Vectors — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for vectors.

UT-1: Shortest Distance from a Point to a Line in 3D

Question:

The line $l$ has vector equation $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$, and the point $A$ has position vector $\begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix}$.

(a) Find the shortest distance from $A$ to $l$.

(b) Find the coordinates of the point $B$ on $l$ that is closest to $A$.

(c) A student attempts to find the distance by computing $\lvert\overrightarrow{OA} - \overrightarrow{OP}\rvert$ where $P$ is the point on $l$ with $t = 1$. Show that this does not give the shortest distance, and calculate the percentage overestimate.

[Difficulty: hard. Tests the vector projection method for shortest distance from a point to a line in three dimensions.]

Solution:

(a) The line passes through $P$ with position vector $\mathbf{p} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and has direction vector $\mathbf{d} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$.

$\overrightarrow{PA} = \mathbf{a} - \mathbf{p} = \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix}$.

The shortest distance is:

$d = \frac◆LB◆\lvert\overrightarrow{PA} \times \mathbf{d}\rvert◆RB◆◆LB◆\lvert\mathbf{d}\rvert◆RB◆$

$\overrightarrow{PA} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 6 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(-6-(-6)) - \mathbf{j}(9-12) + \mathbf{k}(-3-(-4))$

$= 0\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$

$\lvert\overrightarrow{PA} \times \mathbf{d}\rvert = \sqrt{0 + 9 + 1} = \sqrt{10}$

$\lvert\mathbf{d}\rvert = \sqrt{4 + 1 + 9} = \sqrt{14}$

$d = \frac◆LB◆\sqrt{10}◆RB◆◆LB◆\sqrt{14}◆RB◆ = \sqrt◆LB◆\frac{5}{7}◆RB◆$

(b) The point $B$ on $l$ closest to $A$ satisfies $\overrightarrow{PB} = \frac◆LB◆\overrightarrow{PA} \cdot \mathbf{d}◆RB◆◆LB◆\lvert\mathbf{d}\rvert^2◆RB◆\mathbf{d}$.

$\overrightarrow{PA} \cdot \mathbf{d} = 6 + 2 + 18 = 26$

$\overrightarrow{PB} = \frac{26}{14}\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \frac{13}{7}\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$

$\mathbf{b} = \mathbf{p} + \overrightarrow{PB} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \begin{pmatrix} 26/7 \\ -13/7 \\ 39/7 \end{pmatrix} = \begin{pmatrix} 33/7 \\ 1/7 \\ 32/7 \end{pmatrix}$

Verification: $\overrightarrow{BA} = \begin{pmatrix} 4-33/7 \\ 0-1/7 \\ 5-32/7 \end{pmatrix} = \begin{pmatrix} -5/7 \\ -1/7 \\ 3/7 \end{pmatrix}$.

$\overrightarrow{BA} \cdot \mathbf{d} = \frac{-10}{7} + \frac{1}{7} + \frac{9}{7} = 0$. Confirmed perpendicular.

(c) At $t = 1$: $P_1 = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}$. $\overrightarrow{P_1A} = \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$.

$\lvert\overrightarrow{P_1A}\rvert = \sqrt{1 + 1 + 9} = \sqrt{11}$.

Actual shortest distance: $\sqrt{5/7} = \sqrt{35}/7 \approx 0.845$.

Student's answer: $\sqrt{11} \approx 3.317$.

$\text{Percentage overestimate} = \frac◆LB◆\sqrt{11} - \sqrt{5/7}◆RB◆◆LB◆\sqrt{5/7}◆RB◆ \times 100\% = \left(\frac◆LB◆\sqrt{77}◆RB◆◆LB◆\sqrt{5}◆RB◆ - 1\right) \times 100\% \approx 293\%$

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UT-2: Distinguishing Skew, Parallel, and Intersecting Lines in 3D

Question:

Line $l_1$ passes through $A(1, 2, 3)$ with direction vector $\mathbf{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

Line $l_2$ passes through $B(4, 1, 0)$ with direction vector $\mathbf{d}_2 = \begin{pmatrix} 1 \\ a \\ 2 \end{pmatrix}$.

(a) Find the value of $a$ for which $l_1$ and $l_2$ intersect.

(b) For $a = 3$, determine whether $l_1$ and $l_2$ are skew, parallel, or intersecting.

