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This questions regards to doing force calculation using vectors. This image here should display my problem image.

In basic newtonian physics, the normal force = force of gravity * cosine (angle of inclination). $$F_N = F_G \times \cos{\theta}$$

To be clearer:

$$F_N = (x_G, y_G) \times \cos{\theta}$$ $$F_N = (0, 9.8) \times \cos{\theta}$$ $$F_N = (0, 9.8\cos{\theta})$$

The problem is not calculating this value but representing it as a 2D vector format. How is the x-component of $F_N = 0$, when the diagram (according to the direction of where the $F_N$ is pointing) clearly shows that the x-component should have a value? I believe I have messed up my calculations. Could someone please help me? Thanks in advance!

EDIT:

The x-component of $F_N$ should have a value other than $0$, otherwise, if it was = 0, it would be pointing straight up or down. This does not seem to be occurring in my calculations.

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The reason you got wrong answer (x-component of $F_N$ = 0) is that when you worte "$F_N = F_G \times \cos{\theta}$" and calculated further, you actually assumed that two vectors $\vec{F_N}$ and $\vec{F_G}$ are collinear, but they are not!
The expression you wrote - "$F_N = F_G \times \cos{\theta}$" is just a scalar expression!
With this, you can only calculate $\mathbf{magnitude}$ of $\vec{F_N}$.
Knowing magnitude, if you want x-, y- components of $\vec{F_N}$ , you can easily calculate them. $$------------$$ Or if someone still insist upon you to calculate $\vec{F_N}$ using vector operations, you can try this:
unit normal vector of the surface (downward) : $\vec{n}=(sin\theta , -cos\theta)$.
then projection of $\vec{F_G}$ onto $\vec{n}$ is : $(\vec{F_G} \cdot \vec{n}) \cdot \vec{n} $ .
and then the $\vec{F_N}$ is : $-(\vec{F_G} \cdot \vec{n}) \cdot \vec{n} $ .

(where $\vec{F_G} \cdot \vec{n}$ is a dot product of two vectors)

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    $\begingroup$ Welcome to PSE! Note that "check my work" questions are off topic, so one should flag to close the question rather than providing answers to them. $\endgroup$ Commented Aug 25, 2021 at 13:04
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In basic newtonian physics, the normal force = force of gravity * cosine (angle of inclination).

That is not correct. In your problem, the magnitude of the normal force on the block will be equal to the magnitude of weight of the block times the cosine of the inclination angle. (While that happens often, it is not always true.)

The direction of the normal force is not the direction of the weight.

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