Find the components of vector c (3, -4) parallel and perpendicular to the line y=x.
Can somebody help me with solution?

Draw the line y = x and the vector c.

At the starting tip of vector c, draw a line parallel to the line y = x. Make it long enough. Call it line 1

Then, take your ruler and draw a line from the ending tip of vector c, perpendicular. To the line y = x. Make it so that it meets line 1. Call this new line, line 2.

Now, you should get a triangle with vector c, line 1 and line 2 as the sides. Can you go from there? The component parallel to the line y = x is the vector of line 1 from the starting tip of vector c to the point where it touches line 2, the component perpendicular to the line y = x is the vector of the line 2 from the ending tip of vector c to the point where it meets line 1.

Can you do that? :)

Thank you.
In fact, I need to find magnitudes of c components.
(Hints: cross product, unit vector)

Did you manage to get the vector components? When you got that, it would be easier to get the magnitude.

Or if you are allowed to solve it graphically, it's even easier.

Whole content of exercise is: "Find the magnitudes of c components prallel and perpendicular to the line y=x. (Hints: cross product, unit vectors)"
I want to know how to use cross product and unit vectors to find magnitudes, it means, that it can't be solved graphically.

Okay, that is weird since the only place I've encountered cross product is in vectors in 3D.

Dot product, however can be used here. The dot product of the two components should give you 0.

But this route appears to be the most complicated to me, and I am someone who uses the simplest methods where possible (here being trigonometry). After all, you usually are free to make use of a hint, or not, since it didn't instruct you to use those.

Unit vectors can be used to some extent.

One unit vector parallel to the line y = x is \(\sqrt2\\ \sqrt2\) and one perpendicular is \(\sqrt2\\ -\sqrt2\).

Both should add up to the vector c:

a\(\sqrt2\\ \sqrt2\) + b\(\sqrt2\\ -\sqrt2\) = \(3\\ -4\)

a and b are the number of times you use the unit vector to get the component parallel and perpendicular to the line y = x, respectively.

Hopefully, that makes sense to you.