# Projecting a vector component out of vector $a$

Suppose a vector $a$ is given. What is precisely meant by projecting the component of vector $b$ out of $a$? Does that mean that the resulting vector $a_1$ (obtained by "projecting out the component of $b$ from $a$") is orthogonal to $b$? How could one achieve this?

I read that Gram-Schmidt process is used for this; however, it involves the notion of "normalization", and I'm not sure if that means that the vector $a_1$ would be normalized.

-

It means if you decompose $b$ into a vector $b_1$ parallel to $a$ and a vector $b_2$ perpendicular to $a$, $b_1$ would be the answer. Gram Schmidt (GS) is used to find $b_2$, which is normal to $a$, and then $b_1$ can be found by subtracting.
The purpose (context) for which GS is introduced in most textbooks is to take a number of linearly independent vector (a basis for some vector space) and transform them into an orthogonal basis for the same space. In the case of two vectors above, you go from the basis $(a, b)$ to the basis $(a, b_2)$. In the case of a three-basis $(a, b, c)$ you first find $b_2$, so that you have $(a, b_2, c)$. Then you apply GS on $c$ to find a vector $c_2$ normal to the space given by $(a, b_2)$, but so that $(a, b_2, c_2)$ is the same vector space as $(a, b, c)$
I might have gotten the roles of $a$ and $b$ with respect to the original question mixed up. I do not know. – Arthur Aug 11 '12 at 21:33
GS is a bit of overkill here, no? $b_1={b\cdot a\over a\cdot a}a$, and $b_2=b-b_1$. As to which component is "out of $a$", I'm not familiar with the terminology. – Gerry Myerson Aug 11 '12 at 23:29
I'm looking for a transformation on $a$. So, you state that by "projecting out component of $b$" from $a$, one obtains $a_1$ that is parallel to $b$? Well, then GS could not be used, since it is looking for an orthogonal basis. Therefore, I expect that $a_1$ should be orthogonal to $b$. Right? – user506901 Aug 13 '12 at 8:15
@user506901 The trick is that if you use GS to obtain $a_1$ orthogonal, you get $a_2 = a - a_1$ parallel to $b$. – Arthur Aug 14 '12 at 4:10