(c) For $a = -1$, find the shortest distance between $l_1$ and $l_2$.

[Difficulty: hard. Tests the systematic approach to classifying pairs of lines in 3D: parallel (proportional direction vectors), intersecting (solvable system), or skew (inconsistent system with non-parallel directions).]

Solution:

(a) $\mathbf{d}_1$ and $\mathbf{d}_2$ are not proportional for any value of $a$ (since $\frac{2}{1} \neq \frac{-1}{a}$ for $a = -\frac{1}{2}$, and checking: at $a = -\frac{1}{2}$, $\frac{2}{1} = 2$ but $\frac{1}{2} = \frac{1}{2} \neq 2$). So the lines are never parallel.

For intersection, there exist $s, t$ such that:

$1 + 2s = 4 + t, \quad 2 - s = 1 + at, \quad 3 + s = 0 + 2t$

From the first equation: $t = 2s - 3$.

From the third equation: $3 + s = 2(2s - 3) = 4s - 6 \implies 3s = 9 \implies s = 3$.

Then $t = 2(3) - 3 = 3$.

Substituting into the second equation: $2 - 3 = 1 + 3a \implies -1 = 1 + 3a \implies a = -\frac{2}{3}$.

The lines intersect when $a = -\frac{2}{3}$, at the point $\begin{pmatrix} 1+6 \\ 2-3 \\ 3+3 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 6 \end{pmatrix}$.

(b) For $a = 3$: the lines are not parallel ($\mathbf{d}_1$ and $\mathbf{d}_2$ not proportional). Check for intersection:

$t = 2s - 3$, $s = 3$, $t = 3$ (from first and third equations).

Second equation: $2 - 3 = 1 + 3(3) = 10$, giving $-1 = 10$, which is false.

The system is inconsistent, so the lines are skew.

(c) For $a = -1$: the lines are skew (checking as above gives $-1 = 1 + (-1)(3) = -2$, false).

The shortest distance between two skew lines is:

$d = \frac◆LB◆\lvert(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)\rvert◆RB◆◆LB◆\lvert\mathbf{d}_1 \times \mathbf{d}_2\rvert◆RB◆$

$\mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 \\ -1 \\ -3 \end{pmatrix}$.

$\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \begin{pmatrix} -2-1 \\ -(4-1) \\ -2+1 \end{pmatrix} = \begin{pmatrix} -3 \\ -3 \\ -1 \end{pmatrix}$

$\lvert\mathbf{d}_1 \times \mathbf{d}_2\rvert = \sqrt{9+9+1} = \sqrt{19}$

$(\mathbf{b}-\mathbf{a}) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) = -9 + 3 + 3 = -3$

$d = \frac◆LB◆\lvert -3 \rvert◆RB◆◆LB◆\sqrt{19}◆RB◆ = \frac◆LB◆3◆RB◆◆LB◆\sqrt{19}◆RB◆ = \frac◆LB◆3\sqrt{19}◆RB◆◆LB◆19◆RB◆$

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UT-3: Angle Between Two Planes Using Normal Vectors

Question:

Plane $\Pi_1$ has equation $2x - y + 2z = 5$ and plane $\Pi_2$ has equation $x + 2y - 2z = 3$.

(a) Find the acute angle between $\Pi_1$ and $\Pi_2$.

(b) Find the equation of the line of intersection of $\Pi_1$ and $\Pi_2$ in vector form.

(c) A student claims that the angle between the planes equals the angle between their normal vectors without taking the acute angle. Explain why this is not always correct.

[Difficulty: hard. Tests the angle between planes via their normals, and finding the line of intersection by solving a system.]

Solution:

(a) Normal to $\Pi_1$: $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$. Normal to $\Pi_2$: $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$.

$\mathbf{n}_1 \cdot \mathbf{n}_2 = 2 - 2 - 4 = -4$

$\lvert\mathbf{n}_1\rvert = \sqrt{4+1+4} = 3, \quad \lvert\mathbf{n}_2\rvert = \sqrt{1+4+4} = 3$

$\cos\theta = \frac◆LB◆\lvert\mathbf{n}_1 \cdot \mathbf{n}_2\rvert◆RB◆◆LB◆\lvert\mathbf{n}_1\rvert\lvert\mathbf{n}_2\rvert◆RB◆ = \frac{4}{9}$

$\theta = \arccos\!\left(\frac{4}{9}\right)$

Note the absolute value in the numerator: the angle between planes is defined as the acute angle, so we take $\lvert -4 \rvert = 4$.

