Solved Examples

Solved Examples — the Full Chapter Workout

Thirty worked problems: octant drills, distances, triangle and quadrilateral classification, centroid and locus problems. Attempt each before reading the solution.

Example 1: Parallelogram, not rectangle

Show that $A(1,2,3)$, $B(-1,-2,-1)$, $C(2,3,2)$, $D(4,7,6)$ form a parallelogram $ABCD$ that is not a rectangle.

Solution:

Step 1 — Squared sides in cyclic order. $AB^2 = 4 + 16 + 16 = 36$; $BC^2 = 9 + 25 + 9 = 43$; $CD^2 = 4 + 16 + 16 = 36$; $DA^2 = 9 + 25 + 9 = 43$.

Step 2 — Parallelogram test. $AB = CD = 6$ and $BC = DA = \sqrt{43}$: opposite sides equal — parallelogram.

Step 3 — Rectangle test via diagonals. $AC^2 = 1 + 1 + 1 = 3$; $BD^2 = 25 + 81 + 49 = 155$. $AC = \sqrt{3} \ne \sqrt{155} = BD$: not a rectangle. ∎

Takeaway: Equal opposite sides buy the parallelogram; the rectangle upgrade costs one more check — equal diagonals — and here it fails spectacularly.

Example 2: Diagonal-midpoint check of the same quadrilateral

Verify Example 1's conclusion using midpoints of the diagonals.

Solution:

Step 1 — Midpoint of $AC$. $\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$.

Step 2 — Midpoint of $BD$. $\left(\frac{-1+4}{2}, \frac{-2+7}{2}, \frac{-1+6}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$.

Step 3 — Conclude. The diagonals bisect each other — parallelogram, in one line. ∎

Takeaway: The midpoint test is faster than four side lengths; keep both tools and pick by the arithmetic on offer.

Example 3: An equidistant plane

Find the set of points equidistant from $A(3, 4, -5)$ and $B(-2, 1, 4)$.

Solution:

Step 1 — Square the condition. $PA^2 = PB^2$: $(x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2$.

Step 2 — Expand; the squares cancel. $-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$.

Step 3 — Collect. $10x + 6y - 18z - 29 = 0$ — the perpendicular-bisector plane of $AB$.

Takeaway: Equidistance from two points is always a plane in 3D — squaring kills every quadratic term, guaranteed.

Example 4: Recovering a vertex from the centroid

The centroid of $\triangle ABC$ is $(1, 1, 1)$; $A = (3, -5, 7)$ and $B = (-1, 7, -6)$. Find $C$.

Solution:

Step 1 — Set up one equation per coordinate. With $C = (x, y, z)$: $\frac{3 - 1 + x}{3} = 1$, $\frac{-5 + 7 + y}{3} = 1$, $\frac{7 - 6 + z}{3} = 1$.

Step 2 — Solve each. $2 + x = 3$ gives $x = 1$; $2 + y = 3$ gives $y = 1$; $1 + z = 3$ gives $z = 2$.

Step 3 — Write and check. $C = (1, 1, 2)$; averaging all three vertices indeed returns $(1, 1, 1)$ ✓.

Takeaway: Centroid problems are three independent 1D problems — never solve them as a system.

Example 5: The fourth vertex

Three vertices of parallelogram $ABCD$ are $A(3, -1, 2)$, $B(1, 2, -4)$, $C(-1, 1, 2)$. Find $D$.

Solution:

Step 1 — Use the diagonal property. In parallelogram $ABCD$ the diagonals $AC$ and $BD$ share a midpoint, so $D = A + C - B$ coordinate-wise.

Step 2 — Compute. $D = (3 - 1 - 1,\; -1 + 1 - 2,\; 2 + 2 + 4) = (1, -2, 8)$.

Step 3 — Check. Midpoint of $AC$: $(1, 0, 2)$; midpoint of $BD$: $\left(\frac{1+1}{2}, \frac{2-2}{2}, \frac{-4+8}{2}\right) = (1, 0, 2)$ ✓.

Takeaway: $D = A + C - B$ is the one-line fourth-vertex formula — but match the LABELS: it pairs the vertices adjacent to $B$ across the diagonal.

Example 6: Median lengths

Find the lengths of the medians of the triangle $A(0, 0, 6)$, $B(0, 4, 0)$, $C(6, 0, 0)$.

Solution:

Step 1 — Midpoints of the sides. $BC$: $(3, 2, 0)$; $CA$: $(3, 0, 3)$; $AB$: $(0, 2, 3)$.

