if i have a vector $\vec a = (i,j,k) $ projected into vector $\vec d = i_2,j_2,k_2$ and both of them is perpendicular, will the projection vector be 0,0,0 ? if yes, what identity to prove this?
lets say the projection vector is $\vec x$ then $\vec x = \frac{|\vec x|}{|\vec d|}\vec d$ (but doesn't show that perpendicular thing make it zero)
also $|\vec x| = \cos \theta . |\vec a|$ (the $\theta$ here is between $\vec a$ and $\vec x$ whats perpendicular is $\vec a$ and $\vec d$
Yes the projection can be zero and studying when it's zero gives us a lot of information about the linear transformation. The zero vectors is perpendicular to all other vectors and in fact you can project every vector onto it. Typically if a vector $(x,y,z)$ is projected into the zero vector we say it's part of the kernel or nullspace of the linear transformation. All the vectors that are projected onto the zero vector form their own vector space.
To prove this we first use the notation $x^Ty$ to represent the dot product of $x \cdot y$. For two vectors to be orthogonal means that $x^Ty=0$. Now for some non-zero vector $d$ we set $u = d /|d|$ to be the unit vector in the direction of $d$. I now claim that the matrix $P=uu^T$ will be the orthogonal projection onto the span of $d$.
First, we see that for any $x$ that $Px = (uu^T)x=u(u^Tx)$ and since $u^Tx$ is a scalar the vector $Px$ is some scalar multiple of $d$, ie in the span of $d$. For this to be the orthogonal projection we need to show that $Px$ is orthogonal to $x-Px$ since $x-Px + Px = x$. This requires two other small details, first that $P^2=(uu^T)(uu^T) = u(u^Tu)u^T=uu^T=P$ so that we have $P^2=P$ and we need $P^T=(uu^T)^T=(u^T)^Tu^T= uu^T=P$, that is $P$ is symmetric. Finally we take the dot product of these vectors gives us $$(Px)^T(x-Px) = x^TP^T(x-Px)=x^TP^Tx-x^TP^TPx=x^TPx-x^TP^2x=x^TPx-x^TPx=0$$
and so we conclude that they are orthogonal.