Will a vector at 45° to the horizontal be larger or smaller than its horizontal and vertical components?

Answer 1

It will be larger than horizontal and vertical components by amount sqrt (2).

Answer 2

Larger by a factor of the square root of 2.

When seperating a vector into horizontal and vertical components the maximum value of each component can not exceed the original vector, and will be the same value only if the vector is either horizontal or vertical.

Answer 3

Your vector C has magnitude |C| and direction (e.g., 45 deg). Thus, its X component, also a vector, is X = |X| cos(45) and the Y component is Y = |Y| sin(45).

Thus, you have an X,Y,C right triangle with the hypotenuse C, and two sides X and Y. Thus we can invoke the Pythagorean Theorem for |C|^2 = |X|^2 + |Y|^2 and, by inspection, see that |C|, the magnitude of your vector has got to be equal to or greater than either of the sides X and Y. And the only way C could equal either of the sides is for one side to equal zero. (For example, if X = 0, then C = Y is true. Otherwise, C > X and Y.)

Note that we did not need your angle (45 deg) to prove C >= X,Y. That is, C is nearly always greater than the sides of the right triangle it is the hypotenuse for. And the only exception is when one of the two lesser sides is zero; in which case C = a side.

Answer 4

Larger.

Try drawing a line like this at 45 degrees (not sure if this will come out clearly!)

|       /
|     /
|   /
| /____

Now measure the length of your diagonal line. Let's say it is 5cm. Now check how far along the X and Y axis this is. They should both be about 3.5cm, so your vector is longer than either of its components. The actual measurement would be 5 divided by the square root of 2.

Answer 5

Larger