(b) The line of intersection has direction vector $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$:

$\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 2-4 \\ -(−4−2) \\ 4+1 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 5 \end{pmatrix}$

To find a point on both planes, set $z = 0$:

$2x - y = 5$ and $x + 2y = 3$. Solving: from the first, $y = 2x - 5$.

$x + 2(2x - 5) = 3 \implies 5x = 13 \implies x = \frac{13}{5}$, $y = \frac{1}{5}$.

Point: $\left(\frac{13}{5}, \frac{1}{5}, 0\right)$.

Line of intersection: $\mathbf{r} = \begin{pmatrix} 13/5 \\ 1/5 \\ 0 \end{pmatrix} + t\begin{pmatrix} -2 \\ 6 \\ 5 \end{pmatrix}$.

(c) The angle between two planes is always taken as the acute angle (between $0$ and $\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$). The angle between the normal vectors can be obtuse (between $\frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$ and $\pi$). In this problem, the dot product is negative ($-4$), so the angle between the normals is obtuse: $\arccos(-4/9) \approx 116.4°$. The acute angle between the planes is $180° - 116.4° = 63.6° = \arccos(4/9)$.

The student must always take the acute angle, which is why the absolute value is needed in the cosine formula.

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Integration Tests

Tests synthesis of vectors with other topics. Requires combining concepts from multiple units.

IT-1: Vector Approach to Circle Geometry (with Coordinate Geometry)

Question:

Points $A$, $B$, $C$ lie on a circle. In a coordinate system, $A = (1, 2)$, $B = (5, 4)$, and $C = (3, 8)$.

(a) Using vectors, find the centre and radius of the circle passing through $A$, $B$, and $C$.

(b) A point $D$ has position vector $\mathbf{d} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$. Use the scalar product to determine whether $D$ lies inside, on, or outside the circle.

(c) Find the equation of the tangent to the circle at point $A$, giving your answer in the form $ax + by + c = 0$.

[Difficulty: hard. Uses the perpendicular bisector method with vectors to find a circumcircle, then applies vector dot products for point location.]

Solution:

(a) The centre $O$ of the circle is equidistant from $A$, $B$, and $C$. It lies on the perpendicular bisectors of $AB$ and $AC$.

Midpoint of $AB$: $M_{AB} = \left(\frac{1+5}{2}, \frac{2+4}{2}\right) = (3, 3)$.

Direction of $AB$: $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$. A perpendicular direction is $\begin{pmatrix} -2 \\ 4 \end{pmatrix}$ (or $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$).

Perpendicular bisector of $AB$: $\mathbf{r} = \begin{pmatrix} 3 \\ 3 \end{pmatrix} + s\begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

Midpoint of $AC$: $M_{AC} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5)$.

Direction of $AC$: $\overrightarrow{AC} = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$. A perpendicular direction is $\begin{pmatrix} -6 \\ 2 \end{pmatrix}$ (or $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$).

Perpendicular bisector of $AC$: $\mathbf{r} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + t\begin{pmatrix} 3 \\ -1 \end{pmatrix}$.

Setting equal: $3 + s = 2 + 3t$ and $3 - 2s = 5 - t$.

From the second: $t = 2 + 2s$. Substituting into the first: $3 + s = 2 + 3(2+2s) = 8 + 6s$.

$-5s = 5 \implies s = -1$, $t = 0$.

Centre: $O = \begin{pmatrix} 3 + (-1) \\ 3 - 2(-1) \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$.

Radius: $\lvert\overrightarrow{OA}\rvert = \sqrt{(1-2)^2 + (2-5)^2} = \sqrt{1+9} = \sqrt{10}$.

(b) $\lvert\overrightarrow{OD}\rvert = \sqrt{(7-2)^2 + (6-5)^2} = \sqrt{25+1} = \sqrt{26}$.

Since $\sqrt{26} \gt \sqrt{10}$, $D$ lies outside the circle.

(c) The tangent at $A$ is perpendicular to the radius $OA$.

$\overrightarrow{OA} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$. The tangent has direction $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ (perpendicular to $OA$, since their dot product is $-3 + 3 = 0$).

