Vectors

Board	Paper	Notes
AQA	Paper 1, 2	2D vectors in P1; 3D vectors, scalar product in P2
Edexcel	P1, P2	Similar split
OCR (A)	Paper 1, 2	Includes vector equations of lines
CIE (9709)	P1, P2, P3	2D in P1; 3D and lines in P2/P3

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Definition. A vector is a quantity with both magnitude and direction. A scalar is a Quantity with magnitude only.

Vectors in 2D are written as column vectors: $\dbinom{a}{b}$ or $a\mathbf{i} + b\mathbf{j}$.

Vectors in 3D: $\begin{pmatrix}a\\b\\c\end{pmatrix}$ or $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$.

The unit vectors $\mathbf{i} = \dbinom{1}{0}$, $\mathbf{j} = \dbinom{0}{1}$ point along the positive $x$- and $y$-axes respectively. In 3D, $\mathbf{k} = \begin{pmatrix}0\\0\\1\end{pmatrix}$.

The position vector of a point $P$ relative to an origin $O$ is the vector $\overrightarrow{OP}$Written as $\mathbf{r}_P$ or $\mathbf{p}$.

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The magnitude (length) of $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ is

$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

This follows directly from Pythagoras” theorem applied in 3D.

A unit vector has magnitude 1. The unit vector in the direction of $\mathbf{a}$ is

$\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$

The direction cosines of $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ are

$\cos\alpha = \frac{a_1}{|\mathbf{a}|}, \quad \cos\beta = \frac{a_2}{|\mathbf{a}|}, \quad \cos\gamma = \frac{a_3}{|\mathbf{a}|}$

Where $\alpha$, $\beta$, $\gamma$ are the angles between $\mathbf{a}$ and the $x$-, $y$-, $z$-axes Respectively.

Note: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

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To go from $A$ to $C$ via $B$: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.

Theorem. If two vectors $\mathbf{a}$ and $\mathbf{b}$ are represented as adjacent sides of a Parallelogram, then the diagonal $\mathbf{a} + \mathbf{b}$ represents their sum.

Proof. Consider parallelogram $OACB$ where $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Since $OACB$ is a parallelogram, $\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}$.

By the triangle law: $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \mathbf{a} + \mathbf{b}$.

Similarly, $\overrightarrow{BC} = \overrightarrow{OA} = \mathbf{a}$So $\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \mathbf{b} + \mathbf{a}$.

This proves $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$ (vector addition is commutative) and That the diagonal of the parallelogram represents the sum. $\blacksquare$

Vector addition extends to three dimensions. Given $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$:

$\mathbf{a} + \mathbf{b} = \begin{pmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\end{pmatrix}$

The same triangle and parallelogram laws apply. The key properties are:

• Commutativity: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$
• Associativity: $(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})$
• Zero vector: $\mathbf{a} + \mathbf{0} = \mathbf{a}$ where $\mathbf{0} = \begin{pmatrix}0\\0\\0\end{pmatrix}$
• Additive inverse: $\mathbf{a} + (-\mathbf{a}) = \mathbf{0}$

Scalar multiplication also satisfies the distributive laws: $\lambda(\mathbf{a} + \mathbf{b}) = \lambda\mathbf{a} + \lambda\mathbf{b}$ and $(\lambda + \mu)\mathbf{a} = \lambda\mathbf{a} + \mu\mathbf{a}$ for scalars $\lambda, \mu$.

Example. Given points A(1, 2, -1)$$B(4, 0, 3)$$C(2, 5, 1)Find $\overrightarrow{AB} + \overrightarrow{BC}$.

$\overrightarrow{AB} = \begin{pmatrix}3\\-2\\4\end{pmatrix}$ $\overrightarrow{BC} = \begin{pmatrix}-2\\5\\-2\end{pmatrix}$.

$\overrightarrow{AB} + \overrightarrow{BC} = \begin{pmatrix}1\\3\\2\end{pmatrix} = \overrightarrow{AC}$Confirming the triangle law in 3D.

