No, the "i.e." is NOT what the question is asking.

Do you understand what "projection" means? Draw the two vectors, starting at the same point. Draw a line from the tip of v perpendicular to u. The vector from the base of u to the foot of that perpendicular is the "parallel projection of v on u" (typically what is meant by just "projection of v on u". The vector from the foot of that perpendicular to the tip of v is the "perpendicular projection of v on u". Those two vectors add to v.

To find the projetion of u on v, you would draw the perpendicular from the tip of u, a completely different construction.

There is probably a formula in your book but:

You have a right triangle in which v is the hypotenuse, and the length of the projection of v on u is the near side, with angle the angle between the two vectors so the length of the projection is |v|cos(\theta).

You may also recall that the "dot product" of two vectors, u and v, can be defined by so that .

To get a vector with that length in the direction of u, multiply by the unit vector in that direction, [tex]\frac{u}{|u|}.

The (parallel) projection of v on u is .

The perpendicular projection is v minus that vector.