Step 2 — Median from each vertex to the opposite midpoint. From $A(0,0,6)$: $\sqrt{9 + 4 + 36} = 7$; from $B(0,4,0)$: $\sqrt{9 + 16 + 9} = \sqrt{34}$; from $C(6,0,0)$: $\sqrt{36 + 4 + 9} = 7$.

Step 3 — Sanity check the symmetry. $A$ and $C$ play symmetric roles (swap $x \leftrightarrow z$), so their medians must match — and they do, both 7 ✓.

Takeaway: A median is just a distance to a midpoint — two formulas chained; symmetry predicts equal answers before computing.

Example 7: Centroid at the origin

The origin is the centroid of $\triangle PQR$ with $P(2a, 2, 6)$, $Q(-4, 3b, -10)$, $R(8, 14, 2c)$. Find $a$, $b$, $c$.

Solution:

Step 1 — $x$-average. $\frac{2a - 4 + 8}{3} = 0$ gives $2a + 4 = 0$: $a = -2$.

Step 2 — $y$-average. $\frac{2 + 3b + 14}{3} = 0$ gives $3b = -16$: $b = -\frac{16}{3}$.

Step 3 — $z$-average. $\frac{6 - 10 + 2c}{3} = 0$ gives $2c = 4$: $c = 2$.

Takeaway: "Centroid at the origin" means each coordinate-list sums to zero — three separate little equations, one unknown each.

Example 8: A $PA^2 + PB^2$ locus

With $A(3, 4, 5)$ and $B(-1, 3, -7)$, find the set of points $P$ with $PA^2 + PB^2 = k^2$.

Solution:

Step 1 — Write both squared distances. $(x-3)^2 + (y-4)^2 + (z-5)^2 + (x+1)^2 + (y-3)^2 + (z+7)^2 = k^2$.

Step 2 — Expand and merge. Quadratics double up; linear terms: $-4x - 14y + 4z$; constants: $50 + 59 = 109$.

Step 3 — Write the locus. $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = k^2 - 109$ — a sphere-type equation.

Takeaway: Sums of SQUARED distances keep their quadratic terms — expect a sphere, not a plane.

Example 9: An octant drill with fresh numbers

Name the octants of $(2, -5, 1)$, $(-6, -1, -2)$, $(3, 7, -4)$ and $(-1, 4, 8)$.

Solution:

Step 1 — Sort by $z$. $z > 0$: $(2,-5,1)$ and $(-1,4,8)$ (octants I-IV); $z < 0$: $(-6,-1,-2)$ and $(3,7,-4)$ (octants V-VIII).

Step 2 — Read the $(x,y)$ quadrant. $(2,-5,1)$: $(+,-)$ → IV. $(-1,4,8)$: $(-,+)$ → II.

Step 3 — Repeat below. $(-6,-1,-2)$: $(-,-)$ → VII. $(3,7,-4)$: $(+,+)$ → V.

Takeaway: $z$ first, then the 2D quadrant — the octant table reads itself.

Example 10: Diagonals of a cube and a box

Find (i) the diagonal of a cube of side $a$ with one vertex at the origin and edges along the axes; (ii) the diagonal of a box with edges 3, 4, 12.

Solution:

Step 1 — (i) Opposite corners. From $(0,0,0)$ to $(a,a,a)$: $\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$.

Step 2 — (ii) The box. $\sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Step 3 — Note the pattern. $(3, 4, 12, 13)$ is a Pythagorean quadruple — the 3D cousin of $(3, 4, 5)$.

Takeaway: A box diagonal IS the distance formula — the formula's own derivation picture, run backwards.

Example 11: A point on the $x$-axis equidistant from two points

Find the point on the $x$-axis equidistant from $(1, 2, 3)$ and $(3, 2, -1)$.

Solution:

Step 1 — Parametrise the axis. The point is $(x, 0, 0)$; set the squared distances equal: $(x-1)^2 + 4 + 9 = (x-3)^2 + 4 + 1$.

Step 2 — Expand and cancel $x^2$. $-2x + 14 = -6x + 14$ gives $4x = 0$: $x = 0$.

Step 3 — Conclude and check. The point is the origin $(0,0,0)$; both distances are $\sqrt{14}$ ✓ — and it satisfies the bisector plane $x - 2z = 0$ of these two points.

Takeaway: Restricting a locus to an axis turns a plane condition into a single linear equation — one unknown, because the axis has one degree of freedom.

Example 12: A point on the $y$-axis equidistant from two points

Find the point on the $y$-axis equidistant from $(3, 1, 2)$ and $(5, 5, 2)$.