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Definition. The scalar (dot) product of $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ And $\mathbf{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$ is

$\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$

Theorem. $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$Where $\theta$ is the Angle between $\mathbf{a}$ and $\mathbf{b}$.

Proof using the cosine rule. Consider the triangle formed by vectors $\mathbf{a}$, $\mathbf{b}$ And $\mathbf{a} - \mathbf{b}$.

By the cosine rule: $|\mathbf{a}-\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$.

Now compute $|\mathbf{a}-\mathbf{b}|^2$ algebraically:

$$\begin{aligned} |\mathbf{a}-\mathbf{b}|^2 &= (a_1-b_1)^2 + (a_2-b_2)^2 + (a_3-b_3)^2 \\ &= (a_1^2+a_2^2+a_3^2) + (b_1^2+b_2^2+b_3^2) - 2(a_1b_1+a_2b_2+a_3b_3) \\ &= |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,\mathbf{a}\cdot\mathbf{b} \end{aligned}$$

Comparing with the cosine rule:

$|\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta$

$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \quad \blacksquare$

$\mathbf{a} \perp \mathbf{b} \iff \mathbf{a}\cdot\mathbf{b} = 0$ (when neither vector is zero).

This follows since $\cos(\pi/2) = 0$.

Intuition. The dot product $\mathbf{a}\cdot\mathbf{b}$ measures the extent to which $\mathbf{a}$ And $\mathbf{b}$ point in the same direction. It equals the product of the magnitude of $\mathbf{a}$ And the projection of $\mathbf{b}$ onto $\mathbf{a}$: $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}| \cdot (\mathrm{shadow of }\mathbf{b}\mathrm{ on }\mathbf{a})$. If they are perpendicular, the shadow is zero. If they point the same way, the dot product is Positive; if opposite, negative.

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Definition. The vector equation of a line passing through point $A$ with position vector $\mathbf{a}$In the direction of vector $\mathbf{b}$Is

$\mathbf{r} = \mathbf{a} + t\mathbf{b}, \quad t \in \mathbb{R}$

Where $\mathbf{r}$ is the position vector of a general point on the line, and $t$ is a parameter.

If $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$The parametric equations are

$x = a_1 + tb_1, \quad y = a_2 + tb_2, \quad z = a_3 + tb_3$

In 2D, eliminating $t$: $\dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2}$.

### 5.4 Vector equation of a line in 3D

The vector equation of a line in 3D has the same form as in 2D, but now operates in three Dimensions. Given a point $A(x_0, y_0, z_0)$ on the line and a direction vector $\mathbf{d} = \begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}$:

$\mathbf{r} = \begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix} + t\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}, \quad t \in \mathbb{R}$

The parametric form is:

$x = x_0 + td_1, \quad y = y_0 + td_2, \quad z = z_0 + td_3$

**Example.** Find the vector equation of the line through $P(2, -1, 3)$ and $Q(5, 1, -2)$.

Direction: $\overrightarrow{PQ} = \begin{pmatrix}3\\2\\-5\end{pmatrix}$.

$\mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + t\begin{pmatrix}3\\2\\-5\end{pmatrix}$

To check: at $t = 0$ we get $P$; at $t = 1$ we get $\begin{pmatrix}5\\1\\-2\end{pmatrix} = Q$. ✓

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Given $\mathbf{r}_1 = \mathbf{a}_1 + t\mathbf{b}_1$ and $\mathbf{r}_2 = \mathbf{a}_2 + s\mathbf{b}_2$:

Method:

1. Equate the $x$-, $y$-, and $z$-components.
2. Solve two equations for $t$ and $s$.
3. Check the third equation is consistent.

• If consistent: the lines intersect at the point found.
• If inconsistent: the lines are skew (non-parallel and non-intersecting).
• If $\mathbf{b}_1 = k\mathbf{b}_2$ for some scalar $k$: the lines are parallel (coincident if also $\mathbf{a}_2 - \mathbf{a}_1$ is parallel to $\mathbf{b}_1$).

