a.) Write v = 1i + 1j + 1k as the sum of two vectors, one parallel to and one perpendicular to the vector v=4i + 2j + 0k,
i.e - find the projection of u on to v and v on to u.
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Any help would be appreciated. thank you
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 $\displaystyle \theta$ 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 $\displaystyle u\cdot v= |u||v|cos(\theta)$ so that $\displaystyle |v|cos(\theta)= \frac{u\cdot v}{|u|}$.
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 $\displaystyle \frac{u\cdot v}{|u|^2} u$.
The perpendicular projection is v minus that vector.