Solving Geometric Problems with Vectors

Position vectors locate every point in space at once — subtract two of them, and the line joining them falls out for free.

Arithmetic and parallel vectors in 3D

Addition, subtraction, and scalar multiplication of 3D vectors all work exactly as in 2D, applied component-by-component across all three entries. Two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel precisely when one is a scalar multiple of the other, $\mathbf{b}=\lambda\mathbf{a}$ for some scalar $\lambda$ — which in 3D means each corresponding pair of components must share the same scale factor.

Position vectors and displacement in 3D

A position vector locates a point relative to a fixed origin $O$; the point $A=(a_1,a_2,a_3)$ has position vector $\overrightarrow{OA}=\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$. Exactly as in 2D, the displacement vector from one point to another is found by subtracting position vectors:

This single relationship is the engine behind almost every 3D geometric problem: once two points' position vectors are known, the vector connecting them — and hence the direction, and (via the magnitude formula) the distance between them — follows immediately.

Distance between two points in 3D

The distance between points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is the magnitude of $\overrightarrow{AB}$, giving the 3D distance formula:

This is simply the 3D magnitude formula from the previous lesson, applied to the displacement vector $\mathbf{b}-\mathbf{a}$ rather than to a position vector directly.

Worked example 1

Distance between two points in 3D

Points $A$ and $B$ have position vectors $\mathbf{a}=2\mathbf{i}-\mathbf{j}+5\mathbf{k}$ and $\mathbf{b}=6\mathbf{i}+3\mathbf{j}-\mathbf{k}$. Find the exact distance $AB$.

MethodFind the displacement vector, then take its magnitude

1

$\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = (6-2)\mathbf{i}+(3-(-1))\mathbf{j}+(-1-5)\mathbf{k} = 4\mathbf{i}+4\mathbf{j}-6\mathbf{k}$.

Subtract position vectors componentwise: b − a.

2

$AB = |\overrightarrow{AB}| = \sqrt{4^2+4^2+(-6)^2} = \sqrt{16+16+36} = \sqrt{68}$.

Apply the 3D magnitude formula to the displacement vector.

3

$=2\sqrt{17}$.

Simplify the surd: 68 = 4 × 17.

▪ solved

Worked example 2

Finding the fourth vertex of a parallelogram

$ABCD$ is a parallelogram with $A(1,2,-1)$, $B(4,0,3)$, and $C(6,5,2)$. Find the coordinates of $D$.

MethodUse the fact that opposite sides of a parallelogram are equal vectors

1

In parallelogram $ABCD$, side $AB$ is parallel and equal in length to side $DC$, and both are traversed in the same rotational sense, so $\overrightarrow{AB} = \overrightarrow{DC}$.

This is the defining property of a parallelogram, expressed vectorially.

2

$\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = (4-1)\mathbf{i}+(0-2)\mathbf{j}+(3-(-1))\mathbf{k} = 3\mathbf{i}-2\mathbf{j}+4\mathbf{k}$.

Compute the known side AB first.

3

Since $\overrightarrow{DC}=\mathbf{c}-\mathbf{d}$, and this must equal $\overrightarrow{AB}$: $\mathbf{c}-\mathbf{d} = 3\mathbf{i}-2\mathbf{j}+4\mathbf{k} \Rightarrow \mathbf{d} = \mathbf{c}-(3\mathbf{i}-2\mathbf{j}+4\mathbf{k})$.

Rearrange to make d (the unknown position vector) the subject.

4

$\mathbf{d} = (6\mathbf{i}+5\mathbf{j}+2\mathbf{k}) - (3\mathbf{i}-2\mathbf{j}+4\mathbf{k}) = 3\mathbf{i}+7\mathbf{j}-2\mathbf{k}$.

Substitute c and subtract componentwise.

5

$D = (3,7,-2)$.

Read the coordinates from the position vector.

▪ solved

Why not DA = BC?

Either pair of opposite sides can be used ($\overrightarrow{AB}=\overrightarrow{DC}$ or $\overrightarrow{BC}=\overrightarrow{AD}$) — both express the same parallelogram property and give the same point $D$. Choosing the pair that isolates $\mathbf{d}$ most directly just keeps the algebra tidier.

Worked example 3

Finding a missing vertex of a square

$JKLS$ is a square, with $J(1,2,1)$, $K(4,2,5)$ and $L(4,7,5)$. Find the coordinates of $S$, and verify that $JKLS$ is indeed a square by checking that $\overrightarrow{JK}$ and $\overrightarrow{KL}$ are perpendicular and equal in length.

