in the $x,z$ plane is to take the component of $k$ and divide by the component of $i$.

To avoid confusion, I would call the component of $k$ the $z$ component, not the $y$ component. The formula $\arctan(y/x)$ really is meant only for angles measured from the $x$ axis in the $x,y$ plane. It's OK to memorize the formula that way, as long as you remember that if you are in the $u,v$ plane instead, the formula needs to be written $\arctan(v/u)$.

You then calculated that $$\arctan\frac{-0.25}{-3.25} \approx 4.40 \text{ degrees.}$$ That is a correct statement, but it omits one very important fact about the arc tangent. Notice what happens when we apply the formula to a different vector, $3.25i + 0j + 0.25k$: $$\arctan\frac{0.25}{3.25} \approx 4.40 \text{ degrees.}$$

Now consider this: if the direction of a vector $V$ is $4.4$ degrees from the positive $x$ axis, that vector and the positive $x$ axis are pointing almost in the same direction. But the vector $3.25i + 0j + 0.25k$ is exactly opposite the vector $-3.25i + 0j - 0.25k$: $$-3.25i + 0j - 0.25k = -(3.25i + 0j + 0.25k).$$ So those two vectors should be pointing $180$ degrees away from each other. How can both of them be only $4.4$ degrees from the positive $x$ axis?

The clue to the puzzle is that $$\frac{-z}{-x} = \frac{z}{x},$$ that is, when we "flip" the signs of both $x$ and $z$, the ratio $\frac zx$ doesn't change at all, and therefore $\arctan(z/x)$ also doesn't change. You get the same arc tangent for vectors pointing in opposite directions.

The fact is, just taking the arc tangent will sometimes give you the correct answer, but only if the angle you need to calculate is $90$ degrees or less. The arc tangent function will never give you a result outside that range. But it is certainly possible to have a vector whose angle from the positive $x$ axis is between $90$ and $180$ degrees, or between $-90$ and $-180$ degrees (measured counterclockwise).

One way to fix this is to add or subtract $180$ degrees when you need the result to be outside the range $-90$ to $90$ degrees. Compare your problem to the problem in this answer, where the vector had $x$ coordinate $x_1-x_0$ and $y$ coordinate $y_1-y_0$ and the problem was to find the angle in the $x,y$ plane.