Locus problems and applications in 3D

A locus is the set of all points satisfying some geometric condition. In 3D, common loci include planes, spheres, and various curves. Translating a geometric condition into an algebraic equation in $(x, y, z)$ is the central technique.

Standard loci

Plane. $A x + B y + C z + D = 0$ for constants $A, B, C, D$ (not all $A, B, C$ zero). Class XII will study this in depth.

Sphere. All points at a fixed distance $r$ from a fixed centre $(a, b, c)$: $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2.$

Equidistant locus of two points. The perpendicular-bisector plane of the segment joining them.

Locus equidistant from two planes. The pair of planes that bisect the angles between them.

Setting up a locus

1. Let the moving point be $P(x, y, z)$.
2. Translate the geometric condition into an equation involving $x, y, z$.
3. Simplify , usually by squaring distances to eliminate roots.

Worked examples

Example 1. Find the locus of a point that is at distance $5$ from $(1, 2, 3)$.

$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 25$. A sphere centred at $(1, 2, 3)$ with radius $5$.

Example 2. Find the locus equidistant from $(2, 0, 0)$ and $(0, 4, 0)$.

$(x - 2)^2 + y^2 + z^2 = x^2 + (y - 4)^2 + z^2$. $x^2 - 4x + 4 + y^2 = x^2 + y^2 - 8 y + 16$. $-4 x + 4 = -8 y + 16 \Rightarrow 4 x - 8 y + 12 = 0 \Rightarrow x - 2 y + 3 = 0$.

A plane (which extends infinitely in $z$).

Example 3. Find the locus of a point whose distance from the origin equals its distance from the $xy$-plane.

$\sqrt{x^2 + y^2 + z^2} = |z|$. Square: $x^2 + y^2 + z^2 = z^2 \Rightarrow x^2 + y^2 = 0 \Rightarrow x = y = 0$. So the locus is the $z$-axis.

(Geometrically: distance from origin = distance from $xy$-plane only when projection to the plane has length zero , i.e., the point is on the $z$-axis.)

Example 4. A point $P$ moves so that the sum of squares of its distances from $(a, 0, 0)$ and $(-a, 0, 0)$ is $2 a^2 + 2 c$. Find the locus.

$|PA|^2 + |PB|^2 = (x - a)^2 + y^2 + z^2 + (x + a)^2 + y^2 + z^2 = 2 x^2 + 2 a^2 + 2 y^2 + 2 z^2 = 2 a^2 + 2 c$.

So $x^2 + y^2 + z^2 = c$. A sphere of radius $\sqrt{c}$ centred at origin (when $c > 0$).

Example 5. A particle's coordinates at time $t$ are $(t, t^2, t^3)$. Find a point on its trajectory at $t = 2$, and the distance travelled from origin to that point.

At $t = 2$: $(2, 4, 8)$. Distance from origin: $\sqrt{4 + 16 + 64} = \sqrt{84} = 2\sqrt{21}$.

Try it yourself

1. Find the locus of a point at distance $3$ from $(0, 0, 0)$.
2. Find the locus of points equidistant from $(1, 2, 3)$ and $(-1, -2, -3)$.
3. Find the locus of a point whose distance from $(0, 1, 0)$ equals its distance from $(0, -1, 0)$.
4. Find the locus equidistant from the $xy$-plane and the $yz$-plane.
5. A point moves so that the sum of its distances from $(0, 0, 0)$ and $(2, 0, 0)$ is $4$. (Locus is a closed curve , in fact, an ellipsoid.) Set up the equation.
6. A point's distance from the $z$-axis is $5$. Find the locus.
7. Find the locus of points equidistant from the three vertices of triangle $(1, 0, 0), (0, 1, 0), (0, 0, 1)$.
8. The locus of a point whose coordinates satisfy $x = 2 y$ and $y = 3 z$: describe.
9. A point $P$ has $|PA|^2 + |PB|^2 = 26$ where $A = (1, 0, 0), B = (-1, 0, 0)$. Find the locus.
10. Sphere with centre $(2, -1, 3)$ and radius $4$: write its equation.
11. Find the centre and radius of the sphere $x^2 + y^2 + z^2 - 4 x + 6 y - 2 z - 11 = 0$.
12. A point moves on the line through $(1, 1, 1)$ and $(2, 3, 4)$. Express its coordinates parametrically.

Pitfalls / Tricks

• Always simplify by squaring distances first.
• When a locus turns out to be a single point or empty, the original condition may be inconsistent or degenerate , check.
• A sphere's equation has equal coefficients on $x^2, y^2, z^2$.
• Insight. Locus is the bridge between geometry and algebra in 3D. Master the dictionary , "equidistant" → plane, "fixed distance" → sphere , and most problems unlock quickly.