The join of disjoint varieties in disjoint linear spaces has dimension $\dim X +\dim Y+1$ Let $X, Y \subset \Bbb P^n$ be disjoint projective varieties. Their join $J(X,Y)$ is the set of all points on lines joining a point of $X$ to a point of $Y$. It is a projective variety. 
The dimension of the join is always $\dim X +\dim Y+1$. Harris (page 148) writes that in the case where $X$ and $Y$ lie in disjoint linear subspaces of $\Bbb P^n$ this is "Immediate, since no two lines joining points of X and Y can meet, so every point of the join lies on a unique line joining X and Y". I would like help on understanding this:

  
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*Why two lines joining points on two disjoint linear varieties $L_1, L_2$ can never meet?
  
*Why does the set $L$ of lines joining points of $X$ to points of $Y$ have dimension $\dim X + \dim Y$?
  

Assuming these two I know how to finish: looking at the set $$T=\{(l,z) : z\in l, l \text{ intersects both X and }Y\}$$ we see that $J(X,Y)$ is the image of the second projection $\pi_2$, so $\dim J = \dim T $ by (1), and moreover $\dim T = \dim L+1$ since all the fibers of $\pi_1$ have the same dimension 1.
Alternatively, I'd like any approach to showing the following:

Assume $X \subset H_1 \subset \Bbb P^n$ and $Y \subset H_2 \subset \Bbb P^n$ are two projective varieties, where $H_1 \simeq \Bbb P^k$ is the intersection of $n-k$ homogeneous linear polynomials, and likewise $H_2 \simeq \Bbb P^t$, and $H_1 \cap H_2 = \emptyset.$ Why $\dim J(X,Y)=\dim X+\dim Y+1$?

I do know an inequality on one side: $\dim J(X,Y) \leq \dim X+\dim Y+1$ since $J(X,Y)=\pi_3(S)$, where $S=\{(p,q,z): z \in (pq)\} \subset X \times Y \times \Bbb P^n$.
 A: *

*As you have written, it is a bit loose. If $l_1,l_2$ are two lines joining points of $L_1,L_2$, they may meet at a point in $L_1$ or $L_2$. For example, $l_1$ may be the line joining $x\in L_1$ and $y_1\in L_2$ and $l_2$, the line joining the same $x$ to another point $y_2\in L_2$. But, this is the only possibility and can be seen as follows. If $l_1, l_2$ meet, then the linear span of these is a plane $H$. Since $H\cap L_i$ are linear subspaces of $H$ and disjoint, one easily checks that at least one of them must be just a single point and the rest is obvious.

*For this consider the incidence variety, $T=\{(a,b,x)\in L_1\times L_2\times\mathbb{P}^n| x\text{is on the line joining} a,b\}$. By projecting it to $L_1\times L_2$, one notes that this is onto and fibers are lines and thus $\dim T=\dim L_1+\dim L_2+1$. The image of $T$ in $\mathbb{P}^n$ is the join of $L_1,L_2$. Pick an $x$ such that $(a,b,x)\in T$ and $x\neq a, x\neq b$. Then $x$ is in the join and the fiber over $x$ under the projection $T\to\mathbb{P}^n$ is just $(a,b,x)$ from 1). Thus the map is generically finite and thus the image dimension is the same as dimension of $T$, which is what you want to prove.
