Exercises — Section formula in 3D

Ratio $1:1$ means $m=n=1$, which is exactly the midpoint. $x=\frac{1\cdot 8+1\cdot 2}{2}=\frac{10}{2}=5,y=\frac{10+4}{2}=7,z=\frac{12+6}{2}=9.$ $\boxed{P=(5,7,9)}$ This is the midpoint of $AB$. Sanity check: each coordinate of $P$ is halfway between the two endpoints' coordinates. ✓

$m=2,n=1$, so $m+n=3$. Here $A$ is the origin, so $x_1=y_1=z_1=0$ and the $n{x}_1$ terms vanish. $x=\frac{2(6)+1(0)}{3}=\frac{12}{3}=4,y=\frac{2(9)+0}{3}=6,z=\frac{2(-3)+0}{3}=-2.$ $\boxed{P=(4,6,-2)}$ Because $m>n$, $P$ sits nearer $B$ — and indeed $(4,6,-2)$ is two-thirds of the way from the origin to $B$. ✓

External division uses denominator $m-n$.

• Ratio $5:2$: $m=5,n=2\Rightarrow m-n=\boxed{3}$.
• Ratio $2:5$: $m=2,n=5\Rightarrow m-n=\boxed{-3}$. A negative denominator is perfectly fine — it flips the signs in the numerators too, and geometrically it tells you $P$ lands on the other side (beyond $A$ instead of beyond $B$).

$m=2,n=1,m+n=3$. $x=\frac{2(5)+1(-1)}{3}=\frac{10-1}{3}=3,$ $y=\frac{2(-3)+1(3)}{3}=\frac{-6+3}{3}=-1,$ $z=\frac{2(8)+1(2)}{3}=\frac{16+2}{3}=6.$ $\boxed{P=(3,-1,6)}$

External: denominator $m-n=3-2=1$, and subtract in numerators ($n\to -n$). $x=\frac{3(5)-2(3)}{1}=15-6=9,$ $y=\frac{3(1)-2(2)}{1}=3-4=-1,$ $z=\frac{3(4)-2(-1)}{1}=12+2=14.$ $\boxed{P=(9,-1,14)}$ Since $m>n$, $P$ lies beyond $B$ — check: $B=(5,1,4)$ and $P$ overshoots it in every coordinate direction consistent with the $A\to B$ heading. ✓

Midpoint ($1:1$): $M=\left(\frac{7+(-1)}{2},\frac{-4+2}{2},\frac{3+9}{2}\right)=\left(3,-1,6\right).$ Ratio $3:1$ ($m=3,n=1,m+n=4$): $x=\frac{3(-1)+1(7)}{4}=\frac{4}{4}=1,y=\frac{3(2)+1(-4)}{4}=\frac{2}{4}=\frac{1}{2},z=\frac{3(9)+1(3)}{4}=\frac{30}{4}=\frac{15}{2}.$ $\boxed{M=(3,-1,6),P=\left(1,\frac{1}{2},\frac{15}{2}\right)}$

Let the ratio be $k:1$ (so $m=k,n=1$). Use the $x$-coordinate: $4=\frac{k(6)+1(2)}{k+1}\Rightarrow 4k+4=6k+2\Rightarrow 2=2k\Rightarrow k=1.$ So ratio $=1:1$. Verify with $z$: $\frac{1(8)+1(4)}{2}=\frac{12}{2}=6.$ ✓ Since $k=1>0$, the division is internal; in fact $P$ is the midpoint. $\boxed{\text{ratio }1:1\text{ (internal, midpoint)}}$

Let ratio $=k:1$. Use the $x$-coordinate: $1=\frac{k(-3)+1(3)}{k+1}\Rightarrow k+1=-3k+3\Rightarrow 4k=2\Rightarrow k=\frac{1}{2}.$ Verify with $y$: $\frac{\frac{1}{2}(-4)+1(-1)}{\frac{1}{2}+1}=\frac{-2-1}{\frac{3}{2}}=\frac{-3}{3/2}=-2.$ ✓ Verify with $z$: $\frac{\frac{1}{2}(-7)+1(-4)}{3/2}=\frac{-3.5-4}{1.5}=\frac{-7.5}{1.5}=-5.$ ✓ Ratio $=\frac{1}{2}:1=1:2$. The value $k=\frac{1}{2}>0$ is positive, so the division is internal, and $P$ is nearer $A$ (smaller weight on $B$). $\boxed{\text{ratio }1:2\text{ (internal)}}$