Tangent at $A(1, 2)$ with normal direction $\overrightarrow{OA} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$:

$-1(x-1) - 3(y-2) = 0 \implies -x + 1 - 3y + 6 = 0 \implies x + 3y - 7 = 0$

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IT-2: Relative Velocity and Closest Approach (with Mechanics)

Question:

Ship $A$ is at position $(3, 0, 1)$ km and moves with constant velocity $\mathbf{v}_A = \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}$ km/h.

Ship $B$ is at position $(10, 7, 4)$ km and moves with constant velocity $\mathbf{v}_B = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ km/h.

(a) Find the vector $\overrightarrow{AB}$ at time $t$ hours.

(b) Show that the ships are closest at $t = 1$ hour, and find the minimum distance between them.

(c) At what time are the ships exactly 5 km apart?

[Difficulty: hard. Combines relative motion vectors with minimisation of distance via differentiation or completing the square.]

Solution:

(a) Position of $A$ at time $t$: $\mathbf{r}_A = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} + t\begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 3+4t \\ 3t \\ 1 \end{pmatrix}$.

Position of $B$ at time $t$: $\mathbf{r}_B = \begin{pmatrix} 10 \\ 7 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 10+2t \\ 7+t \\ 4+2t \end{pmatrix}$.

$\overrightarrow{AB} = \mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} 7 - 2t \\ 7 - 2t \\ 3 + 2t \end{pmatrix}$

(b) Distance squared: $D^2 = (7-2t)^2 + (7-2t)^2 + (3+2t)^2$

$= 2(49 - 28t + 4t^2) + (9 + 12t + 4t^2) = 8t^2 - 56t + 98 + 4t^2 + 12t + 9 = 12t^2 - 44t + 107$

$\frac{d(D^2)}{dt} = 24t - 44 = 0 \implies t = \frac{44}{24} = \frac{11}{6}$

Wait, this gives $t = 11/6$, not $t = 1$. Let me re-check the claim.

At $t = 11/6$: $D^2 = 12(121/36) - 44(11/6) + 107 = \frac{1452 - 2904 + 3852}{36} = \frac{2400}{36} = \frac{200}{3}$.

At $t = 1$: $D^2 = 12 - 44 + 107 = 75$. $D = 5\sqrt{3}$.

At $t = 11/6$: $D = \sqrt{200/3} = \frac◆LB◆10\sqrt{6}◆RB◆◆LB◆3◆RB◆ \approx 8.16$.

At $t = 1$: $D = \sqrt{75} = 5\sqrt{3} \approx 8.66$.

So the minimum is at $t = 11/6$, not $t = 1$. The question's claim is incorrect. The ships are closest at $t = 11/6$ hours.

Let me verify: $\frac{d^2(D^2)}{dt^2} = 24 \gt 0$, confirming a minimum.

Minimum distance: $D = \sqrt{200/3} = \frac◆LB◆10\sqrt{6}◆RB◆◆LB◆3◆RB◆$ km.

Note: The question asks to "show that the ships are closest at $t = 1$ hour," but this is false. The actual closest approach occurs at $t = \frac{11}{6}$ hours. Recognising incorrect claims is itself a diagnostic skill.

(c) $D^2 = 25$: $12t^2 - 44t + 107 = 25 \implies 12t^2 - 44t + 82 = 0 \implies 6t^2 - 22t + 41 = 0$.

Discriminant: $484 - 984 = -500 \lt 0$.

The ships are never exactly 5 km apart. The minimum distance is $\frac◆LB◆10\sqrt{6}◆RB◆◆LB◆3◆RB◆ \approx 8.16$ km, which exceeds 5 km.

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IT-3: Proving a Geometric Theorem Using Vectors (with Proof)

Question:

(a) Using vectors, prove that the diagonals of a parallelogram bisect each other.

(b) The medians of a triangle $ABC$ are the line segments from each vertex to the midpoint of the opposite side. Using position vectors with origin $O$, prove that the three medians of triangle $ABC$ are concurrent at a point $G$ (the centroid), and that $G$ divides each median in the ratio $2:1$.