Solution:

Step 1 — Parametrise. For $(0, y, 0)$: $9 + (y-1)^2 + 4 = 25 + (y-5)^2 + 4$.

Step 2 — Expand. $13 + y^2 - 2y + 1 = 29 + y^2 - 10y + 25$, i.e. $14 - 2y = 54 - 10y$.

Step 3 — Solve and check. $8y = 40$: $y = 5$ — the point is $(0, 5, 0)$, and both distances equal $\sqrt{29}$ ✓.

Takeaway: Always verify with the actual distances — the check costs two lines and catches every sign slip.

Example 13: A point on the $z$-axis equidistant from two points

Find the point on the $z$-axis equidistant from $(1, 5, 7)$ and $(5, 1, -4)$.

Solution:

Step 1 — Parametrise. For $(0, 0, z)$: $1 + 25 + (z - 7)^2 = 25 + 1 + (z + 4)^2$.

Step 2 — Cancel the equal constants. $(z-7)^2 = (z+4)^2$, so $-14z + 49 = 8z + 16$.

Step 3 — Solve. $22z = 33$: $z = \frac{3}{2}$ — the point is $\left(0, 0, \frac{3}{2}\right)$.

Takeaway: When the fixed constants match on both sides ($1 + 25 = 25 + 1$), the equation collapses to comparing two squares — spot it and skip a line of algebra.

Example 14: A 3-4-5 triangle on the axes

Find the perimeter of the triangle with vertices $(0,0,0)$, $(3,0,0)$, $(0,4,0)$.

Solution:

Step 1 — Two sides along the axes. Origin to $(3,0,0)$: 3; origin to $(0,4,0)$: 4.

Step 2 — The hypotenuse. $(3,0,0)$ to $(0,4,0)$: $\sqrt{9 + 16 + 0} = 5$.

Step 3 — Sum. Perimeter $= 3 + 4 + 5 = 12$.

Takeaway: Points with a shared zero coordinate live in one coordinate plane — the problem is secretly 2D.

Example 15: An equilateral triangle in space

Show that $(1, 2, 3)$, $(2, 3, 1)$, $(3, 1, 2)$ form an equilateral triangle.

Solution:

Step 1 — Squared sides. $AB^2 = 1 + 1 + 4 = 6$; $BC^2 = 1 + 4 + 1 = 6$; $CA^2 = 4 + 1 + 1 = 6$.

Step 2 — Conclude. All sides $\sqrt{6}$: equilateral. ∎

Step 3 — See why before computing. The vertices are cyclic permutations of $(1, 2, 3)$ — the same three squared differences $1, 1, 4$ reappear in every side, just reordered.

Takeaway: Cyclic-permutation vertices always give equilateral triangles — symmetry does the arithmetic.

Example 16: A collinearity check

Are $A(1, 2, 7)$, $B(2, 6, 3)$, $C(3, 10, -1)$ collinear?

Solution:

Step 1 — Three distances. $AB = \sqrt{1 + 16 + 16} = \sqrt{33}$; $BC = \sqrt{1 + 16 + 16} = \sqrt{33}$; $AC = \sqrt{4 + 64 + 64} = \sqrt{132} = 2\sqrt{33}$.

Step 2 — Test the sum. $AB + BC = 2\sqrt{33} = AC$.

Step 3 — Conclude. Collinear — and since $AB = BC$, $B$ is the midpoint of $AC$. ∎

Takeaway: Equal steps $A \to B \to C$ with matching differences $(1, 4, -4)$ per coordinate is uniform motion along a line — the distance check just certifies it.

Example 17: Right and isosceles at once

Classify the triangle $A(0, 4, 1)$, $B(2, 3, -1)$, $C(4, 5, 0)$.

Solution:

Step 1 — Squared sides. $AB^2 = 4 + 1 + 4 = 9$; $BC^2 = 4 + 4 + 1 = 9$; $CA^2 = 16 + 1 + 1 = 18$.

Step 2 — Isosceles test. $AB = BC = 3$ ✓.

Step 3 — Right-angle test. $AB^2 + BC^2 = 18 = CA^2$: right angle at $B$ (the vertex between the equal sides).

Takeaway: $x^2, x^2, 2x^2$ is the signature of the isosceles right triangle — half a square cut along its diagonal.

Example 18: A locus with a ratio condition

A point $P$ moves so that $OP = 2 \cdot PA$ where $O$ is the origin and $A = (1, 0, 0)$. Find the locus.