Definition. Two lines in 3D are skew if they are not parallel and do not intersect.

To verify skewness, show that the system of equations for $t$ and $s$ is inconsistent.

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From the dot product formula:

$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

The angle between two lines is found using the direction vectors.

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To find the shortest distance from point $P$ to line $\mathbf{r} = \mathbf{a} + t\mathbf{b}$:

1. Let $Q$ be the closest point on the line to $P$.
2. $\overrightarrow{PQ}$ is perpendicular to $\mathbf{b}$So $\overrightarrow{PQ}\cdot\mathbf{b} = 0$.
3. If $P$ has position vector $\mathbf{p}$Then $\overrightarrow{PQ} = \mathbf{a} + t\mathbf{b} - \mathbf{p}$.
4. $(\mathbf{a} + t\mathbf{b} - \mathbf{p})\cdot\mathbf{b} = 0$ gives $t$.
5. Substitute back to find $Q$ and compute $|\overrightarrow{PQ}|$.

The above procedure yields the general formula. For a line through $A$ with direction $\mathbf{d}$ And a point $P$ with position vector $\mathbf{p}$:

$d = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{d}|}{|\mathbf{d}|}$

This uses the cross product (vector product), which gives a vector perpendicular to both $\overrightarrow{AP}$and $\mathbf{d}$ whose magnitude equals the area of the parallelogram they Span. Dividing by $|\mathbf{d}|$ (the base) gives the perpendicular height, i.e. The shortest Distance.

**Example using the dot-product method.** Find the distance from $P(4, 1, -1)$ to the line $\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}$.

$\overrightarrow{PQ} = \begin{pmatrix}1+2t\\t\\2-t\end{pmatrix} - \begin{pmatrix}4\\1\\-1\end{pmatrix} = \begin{pmatrix}2t-3\\t-1\\3-t\end{pmatrix}$.

Set $\overrightarrow{PQ}\cdot\begin{pmatrix}2\\1\\-1\end{pmatrix} = 0$:

$2(2t-3) + (t-1) - (3-t) = 0 \implies 4t - 6 + t - 1 - 3 + t = 0 \implies 6t = 10 \implies t = \dfrac{5}{3}$.

$Q = \begin{pmatrix}1+10/3\\5/3\\2-5/3\end{pmatrix} = \begin{pmatrix}13/3\\5/3\\1/3\end{pmatrix}$.

$d = \left|\begin{pmatrix}1/3\\2/3\\-2/3\end{pmatrix}\right| = \sqrt{\dfrac{1}{9} + \dfrac{4}{9} + \dfrac{4}{9}} = \sqrt{\dfrac{9}{9}} = 1$.

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Definition. The scalar triple product of three vectors $\mathbf{a}$, $\mathbf{b}$ $\mathbf{c}$ is

$[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$

In component form, this equals the determinant:

$[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

Theorem. The absolute value of the scalar triple product equals the volume of the parallelepiped With edges defined by $\mathbf{a}$, $\mathbf{b}$And $\mathbf{c}$.

$V = |\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$

Proof. The vector $\mathbf{b}\times\mathbf{c}$ has magnitude $|\mathbf{b}||\mathbf{c}|\sin\theta$ equal to the area of the parallelogram with sides $\mathbf{b}$ And $\mathbf{c}$And direction perpendicular to both. The height of the parallelepiped is the Projection of $\mathbf{a}$ onto $\mathbf{b}\times\mathbf{c}$Which is $|\mathbf{a}|\cos\phi$ where $\phi$ is the angle between $\mathbf{a}$ and $\mathbf{b}\times\mathbf{c}$.

