Exercises — Distance formula in 3D

WHAT we do: subtract matching coordinates to get the three box edges. $a=5-2=3,b=7-3=4,c=13-1=12$ WHY: these gaps are exactly the edges $a,b,c$ of the box whose body-diagonal is $PQ$ (see the figure above). Now apply the formula: $PQ=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$ Answer: $13$ units.

When one point is the origin, every "$x_1$" etc. is just the point's own coordinate, because subtracting $0$ changes nothing. $OA=\sqrt{3^2+(-4{)}^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$ Answer: $13$ units. (Notice $-4$ squared becomes $+16$ — sign disappears.)

WHAT we do: write the squared distance so the outer root disappears, then solve for $x$. $P{Q}^2=(x-1{)}^2+(-1-1{)}^2+(4-2{)}^2=14$ $(x-1{)}^2+4+4=14\Rightarrow (x-1{)}^2=6$ WHY two answers: a square equal to $6$ has two roots — the point can sit on either side of $Q$ along the $x$-direction and still be the same distance away. $x-1=\pm \sqrt{6}\Rightarrow x=1+\sqrt{6}\text{ or }x=1-\sqrt{6}$ Answer: $x=1\pm \sqrt{6}$.

WHAT "equidistant" means: $AB=AC$. Squaring both sides removes the roots and keeps the equation clean. $A{B}^2=(0-2{)}^2+(y-1{)}^2+(0+3{)}^2=4+(y-1{)}^2+9$ $A{C}^2=(0+1{)}^2+(y-2{)}^2+(0-1{)}^2=1+(y-2{)}^2+1$ Set them equal: $4+(y-1{)}^2+9=1+(y-2{)}^2+1$ Expand the squares: $(y-1{)}^2=y^2-2y+1$ and $(y-2{)}^2=y^2-4y+4$. $13+y^2-2y+1=2+y^2-4y+4$ $y^2-2y+14=y^2-4y+6$ The $y^2$ cancels (good — that always happens for equidistant problems, leaving a linear equation): $-2y+14=-4y+6\Rightarrow 2y=-8\Rightarrow y=-4$ Answer: $y=-4$.

The test: three points are collinear iff the largest pairwise distance equals the sum of the other two — because a straight path has no shortcut through space, while a bent path (a triangle) is always shorter along the direct route than around two sides. $AB=\sqrt{2^2+2^2+4^2}=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}$ $BC=\sqrt{2^2+2^2+4^2}=\sqrt{24}=2\sqrt{6}$ $AC=\sqrt{4^2+4^2+8^2}=\sqrt{16+16+64}=\sqrt{96}=4\sqrt{6}$ Check: $AB+BC=2\sqrt{6}+2\sqrt{6}=4\sqrt{6}=AC.$ ✓ Since the two shorter distances sum to the longest, the points are collinear, and because $AB+BC=AC$, point $B$ lies between $A$ and $C$.

WHAT we do: compute the three squared sides, then test Pythagoras. We use squared lengths because Pythagoras compares ${\text{side}}^2$ values — taking roots would only add avoidable work. $A{B}^2=(-1-0{)}^2+(6-7{)}^2+(6-10{)}^2=1+1+16=18$ $B{C}^2=(-4+1{)}^2+(9-6{)}^2+(6-6{)}^2=9+9+0=18$ $A{C}^2=(-4-0{)}^2+(9-7{)}^2+(6-10{)}^2=16+4+16=36$ Test: the two smallest squares are $A{B}^2=18$ and $B{C}^2=18$; their sum is $36=A{C}^2$. $A{B}^2+B{C}^2=A{C}^2\Rightarrow \text{right angle opposite the longest side }AC.$ The right angle sits at the vertex common to $AB$ and $BC$, namely $B$. Also $AB=BC=\sqrt{18}$, so it is a right-angled isosceles triangle.

