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In vectors, if a vector is broken down into its components then can one of the components of the vector have the same magnitude of the vector itself??

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closed as off-topic by stafusa, AccidentalFourierTransform, ZeroTheHero, user191954, Jon Custer Sep 11 '18 at 13:06

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    $\begingroup$ Hint: When can $x = \sqrt{x^2+y^2+z^2}$ $\endgroup$ – ja72 Sep 10 '18 at 16:49
  • $\begingroup$ Yes if the components are orthogonal and if all the other components have zero magnitude (need to have 'orthogonal'). $\endgroup$ – user45664 Sep 10 '18 at 17:49
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If a component $a$ of the vector has the same magnitude of the vector $\vec{v}$, it means that it is decomposed in a direction parallel to it:

$a = \|\vec{v}\|\cdot \cos{\theta}$

$a = \|\vec{v}\|$

if $\cos{\theta} = 1$

so $\theta = 0$

it means that there is no angle of difference by the direction of the vector and the line on which the vector is decomposed in its component.

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  • $\begingroup$ can you answer my next question its related to this question and friction???? $\endgroup$ – Mad Dawg Sep 10 '18 at 15:52
  • $\begingroup$ @MadDawg This is not true unless more assumptions are made. See my answer below. $\endgroup$ – Aaron Stevens Sep 10 '18 at 16:03
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In opposition to Costantino's answer, I am going to say it actually is possible if you are not using orthogonal components.

An example: $$\langle1,0\rangle=\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle+\langle1-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\rangle$$

As you can see, the unit vector in the x-direction can be expressed as having a component along the vector $\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle$, which also has unit length, and the vector $\langle1-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\rangle$. Of course, our two components are not orthogonal, since $\langle\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\rangle\cdot\langle1-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\rangle\neq 0$. But orthogonal components was not asked for in the question.

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Usually we represent vectors as a tuple of coordinates $\mathbf{V}=(x,y,z,\ldots)$ in some coordinate system, but the vector $\mathbf{V}$ is not really a tuple. One can represent it in other ways (such as a derivative along a certain curve, popular in differential geometry).

So treating it as $\mathbf{V}=[\text{magnitude},\text{direction}]$ is entirely OK. In 2D this is fairly easy, one can think of it as $[r,\theta]$. In 3D the direction part now is a 3D direction, which can be thought of as a unit vector, a point on the unit sphere, or the complex parts of a quaternion.

However, when deciding how to represent a vector one also needs to consider what makes it easy to calculate with. While vector addition is well defined, you will not get the right answers if you do a component-wise addition in the above examples (the two orthogonal unit vectors along the x- and y-axis can be represented as $[1,0]=(1,0)$ and $[1,\pi/2]=(0,1)$, but their sum is not $[2,\pi/2]=(0,2)$!) So when inventing another representation you need to find the right formulas for how the components change under elementary operations.

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    $\begingroup$ I don't see how this answers the question $\endgroup$ – Kyle Kanos Sep 11 '18 at 1:47
  • $\begingroup$ I answered the question as it was originally stated. An edit changed it into a totally different one. $\endgroup$ – Anders Sandberg Sep 11 '18 at 23:32

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