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So this question came to my head this morning. When analyzing such a problem, its easier to break the W component into 2 seperate components, here I've marked the components in Red & Pink arrows.

I've been doing this type of questions a lot but never thought of it. Why should those 2 components of W (Red & Pink) be Perpendicular to each other? Why can't it be like the first (top) diagram I made.

  • $\begingroup$ The only requirement is that the red and pink arrow, when added together make the black arrow, which means the red arrow in the lower diagram is too long $\endgroup$ – Triatticus May 18 at 12:43
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    $\begingroup$ @Triatticus The OP says the drawing is not to scale. I would assume that includes the lengths of the arrow. But your statement is correct. $\endgroup$ – Bob D May 18 at 12:55
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    $\begingroup$ While there is no fundamental requirement from linear algebra that the basis vectors you use have a vanishing inner product, almost all manipulations are easier when that is the case (and it is generally the only way we teach the work). Thankfully it is easy to construct an orthonormal basis from an arbitrary one (Gram-Schmitt procedure). $\endgroup$ – dmckee May 18 at 16:41

Components at right angles to each other are especially useful because they are so easy to calculate.

If vector $\vec V$ has components $\vec{V_1}$ and $\vec{V_2}$ at right angles to each other and in the same plane as $\vec V$, then their magnitudes are $$V_1=V \cos \theta_1\ \ \ \text{and}\ \ \ 𝑉_2=𝑉 \cos𝜃_2$$ in which $V$ is the magnitude of $\vec V$ and $\theta_1$ and $\theta_2$ are the angles between $\vec V$ and the components.

If the components are not at right angles to each other, their magnitudes are not given by so simple a rule – as you can see by drawing the vector addition diagram for $\vec V= \vec{V_1} +\vec{V_2}$. This yields $$V_1=V \frac{\sin \theta_2}{\sin (\theta_1 +\theta_2)} \ \ \ \text{and}\ \ \ 𝑉_2=𝑉 \frac{\sin \theta_1}{\sin (\theta_1 +\theta_2)}$$


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