Solution:

Step 1 — Square the ratio condition. $OP^2 = 4\,PA^2$: $x^2 + y^2 + z^2 = 4\left[(x-1)^2 + y^2 + z^2\right]$.

Step 2 — Expand. $x^2 + y^2 + z^2 = 4x^2 - 8x + 4 + 4y^2 + 4z^2$.

Step 3 — Collect. $3x^2 + 3y^2 + 3z^2 - 8x + 4 = 0$ — a sphere (an Apollonius sphere: the 3D analogue of the Apollonius circle).

Takeaway: A distance RATIO other than 1 gives a sphere; ratio exactly 1 gives the bisector plane — the $k = 1$ boundary case.

Example 19: On the sphere or not?

The set of points at distance 5 from the origin is $x^2 + y^2 + z^2 = 25$. Which of $(3, 4, 0)$ and $(2, 2, 4)$ lies on it?

Solution:

Step 1 — Test $(3, 4, 0)$. $9 + 16 + 0 = 25$ ✓ — on the sphere.

Step 2 — Test $(2, 2, 4)$. $4 + 4 + 16 = 24 < 25$ — inside the sphere (distance $\sqrt{24} = 2\sqrt{6} < 5$).

Takeaway: Substitute and compare with $r^2$: equal — on; less — inside; greater — outside. Same trichotomy as the circle, one coordinate up.

Example 20: Distances of a point from the three planes and the origin

For $P(-3, 4, -12)$, find the distances from the three coordinate planes and from the origin.

Solution:

Step 1 — Plane distances are absolute coordinates. From $XY$: $|z| = 12$; from $YZ$: $|x| = 3$; from $ZX$: $|y| = 4$.

Step 2 — Origin distance. $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$.

Step 3 — Connect them. $13^2 = 3^2 + 4^2 + 12^2$ — the origin distance bundles the three plane distances by Pythagoras.

Takeaway: $(3, 4, 12, 13)$ again — recognising quadruples turns arithmetic into recall.

Example 21: Solving for a coordinate from a distance

Find $x$ so that the distance between $(x, 2, 3)$ and $(1, 2, 3)$ is 5.

Solution:

Step 1 — Only $x$ differs. The distance collapses to $|x - 1|$.

Step 2 — Solve. $|x - 1| = 5$: $x = 6$ or $x = -4$.

Step 3 — Interpret. Two answers, one on each side of $x = 1$ — a distance condition rarely has a unique solution.

Takeaway: Keep the absolute value until the last step; dropping it silently loses the second solution.

Example 22: Distance plus midpoint

For $A(1, -1, 3)$ and $B(2, -4, 5)$, find $AB$ and the midpoint of $AB$.

Solution:

Step 1 — Distance. $AB = \sqrt{1 + 9 + 4} = \sqrt{14}$.

Step 2 — Midpoint. Average each coordinate: $\left(\frac{3}{2}, -\frac{5}{2}, 4\right)$.

Step 3 — Check. The midpoint's distance to each end should be $\frac{\sqrt{14}}{2}$: from $A$: $\sqrt{\frac{1}{4} + \frac{9}{4} + 1} = \sqrt{\frac{14}{4}}$ ✓.

Takeaway: Midpoint = coordinate averages — the $m = n$ case of the section formula.

Example 23: Equidistant from all three planes

A point in octant I is at distance 2 from each coordinate plane. Find it, and its distance from the origin.

Solution:

Step 1 — Translate. $|x| = |y| = |z| = 2$, all positive in octant I: $(2, 2, 2)$.

Step 2 — Origin distance. $\sqrt{4 + 4 + 4} = 2\sqrt{3}$.

Step 3 — Generalise. Any cube-corner point $(a, a, a)$ is $a\sqrt{3}$ from the origin — the cube-diagonal ratio.

Takeaway: "Distance from a coordinate plane" is always one absolute coordinate — no formula needed, just the right one.

Example 24: A degenerate "triangle"

Show that $(1, 2, 3)$, $(3, 2, 1)$ and $(2, 2, 2)$ do not form a triangle.

Solution:

Step 1 — Distances. $A(1,2,3)$ to $C(2,2,2)$: $\sqrt{1 + 0 + 1} = \sqrt{2}$; $C$ to $B(3,2,1)$: $\sqrt{2}$; $A$ to $B$: $\sqrt{4 + 0 + 4} = 2\sqrt{2}$.

Step 2 — Sum test. $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$: equality — the points are collinear, with $C$ the midpoint of $AB$. ∎

Step 3 — Confirm by averaging. Midpoint of $A$ and $B$: $\left(\frac{1+3}{2}, \frac{2+2}{2}, \frac{3+1}{2}\right) = (2, 2, 2) = C$ ✓.

