Just a small thing 762 here,the question is about projection to the plane $T: x+2z-3y+3=0$ in the direction of the vector $\bar{i}+\bar{j}-\bar{k}$. Now I understand this in a way
where $k\in\mathbb R$ but I got into the dead-end with 3. Initially, I thought $\mathbb R^{4}$ but it mentions $f: \mathbb R^{3}\mapsto \mathbb R^{3}$. Now I need some mapping $(1\times3) (3\times1)$ to get scalar, thinking. I must be misunderstanding something with "projection to the plane in the direction of the vector"
. Perhaps, this way?
$\bar{n}* (\bar{r}-\bar{r}_{0})=-3$
where $\bar{n}=(1,-3,2)$ is the normal vector to the plane. Now
$\bar{n}*\left(\bar{x}_{1}+k (1,1,-1)\right)=-3$
where is the $\bar{r}_{0}=k(1,1,-1)$ so $1(a+k)-3(b+k)+2(c-k)=-3$. Not sure yet whether this was the correct way of doing, thinking...trying to visualize...
Perhaps Related