Vectors -- Diagnostic Tests

Vectors are like arrows with both direction and magnitude — they point where you need to go and tell you how far.

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

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:

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

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

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

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

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

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

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

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

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

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

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: $\sqrt{11} \approx 3.317$.

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

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:

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

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

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

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

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

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

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 $\Pi_2$: $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$.

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

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

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

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

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 $\frac{\pi}{2}$). The angle between the normal vectors can be obtuse (between $\frac{\pi}{2}$ 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.

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

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$$BAnd $C$.

(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$$BAnd $C$. 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 $\overrightarrow{OA} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}$:

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

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

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: $D = \sqrt{200/3} = \frac{10\sqrt{6}}{3}$ 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 $\frac{10\sqrt{6}}{3} \approx 8.16$ km, which exceeds 5 km.

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 $\frac{\mathbf{p}+\mathbf{q}}{2}$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 $AC$: $\frac{\mathbf{a} + \mathbf{c}}{2}$.

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

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 $\frac{\mathbf{b}+\mathbf{c}}{2}$.

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

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

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

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

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

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

By the cyclic symmetry of $\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}$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 $\frac{\mathbf{p}+\mathbf{q}}{2}$.

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

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

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

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

Confusing the scalar (dot) product with the vector (cross) product: The dot product $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ gives a scalar and is used to find angles and test perpendicularity. The cross product $\mathbf{a} \times \mathbf{b}$ gives a vector perpendicular to both and is used to find areas and the direction perpendicular to two vectors. Using the wrong one gives fundamentally different results.

Forgetting that two vectors are parallel if one is a scalar multiple of the other: $\mathbf{a} \parallel \mathbf{b}$ if and only if $\mathbf{a} = k\mathbf{b}$ for some scalar $k$. Students often try to use the dot product to test for parallelism (which actually tests perpendicularity). For parallelism, check that the ratios of corresponding components are equal.

Misidentifying the direction vector of a line of intersection: The line where two planes intersect has direction vector $\mathbf{n}_1 \times \mathbf{n}_2$ (the cross product of the normals). Students sometimes use the dot product instead, which gives a scalar, not a direction vector.

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