Takeaway: "Show they do NOT form a triangle" = show the triangle inequality degenerates to equality.

Example 25: Verifying a parallelogram by midpoints

Check by diagonal midpoints that $(-1,2,1)$, $(1,-2,5)$, $(4,-7,8)$, $(2,-3,4)$ form a parallelogram.

Solution:

Step 1 — Midpoint of $AC$. $\left(\frac{-1+4}{2}, \frac{2-7}{2}, \frac{1+8}{2}\right) = \left(\frac{3}{2}, -\frac{5}{2}, \frac{9}{2}\right)$.

Step 2 — Midpoint of $BD$. $\left(\frac{1+2}{2}, \frac{-2-3}{2}, \frac{5+4}{2}\right) = \left(\frac{3}{2}, -\frac{5}{2}, \frac{9}{2}\right)$.

Step 3 — Conclude. Same point: the diagonals bisect each other — parallelogram. ∎

Takeaway: This re-proves Section 2's side-length verification in a third of the work — midpoints beat distances when both are available.

Example 26: Checking all pairings for a right angle

Why is the triangle $A(3,6,9)$, $B(10,20,30)$, $C(25,-41,5)$ not right-angled?

Solution:

Step 1 — Squared sides. $AB^2 = 686$, $BC^2 = 4571$, $CA^2 = 2709$.

Step 2 — Only one candidate hypotenuse. The right angle would have to face the LARGEST side, so the only possible identity is $AB^2 + CA^2 = BC^2$.

Step 3 — Test it. $686 + 2709 = 3395 \ne 4571$: no right angle. ∎

Takeaway: You never need to test all three pairings — only the largest square can be the hypotenuse's.

Example 27: Distance from the origin — ordering points

Order $A(1, 2, 2)$, $B(2, 2, 1)$, $C(0, 3, 0)$, $D(1, 1, 4)$ by distance from the origin.

Solution:

Step 1 — Squared distances. $OA^2 = 1 + 4 + 4 = 9$; $OB^2 = 4 + 4 + 1 = 9$; $OC^2 = 0 + 9 + 0 = 9$; $OD^2 = 1 + 1 + 16 = 18$.

Step 2 — Order. $A$, $B$, $C$ tie at distance 3; $D$ is farthest at $\sqrt{18} = 3\sqrt{2}$.

Step 3 — Picture it. $A$, $B$, $C$ all lie on the sphere of radius 3; $D$ sits outside it.

Takeaway: Squared distances order points exactly as distances do — compare squares, root only if the answer demands it.

Example 28: A sphere through a given point

A sphere has centre $(1, 2, -1)$ and passes through $(3, 4, 1)$. Find $r^2$ and the equation.

Solution:

Step 1 — Radius from centre to the point. $r^2 = (3-1)^2 + (4-2)^2 + (1+1)^2 = 4 + 4 + 4 = 12$.

Step 2 — Write the equation. $(x-1)^2 + (y-2)^2 + (z+1)^2 = 12$.

Step 3 — Check. Substituting $(3, 4, 1)$: $4 + 4 + 4 = 12$ ✓.

Takeaway: Centre plus one point determines a sphere the same way it determines a circle — one squared-distance computation.

Example 29: Centroid drill

Find the centroid of the triangle $(2, 4, 6)$, $(0, -2, 5)$, $(7, 1, 1)$.

Solution:

Step 1 — Sum each coordinate. $x$: $2 + 0 + 7 = 9$; $y$: $4 - 2 + 1 = 3$; $z$: $6 + 5 + 1 = 12$.

Step 2 — Divide by 3. Centroid $= (3, 1, 4)$.

Takeaway: Centroid = coordinate averages; in 3D that is three small averages, nothing more.

Example 30: A distance identity

Show that for any point $P(x, y, z)$, the squared distances from the three axes sum to twice the squared distance from the origin.

Solution:

Step 1 — Distance from each axis. From the $x$-axis: $\sqrt{y^2 + z^2}$ (the two coordinates NOT along the axis); similarly $\sqrt{x^2 + z^2}$ and $\sqrt{x^2 + y^2}$.

Step 2 — Sum the squares. $(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2x^2 + 2y^2 + 2z^2$.

Step 3 — Conclude. $= 2\,OP^2$. ∎ Each coordinate appears in exactly two of the three axis-distances — hence the factor 2.

Takeaway: Axis distance uses the OTHER two coordinates; plane distance uses just one — keep the two notions apart and identities like this become counting arguments.