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To increase readability, I use $\oplus$ for vector addition, and $\lnot$ for vector subtraction.

Given two vectors: $$\begin{align} \vec D &= (6 \hat{i} \oplus 3 \hat{j}) \\ \vec E &= (4 \hat{i} \; \lnot \; 5 \hat{j}) \\ \end{align}$$

To calculate $2 \vec{D} \; \lnot \; \vec{E}$, denoted as $\vec F$: $$\begin{align} \vec{F} &= 2 \, (6 \hat{i} \oplus 3 \hat{j}) \; \lnot \; (4 \hat{i} \; \lnot \; 5 \hat{j}) \\ &= (12 \hat{i} \oplus 6 \hat{j}) \; \lnot \; (4 \hat{i} \oplus (-5) \hat{j}) \\ &= (12 \hat{i} \oplus 6 \hat{j}) \oplus ((-4) \hat{i} \oplus (+5) \hat{j}) \\ &= 8 \hat{i} \oplus 11 \hat{j} \end{align}$$

Are these steps correct? Why it's true that $\lnot \; 5 \hat{j}$ is equal to $\oplus \, (-5) \hat{j}$? What's the relation between $\lnot$ and $-$?

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Your steps are correct: to flip a vector, you flip all of its components. In general, to multiply a vector by any scale factor, you multiply all its components by that scale factor. If you think of a scalar as a 1D vector, then there is no longer a difference between "minus sign for vector" and "minus sign for scalar" - the operation is identical.

I think you are worrying when you don't need to.

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There are two operations in vector space definition: addition (of vectors) and multiplication (by scalar). Note there is no subtraction. Every time you see a minus in front of a vector it is only an abbreviation for $-1$ scalar factor.

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