Vectors — Diagnostic Tests

Unit Tests

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

UT-1: Skew Lines — Classification and Shortest Distance

Question:

Two lines are given by:

$L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$

$L_2: \mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$

(a) Determine whether $L_1$ and $L_2$ are parallel, intersecting, or skew.

(b) If skew, find the shortest distance between them.

(c) A student claims: “Since the direction vectors are not scalar multiples, the lines must intersect.” Explain why this reasoning is wrong.

[Difficulty: hard. Tests classification of lines in 3D and computation of shortest distance between skew lines.]

Solution:

(a) The direction vectors are $\mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$.

These are not scalar multiples, so the lines are not parallel.

To check for intersection, set the position vectors equal:

$1 + 2\lambda = 3 + \mu \implies 2\lambda - \mu = 2 \quad \text{(i)}$

$2 + \lambda = 1 - \mu \implies \lambda + \mu = -1 \quad \text{(ii)}$

$0 - \lambda = 4 + 2\mu \implies -\lambda - 2\mu = 4 \quad \text{(iii)}$

From (i) and (ii): adding gives $3\lambda = 1$So $\lambda = \frac{1}{3}$Then $\mu = -\frac{4}{3}$.

Check (iii): $-\frac{1}{3} - 2\!\left(-\frac{4}{3}\right) = -\frac{1}{3} + \frac{8}{3} = \frac{7}{3} \neq 4$.

The system is inconsistent, so the lines do not intersect. They are skew.

(b) The shortest distance between skew lines is:

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

$\mathbf{a}_2 - \mathbf{a}_1 = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$

$\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -1 \\ 1 & -1 & 2 \end{vmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}$

$\lvert \mathbf{d}_1 \times \mathbf{d}_2 \rvert = \sqrt{1 + 25 + 9} = \sqrt{35}$

$(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2) = 2(1) + (-1)(-5) + 4(-3) = 2 + 5 - 12 = -5$

$d = \frac{5}{\sqrt{35}} = \frac{5\sqrt{35}}{35} = \frac{\sqrt{35}}{7}$

(c) The student”s error is that in three dimensions, two lines that are not parallel can still fail to intersect. In 2D, non-parallel lines always intersect, but in 3D they can be skew — they pass at different “heights” and never meet. The student has incorrectly generalised the 2D result.

UT-2: Point-to-Plane Distance from Three Points

Question:

Three points are given: $A(1, 0, 2)$, $B(3, 1, -1)$And $C(2, 2, 3)$.

(a) Find the Cartesian equation of the plane $\Pi$ passing through $A$, $B$And $C$.

(b) Find the perpendicular distance from the origin to $\Pi$.

(c) A student computes $\overrightarrow{AB} \times \overrightarrow{AC}$and gets $\begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$Concluding the plane equation is $x + y + z = 3$. Identify the error.

[Difficulty: hard. Tests plane from three points, normal vector computation, and point-to-plane distance.]

Solution:

(a)

$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$

$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & -3 \\ 1 & 2 & 1 \end{vmatrix}$

$= \mathbf{i}(1 + 6) - \mathbf{j}(2 + 3) + \mathbf{k}(4 - 1) = \begin{pmatrix} 7 \\ -5 \\ 3 \end{pmatrix}$

Using point $A(1, 0, 2)$:

$7(x - 1) - 5(y - 0) + 3(z - 2) = 0$

$7x - 7 - 5y + 3z - 6 = 0$

$7x - 5y + 3z = 13$

(b) The distance from the origin $(0, 0, 0)$ to the plane $7x - 5y + 3z = 13$ is:

$d = \frac{|7(0) - 5(0) + 3(0) - 13|}{\sqrt{49 + 25 + 9}} = \frac{13}{\sqrt{83}}$

(c) The student’s cross product computation is wrong. The correct cross product is $\begin{pmatrix} 7 \\ -5 \\ 3 \end{pmatrix}$Not $\begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}$. Specifically:

• The $x$-component: $1 \times 1 - (-3) \times 2 = 1 + 6 = 7$Not $3$.
• The $y$-component: $-(2 \times 1 - (-3) \times 1) = -(2 + 3) = -5$Not $3$.
• The $z$-component: $2 \times 2 - 1 \times 1 = 4 - 1 = 3$Which coincidentally matches.

Integration Tests

Tests synthesis of vectors with other topics.

IT-1: Volume of a Tetrahedron — Scalar Triple Product

Question:

Four points are given: O(0, 0, 0)$$A(2, 1, 0)$$B(1, 3, 2)And $C(0, 1, 4)$.

(a) Find the volume of tetrahedron $OABC$.

