7F - Vector intersections and concurrency

How can you prove that three lines meet at one point when a sketch is too imprecise to decide? You will describe that shared point, solve for its position and ratios, then test whether a third line passes through exactly the same point.

Two vector descriptions of a point

A point on a straight line is reached by starting at a known point and taking a scalar multiple of the line's direction vector.

For a point $X$ on the line through distinct points $A,B$ , write $\overrightarrow{AX}=t\overrightarrow{AB}$ . The real number $t$ is a parameter: it describes how far along that directed line the point lies. Adding the starting position gives

\overrightarrow{OX}=\mathbf{a}+t(\mathbf{b}-\mathbf{a})

When $t=0$ , $X=A$ ; when $t=1$ , $X=B$ . Values $0<t<1$ lie inside the segment, $t>1$ goes beyond $B$ , and $t<0$ goes backwards from $A$ . Unless the question restricts $X$ to a segment, keep all real values available.

Key idea

At an intersection, both routes give the same position vector. Use a separate parameter for each line: equal positions do not usually mean equal fractions along the two lines.

Consider a nondegenerate triangle $OAB$ , with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ . These reference vectors are nonparallel, so coefficient comparison will be available after the routes are written.

Worked example 1: two midpoint routes

Question. Let $M$ be the midpoint of $AB$ , and $N$ the midpoint of $OB$ . The lines $OM$ and $AN$ meet at $X$ . Write two expressions for $\overrightarrow{OX}$ , using $\overrightarrow{OX}=t\overrightarrow{OM}$ and $\overrightarrow{AX}=s\overrightarrow{AN}$ .

Plan. Find the midpoint positions, then follow each route to $X$ .

\begin{aligned} \overrightarrow{OM}&=\frac12(\mathbf{a}+\mathbf{b}),\qquad \overrightarrow{ON}=\frac12\mathbf{b}\\ \overrightarrow{OX}&=\frac t2\mathbf{a}+\frac t2\mathbf{b}\\ \overrightarrow{OX}&=\mathbf{a}+s\left(\frac12\mathbf{b}-\mathbf{a}\right)\\ &=(1-s)\mathbf{a}+\frac s2\mathbf{b} \end{aligned}

Answer. $\overrightarrow{OX}=\frac t2\mathbf{a}+\frac t2\mathbf{b}=(1-s)\mathbf{a}+\frac s2\mathbf{b}$ .

Check. The first route gives $O$ at $t=0$ and $M$ at $t=1$ . The second gives $A$ at $s=0$ and $N$ at $s=1$ . Each expression therefore follows its intended line.

Change just one midpoint to a different division point. The first direction vector changes, while the route from $A$ stays the same.

Worked example 2: changing the division point

Question. Now let $AM:MB=2:1$ , keeping $N$ as the midpoint of $OB$ . The lines $OM$ and $AN$ meet at $X$ . Again write two expressions for $\overrightarrow{OX}$ , with parameters $t$ on $OM$ and $s$ from $A$ towards $N$ .

Plan. Use internal division to find $M$ , then substitute into the two routes.

\begin{aligned} \overrightarrow{OM}&=\frac{\mathbf{a}+2\mathbf{b}}3\\ \overrightarrow{OX}&=\frac t3\mathbf{a}+\frac{2t}3\mathbf{b}\\ \overrightarrow{OX}&=\mathbf{a}+s\left(\frac12\mathbf{b}-\mathbf{a}\right)\\ &=(1-s)\mathbf{a}+\frac s2\mathbf{b} \end{aligned}

Answer. $\overrightarrow{OX}=\frac t3\mathbf{a}+\frac{2t}3\mathbf{b}=(1-s)\mathbf{a}+\frac s2\mathbf{b}$ .

Check. At $t=1$ , the first route reaches the correct ratio point $(\mathbf{a}+2\mathbf{b})/3$ . The second route still reaches $N$ at $s=1$ .

The same starting-position-plus-movement idea models a constant rate. For example, a moving point starting at $2\mathbf{i}+\mathbf{j}$ , with displacement $\mathbf{i}+2\mathbf{j}$ each second, has position $2\mathbf{i}+\mathbf{j}+t(\mathbf{i}+2\mathbf{j})$ after $t$ seconds. Here $t\ge0$ represents elapsed time. A geometric intersection of paths may use different parameters; two moving objects meet only if they occupy that position at the same time.

Finding intersection ratios

Once the two position expressions agree, compare their coefficients in the independent reference directions and solve the resulting simultaneous equations.

Use the parameter values to interpret the geometry, not just to calculate a position vector. If $\overrightarrow{AX}=s\overrightarrow{AN}$ with $0<s<1$ , then $AX:XN=s:(1-s)$ . If $s>1$ , $X$ is beyond $N$ , so it is outside the segment even though it remains on the line.

Worked example 3: intersecting midpoint lines

Question. Return to Example 1, where $M$ and $N$ are midpoints. Find $\overrightarrow{OX}$ and $OX:XM$ .

Plan. Equate coefficients in the two descriptions of $X$ . This is valid because $\mathbf{a},\mathbf{b}$ are nonparallel.

\begin{aligned} \frac t2&=1-s & \frac t2&=\frac s2\\ t&=s\\ \frac t2&=1-t\\ t&=\frac23,\qquad s=\frac23\\ \overrightarrow{OX}&=\frac13\mathbf{a}+\frac13\mathbf{b} \end{aligned}

Answer. $\overrightarrow{OX}=(\mathbf{a}+\mathbf{b})/3$ . Since $t=2/3$ , $OX:XM=(2/3):(1/3)=2:1$ .

Check. The route from $A$ gives $\mathbf{a}+\frac23(\frac12\mathbf{b}-\mathbf{a})=\frac13\mathbf{a}+\frac13\mathbf{b}$ , confirming the same point on the second line.

Now solve Example 2. Changing the division of $AB$ changes the fractions along both intersecting lines.

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7F - Vector intersections and concurrency

Two vector descriptions of a point

Finding intersection ratios

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