Ratio $k:1$. Use $x$: $7=\frac{5k+3}{k+1}\Rightarrow 7k+7=5k+3\Rightarrow 2k=-4\Rightarrow k=-2.$ Verify with $y$: $\frac{-2(1)+1(2)}{-2+1}=\frac{0}{-1}=0.$ ✓ Verify with $z$: $\frac{-2(0)+1(1)}{-1}=\frac{1}{-1}=-1.$ ✓ A negative $k$ means the division is external: $P$ lies on the line but outside segment $AB$. Writing $k=-2$ as $2:1$ external, $P$ lies beyond $B$. $\boxed{k=-2\Rightarrow \text{external division, ratio }2:1\text{ (beyond }B)}$

Let $P$ divide $AB$ in ratio $k:1$. Its $z$-coordinate must equal $6$: $6=\frac{k(10)+1(4)}{k+1}\Rightarrow 6k+6=10k+4\Rightarrow 2=4k\Rightarrow k=\frac{1}{2}.$ So the ratio is $1:2$ (internal). Now find $x,y$ with $m=1,n=2,m+n=3$: $x=\frac{1(8)+2(2)}{3}=\frac{12}{3}=4,y=\frac{1(0)+2(-3)}{3}=\frac{-6}{3}=-2.$ $\boxed{P=(4,-2,6),\text{ratio }1:2}$ Check: $z=6$ as required. ✓ (This is the line-meets-plane idea — one coordinate pins the ratio, the rest follow.)

The midpoint of $BC$ is $M=\left(\frac{2+4}{2},\frac{1+5}{2},\frac{3-1}{2}\right)=(3,3,1).$ The centroid $G$ divides $AM$ (from vertex $A$ to midpoint $M$) internally in ratio $2:1$, with the weight $2$ on the far point $M$: $G=\frac{2M+1A}{3}\Rightarrow A=3G-2M.$ $A_x=3(3)-2(3)=3,A_y=3(2)-2(3)=0,A_z=3(1)-2(1)=1.$ $\boxed{A=(3,0,1)}$ Verify with the centroid formula (average of vertices): $\left(\frac{3+2+4}{3},\frac{0+1+5}{3},\frac{1+3-1}{3}\right)=(3,2,1)=G.✓$

The $xy$-plane is $z=0$. Ratio $k:1$, set $z=0$: $0=\frac{k(-3)+1(5)}{k+1}\Rightarrow -3k+5=0\Rightarrow k=\frac{5}{3}.$ Positive, so internal, ratio $5:3$. Now $m=5,n=3,m+n=8$: $x=\frac{5(-1)+3(3)}{8}=\frac{4}{8}=\frac{1}{2},y=\frac{5(4)+3(-2)}{8}=\frac{14}{8}=\frac{7}{4}.$ $\boxed{P=\left(\frac{1}{2},\frac{7}{4},0\right),\text{ratio }5:3}$ Shortcut worth knowing: the plane $z=0$ divides $AB$ in ratio $z_1:(-z_2)$... let's not memorise; the $z=0$ substitution above always works.

If $C$ lies on line $AB$, then $C$ divides $AB$ in some ratio $k:1$, and that same $k$ must satisfy all three coordinates. Solve from $x$: $2=\frac{4k+1}{k+1}\Rightarrow 2k+2=4k+1\Rightarrow k=\frac{1}{2}.$ Test $y$: $\frac{\frac{1}{2}(5)+2}{3/2}=\frac{4.5}{1.5}=3.$ ✓ (matches $C_y=3$). Test $z$: $\frac{\frac{1}{2}(6)+3}{3/2}=\frac{6}{1.5}=4.$ But $C_z=5\ne 4$. ✗ The three coordinates demand different ratios, so no single point of $AB$ equals $C$. $\boxed{\text{Not collinear.}}$ This is the collinearity test in disguise: on a straight line one ratio must serve every coordinate.

Algebra: external denominator is $m-n=1-1=0$. Division by zero → undefined; no finite point exists. Geometry: external division in $1:1$ demands $AP=PB$ with $P$ outside the segment. The only "solution" is a point infinitely far away — the direction of the line, not a location. So external $1:1$ = point at infinity. E9 reinterpretation: we found $k=-2$, i.e. $P$ divides $AB$ internally in ratio $-2:1$. A negative internal ratio is exactly an external division: $-2:1$ means external ratio $2:1$. Confirm with the external formula ($m=2,n=1,m-n=1$) on $A(3,2,1),B(5,1,0)$: $x=\frac{2(5)-1(3)}{1}=7,y=\frac{2(1)-1(2)}{1}=0,z=\frac{2(0)-1(1)}{1}=-1,$ giving $P=(7,0,-1)$ — the very point in E9. ✓ Negative ratio and external division are two names for the same geometry. $\boxed{\text{External }1:1\Rightarrow \text{point at infinity; negative }k\iff \text{external division.}}$