Prove an if and only if condition for two lines to intersect in Rⁿ

by RoRi

April 29, 2016

Let $L(P;A)$ and $L(Q;B)$ be two lines in $\mathbb{R}^n$ . Prove that they intersect if and only if $P-Q$ is in the linear span of $A$ and $B$ .
Consider the two lines in $\mathbb{R}^3$ ,
$L = \{ (1,1,-1) + t(-2,1,3) \}, \qquad L' = \{ (3,-4,1) + t(-1,5,2) \}.$

Determine whether they intersect.

Proof. Assume $L(P;A)$ and $L(Q;B)$ intersect. Since $L(P;A) = \{ P+tA \}$ and $L(Q;B) = \{ Q + t'B\}$ we know there exists an $X \in \mathbb{R}^n$ such that $X = P+tA = Q + t'B$ . This implies
$P-Q = t'B - tA = t'B + (-t)A.$

This implies $P-Q$ is in the linear span of $A,B$ .

Conversely, assume $P-Q$ is in the linear span of $A,B$ . Then there exist $t, t' \in \mathbb{R}$ such that

$\begin{align*} P-Q = tA + t'B && \implies && P-t'B &= Q+ tA \\ && \implies && P + (-t)A &= Q + t'B. \end{align*}$

Thus, there is some point in both $L(P;A)$ and $L(Q;B)$ so they intersect $. \qquad \blacksquare$
The two given lines do not intersect. We know from part (a) that two lines $L(P;A)$ and $L(Q;B)$ intersect if and only if $P-Q$ is in the linear span of $A,B$ . In this case we have
$P - Q = (1,1,-1) - (3,-4,1) = (-2,5,-2)$

and $A = (-2,1,3)$ , $B = (-1,5,2)$ . For $P-Q$ to be in the linear span of $A,B$ we must have $r,s \in \mathbb{R}$ such that

$\begin{align*} -2r-s &= -2 \\ r + 5s &= 5 \\ 3r + 2s &= -2. \end{align*}$

But the second equation implies $r = 5-5s$ . The third equation would then require $15 - 15s + 2s = -2$ which gives $s = \frac{17}{13}$ . Then, $r = -\frac{20}{13}$ . But then from the first equation

$-2 \left(-\frac{20}{13} \right) - \left( \frac{17}{13} \right) = \frac{23}{13} \neq -2.$

Hence, there are no such $r,s$ so $P-Q$ is not in the linear span of $A,B$ . Hence, these two lines do not intersect.

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