Calculating Vector Components at 45 Degrees

[thehatchetmen]

the question i got was:
will the horizontal and vertical components of a vector at 45degrees to the horizontal be larger or smaller than the vector? By how much?

so far this is what i understand the vertical and horizontal components form a right angle and the vector is the line in the middle. and what I am guessing is that the horizontal component and the vector form a 45 degree angle..but that's all i understood so far...please help!

 

[BishopUser]

if you imagine an XY coordinate system, imagine laying the vector on the x-axis then moving it up 45 degrees (its tail at the origin and arrow pointing down the positive x-axis). that is all it is. the way to represent the x/y components of this vector is to make a triangle out of it (with the vector being the hypotenuse). The x component is simply the base of the triangle and the Y component is the height. With a 45 degree angle the x/y components will be the same.

to find the x component of the vector take the cos(angle)*vector or sin(angle)*vector for the y component. In the case of 45 degrees the x/y components will be about 70.7% of what the vector will be

 

[thehatchetmen]

thanks man

 

[science_geek]

yeah...sorry, but i don't get this..a little help?
(im only in 7th grade tho, so lamens terms please)

 

[i_luv_science]

the angle is at 45°, then the resultant is the square root of 2 which equals 1.41 units. Since both sides will be 1 unit long the resultant will be 0.41 units LARGER.
hope this helps! :)