MethodUse JS = KL to locate S, then check perpendicularity and length with the two known sides

1

$\overrightarrow{JK} = \mathbf{k}-\mathbf{j} = (4-1)\mathbf{i}+(2-2)\mathbf{j}+(5-1)\mathbf{k} = 3\mathbf{i}+0\mathbf{j}+4\mathbf{k}$.

Find one known side of the square as a displacement vector.

2

$\overrightarrow{KL} = \mathbf{l}-\mathbf{k} = (4-4)\mathbf{i}+(7-2)\mathbf{j}+(5-5)\mathbf{k} = 0\mathbf{i}+5\mathbf{j}+0\mathbf{k}$.

Find the other known side.

3

In square $JKLS$ (vertices in order), side $\overrightarrow{JS}$ is equal and parallel to side $\overrightarrow{KL}$, so $\mathbf{s} = \mathbf{j}+\overrightarrow{KL} = (\mathbf{i}+2\mathbf{j}+\mathbf{k})+(5\mathbf{j}) = \mathbf{i}+7\mathbf{j}+\mathbf{k}$.

Exactly the same "opposite sides are equal vectors" idea used to find D in the parallelogram above.

4

$S = (1,7,1)$.

Read the coordinates from the position vector.

5

Check perpendicular: $\overrightarrow{JK}\cdot\overrightarrow{KL} = (3)(0)+(0)(5)+(4)(0) = 0$. ✓

A zero dot product confirms the two sides meet at a right angle.

6

Check equal length: $|\overrightarrow{JK}| = \sqrt{9+0+16}=\sqrt{25}=5$; $|\overrightarrow{KL}| = \sqrt{0+25+0}=\sqrt{25}=5$. ✓

Equal adjacent side lengths, together with the right angle, confirm JKLS really is a square (not just a rectangle or rhombus).

▪ solved

Proving a shape, not just locating it

Finding a missing vertex only uses the "opposite sides equal" property, which alone just proves a parallelogram. Confirming a right angle (via a zero dot product) and equal adjacent side lengths (via the magnitude formula) are what upgrade that parallelogram to a genuine square — the same two checks distinguish a rectangle (right angles, unequal sides) or a rhombus (equal sides, no right angle) from a square.

Worked example 4

Worked example 5

Worked example 6

Midpoint and an isosceles triangle condition · Challenge

Triangle $ABC$ has $A(2,-1,4)$, $B(0,3,2)$, $C(6,1,-2)$. Let $M$ be the midpoint of $BC$. Find the coordinates of $M$, and determine whether triangle $ABM$ is isosceles.

MethodFind M using the midpoint formula, then compare side lengths using the distance formula

1

$\mathbf{m} = \dfrac{\mathbf{b}+\mathbf{c}}{2} = \dfrac{(0+6)\mathbf{i}+(3+1)\mathbf{j}+(2+(-2))\mathbf{k}}{2} = 3\mathbf{i}+2\mathbf{j}+0\mathbf{k}$.

The midpoint's position vector is the average of the two endpoints' position vectors — this extends unchanged to 3D.

2

$M = (3,2,0)$.

Read off coordinates.

3

$AM = |\mathbf{m}-\mathbf{a}| = |(3-2)\mathbf{i}+(2-(-1))\mathbf{j}+(0-4)\mathbf{k}| = |\mathbf{i}+3\mathbf{j}-4\mathbf{k}| = \sqrt{1+9+16}=\sqrt{26}$.

Distance from A to M using the 3D distance formula.

4

$AB = |\mathbf{b}-\mathbf{a}| = |(-2)\mathbf{i}+4\mathbf{j}-2\mathbf{k}| = \sqrt{4+16+4}=\sqrt{24}$.

Distance from A to B.

5

$BM = |\mathbf{m}-\mathbf{b}| = |3\mathbf{i}-\mathbf{j}-2\mathbf{k}| = \sqrt{9+1+4}=\sqrt{14}$.

Distance from B to M, to complete all three sides of triangle ABM.

6

The three side lengths $\sqrt{26}$, $\sqrt{24}$, $\sqrt{14}$ are all different, so triangle $ABM$ is not isosceles.

An isosceles triangle needs at least two equal side lengths — none match here.

▪ solved

Look deeper — the midpoint formula in 3D

Just as with magnitude, the midpoint formula extends from 2D to 3D by simply averaging the extra component — no new theory is required, only the willingness to carry an established 2D idea one dimension further. This pattern (extend componentwise, apply Pythagoras twice for lengths) covers the vast majority of 3D vector geometry problems.