(b) Show that the four points are coplanar if and only if the scalar triple product $[\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}] = 0$.

[Difficulty: hard. Combines scalar triple product with geometric interpretation.]

Solution:

(a) The volume of a tetrahedron with vertices O$$A$$B$$C is:

$V = \frac{1}{6}\lvert [\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}] \rvert$

$\overrightarrow{OA} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}, \quad \overrightarrow{OC} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$

$[\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}] = \overrightarrow{OA} \cdot (\overrightarrow{OB} \times \overrightarrow{OC})$

$\overrightarrow{OB} \times \overrightarrow{OC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 2 \\ 0 & 1 & 4 \end{vmatrix} = \begin{pmatrix} 10 \\ -4 \\ 1 \end{pmatrix}$

$[\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}] = 2(10) + 1(-4) + 0(1) = 20 - 4 = 16$

$V = \frac{16}{6} = \frac{8}{3}$

(b) The scalar triple product $[\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ equals the volume of the parallelepiped spanned by \mathbf{a}$$\mathbf{b}And $\mathbf{c}$.

If the four points are coplanar, the three vectors \overrightarrow{OA}$$\overrightarrow{OB}$$\overrightarrow{OC} all lie in the same plane, meaning the parallelepiped they span has zero volume (it is flat). Therefore $[\overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC}] = 0$.

Conversely, if the scalar triple product is zero, then $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$Which means $\mathbf{a}$ is perpendicular to $\mathbf{b} \times \mathbf{c}$. Since $\mathbf{b} \times \mathbf{c}$ is perpendicular to the plane containing $\mathbf{b}$ and $\mathbf{c}$It follows that $\mathbf{a}$ lies in the same plane as $\mathbf{b}$ and $\mathbf{c}$. Hence the four points are coplanar.

IT-2: Reflection of a Point in a Plane

Question:

The plane $\Pi$ has equation $2x - y + 2z = 5$. The point $P$ has coordinates $(1, 3, -1)$.

(a) Find the coordinates of the reflection $P'$ of $P$ in the plane $\Pi$.

(b) Find the equation of the line $PP'$ and verify that the midpoint of $PP'$ lies on $\Pi$.

[Difficulty: hard. Combines line-plane intersection with vector geometry.]

Solution:

(a) The normal to $\Pi$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

The line through $P$ perpendicular to $\Pi$ is:

$\mathbf{r} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$

Find the foot of the perpendicular (intersection with $\Pi$):

$2(1 + 2t) - (3 - t) + 2(-1 + 2t) = 5$

$2 + 4t - 3 + t - 2 + 4t = 5$

$9t - 3 = 5 \implies 9t = 8 \implies t = \frac{8}{9}$

Foot of perpendicular:

$M = \begin{pmatrix} 1 + \frac{16}{9} \\ 3 - \frac{8}{9} \\ -1 + \frac{16}{9} \end{pmatrix} = \begin{pmatrix} \frac{25}{9} \\ \frac{19}{9} \\ \frac{7}{9} \end{pmatrix}$

The reflection $P'$ is such that $M$ is the midpoint of $PP'$:

$M = \frac{P + P'}{2} \implies P' = 2M - P = 2\begin{pmatrix} \frac{25}{9} \\ \frac{19}{9} \\ \frac{7}{9} \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{41}{9} \\ \frac{11}{9} \\ \frac{23}{9} \end{pmatrix}$

(b) The line $PP'$ passes through $P(1, 3, -1)$ and $P'\!\left(\frac{41}{9}, \frac{11}{9}, \frac{23}{9}\right)$:

Direction: $\begin{pmatrix} \frac{41}{9} - 1 \\ \frac{11}{9} - 3 \\ \frac{23}{9} + 1 \end{pmatrix} = \begin{pmatrix} \frac{32}{9} \\ -\frac{16}{9} \\ \frac{32}{9} \end{pmatrix} = \frac{16}{9}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$

This is parallel to $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$Confirming $PP'$ is perpendicular to $\Pi$.

Verify midpoint: $\frac{1}{2}\!\left[\begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix} + \begin{pmatrix} \frac{41}{9} \\ \frac{11}{9} \\ \frac{23}{9} \end{pmatrix}\right] = \begin{pmatrix} \frac{25}{9} \\ \frac{19}{9} \\ \frac{7}{9} \end{pmatrix}$.

Check on $\Pi$: $2\!\left(\frac{25}{9}\right) - \frac{19}{9} + 2\!\left(\frac{7}{9}\right) = \frac{50 - 19 + 14}{9} = \frac{45}{9} = 5$. Confirmed.