Projection of vectors?

Question

Suppose I want to find the projection of vector $u$ onto $v$ . To me it makes sense to just make a right triangle with $u$ as the hypotenuse, and the problem is to find the base of this triangle. The base would be given by $||u|| \cos\theta$, and since we can relate $\cos\theta$ to the dot product, the final answer should be $||u|| \dfrac{u v}{||u|| ||v||} = \dfrac{u v}{||v||}$. However, the book says that the projection of $u$ onto $v$ is actually given by $\dfrac{u v}{||v||^2} v$. What am I doing wrong?

Answer 1

The answer by the book is the projection as a vector. Your answer is the length of such vector. You only have to add "in the direction of $v$" to complete your answer.

Answer 2

HINT: Projection of the vector $u$ onto the vector $v$ is $$=|u|\cos \theta$$

But from dot product, $u\cdot v=|u||v|\cos\theta\iff \cos\theta=\frac{u\cdot v}{|u||v|} $ $$=|u|\frac{(u\cdot v)}{|u||v|} $$ $$=\color{red}{\frac{u\cdot v}{|v|}}$$