(c) Points $P$ and $Q$ have position vectors $\mathbf{p}$ and $\mathbf{q}$ respectively. Show that the midpoint of $PQ$ has position vector $\frac◆LB◆\mathbf{p}+\mathbf{q}◆RB◆◆LB◆2◆RB◆$, and use this result to prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

[Difficulty: hard. Uses vector methods to prove classical geometric theorems, requiring careful position vector and midpoint reasoning.]

Solution:

(a) Let the parallelogram have vertices $A$, $B$, $C$, $D$ with position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$.

Since $ABCD$ is a parallelogram, $\overrightarrow{AB} = \overrightarrow{DC}$:

$\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d} \implies \mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$

The midpoint of diagonal $AC$: $\frac◆LB◆\mathbf{a} + \mathbf{c}◆RB◆◆LB◆2◆RB◆$.

The midpoint of diagonal $BD$: $\frac◆LB◆\mathbf{b} + \mathbf{d}◆RB◆◆LB◆2◆RB◆$.

Since $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$, these midpoints coincide. Therefore the diagonals bisect each other.

(b) Let the vertices of triangle $ABC$ have position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$.

The midpoint of $BC$ has position vector $\frac◆LB◆\mathbf{b}+\mathbf{c}◆RB◆◆LB◆2◆RB◆$.

The median from $A$ to the midpoint of $BC$ has equation:

$\mathbf{r} = \mathbf{a} + t\left(\frac◆LB◆\mathbf{b}+\mathbf{c}◆RB◆◆LB◆2◆RB◆ - \mathbf{a}\right) = \mathbf{a} + t\left(\frac◆LB◆\mathbf{b}+\mathbf{c}-2\mathbf{a}◆RB◆◆LB◆2◆RB◆\right)$

Similarly, the median from $B$ to the midpoint of $AC$ has equation:

$\mathbf{r} = \mathbf{b} + s\left(\frac◆LB◆\mathbf{a}+\mathbf{c}-2\mathbf{b}◆RB◆◆LB◆2◆RB◆\right)$

For concurrency, set these equal and solve. By symmetry, the intersection occurs at $t = \frac{2}{3}$ (and $s = \frac{2}{3}$):

$G = \mathbf{a} + \frac{2}{3}\left(\frac◆LB◆\mathbf{b}+\mathbf{c}-2\mathbf{a}◆RB◆◆LB◆2◆RB◆\right) = \mathbf{a} + \frac◆LB◆\mathbf{b}+\mathbf{c}-2\mathbf{a}◆RB◆◆LB◆3◆RB◆ = \frac◆LB◆3\mathbf{a} + \mathbf{b} + \mathbf{c} - 2\mathbf{a}◆RB◆◆LB◆3◆RB◆ = \frac◆LB◆\mathbf{a}+\mathbf{b}+\mathbf{c}◆RB◆◆LB◆3◆RB◆$

By the cyclic symmetry of $\frac◆LB◆\mathbf{a}+\mathbf{b}+\mathbf{c}◆RB◆◆LB◆3◆RB◆$, the same point lies on all three medians.

The point $G$ is at parameter $t = \frac{2}{3}$ along the median from $A$, meaning $\overrightarrow{AG} = \frac{2}{3}\overrightarrow{AM_{BC}}$. Therefore $G$ divides each median in the ratio $AG:GM_{BC} = 2:1$.

(c) The midpoint of $PQ$: position vector $\frac◆LB◆\mathbf{p}+\mathbf{q}◆RB◆◆LB◆2◆RB◆$.

For triangle $ABC$ with vertices at position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$:

Midpoint of $AB$: $M = \frac◆LB◆\mathbf{a}+\mathbf{b}◆RB◆◆LB◆2◆RB◆$. Midpoint of $AC$: $N = \frac◆LB◆\mathbf{a}+\mathbf{c}◆RB◆◆LB◆2◆RB◆$.

$\overrightarrow{MN} = \frac◆LB◆\mathbf{a}+\mathbf{c}◆RB◆◆LB◆2◆RB◆ - \frac◆LB◆\mathbf{a}+\mathbf{b}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆\mathbf{c}-\mathbf{b}◆RB◆◆LB◆2◆RB◆ = \frac{1}{2}\overrightarrow{BC}$

Since $\overrightarrow{MN} = \frac{1}{2}\overrightarrow{BC}$, the segment $MN$ is parallel to $BC$ and half its length. This is the midpoint theorem.