Set-up: any point on the $z$-axis has the form $P(0,0,z)$ — its $x$ and $y$ are forced to $0$. Only $z$ is unknown. Condition $PA=PB$, squared: $P{A}^2=(0-1{)}^2+(0-5{)}^2+(z-7{)}^2=1+25+(z-7{)}^2$ $P{B}^2=(0-5{)}^2+(0-1{)}^2+(z+4{)}^2=25+1+(z+4{)}^2$ Setting equal and cancelling the common $26$: $(z-7{)}^2=(z+4{)}^2$ Expand: $z^2-14z+49=z^2+8z+16$. The $z^2$ cancels: $-14z+49=8z+16\Rightarrow 33=22z\Rightarrow z=\frac{33}{22}=\frac{3}{2}$ Answer: the point is $\left(0,0,\frac{3}{2}\right)$.

Centre = midpoint of the diameter. The midpoint is the average of matching coordinates (the section formula with ratio $1:1$): $C=\left(\frac{2+4}{2},\frac{-1+3}{2},\frac{3+(-1)}{2}\right)=(3,1,1)$ Radius = half the diameter length. First the full diameter: $AB=\sqrt{(4-2{)}^2+(3+1{)}^2+(-1-3{)}^2}=\sqrt{4+16+16}=\sqrt{36}=6$ So the radius is $r=\frac{AB}{2}=3$. Answer: centre $(3,1,1)$, radius $3$. (Sanity check: distance from $C$ to $A$ is $\sqrt{1+4+4}=\sqrt{9}=3=r$ ✓.)

Condition $PA=PB$, squared (squaring is safe: both distances are non-negative, so $PA=PB⟺P{A}^2=P{B}^2$): $(x-1{)}^2+(y-2{)}^2+(z-3{)}^2=(x-3{)}^2+(y-2{)}^2+(z+1{)}^2$ The $(y-2{)}^2$ term is identical on both sides — it cancels immediately. Expand the rest: $x^2-2x+1+z^2-6z+9=x^2-6x+9+z^2+2z+1$ The $x^2$ and $z^2$ cancel (always, in an equidistance locus — that is why the surface is flat, not curved): $-2x+1-6z+9=-6x+9+2z+1$ $-2x-6z+10=-6x+2z+10$ $4x-8z=0\Rightarrow x=2z$ Answer: $x-2z=0$, a plane. Geometrically it is the perpendicular bisector plane of segment $AB$ — every point on it is exactly as far from $A$ as from $B$. (Check the midpoint $M=(2,2,1)$: it satisfies $2-2(1)=0$ ✓.)

(a) "Distance from a fixed point is constant" is the definition of a sphere. Squaring the distance $=5$: $(x-1{)}^2+(y+2{)}^2+(z-3{)}^2=25$ This is a sphere centred at $Q(1,-2,3)$ with radius $5$. (b) Compare the actual distance $QR$ with the radius $5$: $QR=\sqrt{(4-1{)}^2+(2+2{)}^2+(3-3{)}^2}=\sqrt{9+16+0}=\sqrt{25}=5$ Since $QR=5=$ radius, the point $R$ lies exactly on the sphere. (If $QR<5$ it would be inside; if $QR>5$, outside — we tested all three cases and landed on the boundary.)

Compute all three side-lengths: $AB=\sqrt{(10-3{)}^2+(20-6{)}^2+(30-9{)}^2}=\sqrt{49+196+441}=\sqrt{686}=7\sqrt{14}$ $BC=\sqrt{(25-10{)}^2+(-41-20{)}^2+(5-30{)}^2}=\sqrt{225+3721+625}=\sqrt{4571}$ $AC=\sqrt{(25-3{)}^2+(-41-6{)}^2+(5-9{)}^2}=\sqrt{484+2209+16}=\sqrt{2709}$ Isosceles test: we need two equal sides. Compare the squared lengths: $A{B}^2=686$, $B{C}^2=4571$, $A{C}^2=2709$ — all different, so with these coordinates the triangle is scalene, not isosceles. What to do: trust the computation over the claim. The distances are unequal, so the stated "isosceles" is false; the triangle is scalene with perimeter $AB+BC+AC=7\sqrt{14}+\sqrt{4571}+\sqrt{2709}\text{units}.$ Lesson: a distance calculation is a proof. When numbers contradict a claim, the numbers win — always double-check the claim itself, never bend the arithmetic to fit it.

collinearity, and which proves a right angle? Collinearity: the three lengths add — $AB+BC=AC$. Right angle: the three squared lengths obey Pythagoras — $A{B}^2+B{C}^2=A{C}^2$.