$V = \mathrm{base area} \times \mathrm{height} = |\mathbf{b}\times\mathbf{c}| \cdot |\mathbf{a}|\cos\phi = |\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})| \quad \blacksquare$

• Cyclic permutation: $[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = [\mathbf{b},\, \mathbf{c},\, \mathbf{a}] = [\mathbf{c},\, \mathbf{a},\, \mathbf{b}]$
• Anti-symmetry: Swapping any two vectors changes the sign: $[\mathbf{a},\, \mathbf{c},\, \mathbf{b}] = -[\mathbf{a},\, \mathbf{b},\, \mathbf{c}]$
• Coplanarity test: \mathbf{a}$$\mathbf{b}$$\mathbf{c} are coplanar if and only if $[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = 0$ (the parallelepiped has zero volume).
• Volume of a tetrahedron: $V_{\mathrm{tet}} = \dfrac{1}{6}|\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})|$Since a tetrahedron is $\dfrac{1}{6}$ of a parallelepiped.

Example. Find the volume of the parallelepiped with edges $\mathbf{a} = \begin{pmatrix}2\\0\\1\end{pmatrix}$ $\mathbf{b} = \begin{pmatrix}1\\3\\-1\end{pmatrix}$ $\mathbf{c} = \begin{pmatrix}0\\2\\4\end{pmatrix}$.

$[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = \begin{vmatrix} 2 & 0 & 1 \\ 1 & 3 & -1 \\ 0 & 2 & 4 \end{vmatrix}$

$= 2\begin{vmatrix}3 & -1\\2 & 4\end{vmatrix} - 0\begin{vmatrix}1 & -1\\0 & 4\end{vmatrix} + 1\begin{vmatrix}1 & 3\\0 & 2\end{vmatrix}$

$= 2(12+2) + 0 + 1(2) = 28 + 2 = 30$.

Volume $= |30| = 30$ cubic units.

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Points A$$B$$C are collinear if and only if $\overrightarrow{AB}$is parallel to $\overrightarrow{BC}$I.e. $\overrightarrow{AB} = k\,\overrightarrow{BC}$for some scalar $k$.

Equivalently, $\overrightarrow{AB} \times \overrightarrow{BC} = \mathbf{0}$ (zero vector).

Method:

1. Compute $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ and $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$.
2. Check if one is a scalar multiple of the other.
3. Alternatively, check if $\overrightarrow{AC}$is parallel to $\overrightarrow{AB}$.

Example. Show that A(1, 2, 3)$$B(3, 4, 5)$$C(5, 6, 7) are collinear.

$\overrightarrow{AB} = \begin{pmatrix}2\\2\\2\end{pmatrix}$ $\overrightarrow{BC} = \begin{pmatrix}2\\2\\2\end{pmatrix}$.

Since $\overrightarrow{AB} = \overrightarrow{BC}$(i.e. $k = 1$), the points are collinear. $\blacksquare$

Two lines are perpendicular if and only if their direction vectors have dot product zero.

Method:

1. Identify the direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$ of the two lines.
2. Compute $\mathbf{d}_1\cdot\mathbf{d}_2$.
3. If the result is zero (and neither direction vector is zero), the lines are perpendicular.

Example. Show that the lines $\mathbf{r}_1 = \begin{pmatrix}0\\1\\2\end{pmatrix} + t\begin{pmatrix}1\\2\\-2\end{pmatrix}$ and $\mathbf{r}_2 = \begin{pmatrix}3\\0\\-1\end{pmatrix} + s\begin{pmatrix}2\\-1\\0\end{pmatrix}$ are Perpendicular.

$\mathbf{d}_1\cdot\mathbf{d}_2 = (1)(2) + (2)(-1) + (-2)(0) = 2 - 2 + 0 = 0$.

Since the dot product is zero, the lines are perpendicular. $\blacksquare$

Points A$$B$$C$$D form a parallelogram (in order) if and only if $\overrightarrow{AB} = \overrightarrow{DC}$(or equivalently $\overrightarrow{AD} = \overrightarrow{BC}$).

Method:

1. Compute the relevant displacement vectors.
2. Show opposite sides are equal as vectors (same components).

### 10.4 Using vectors in geometric proofs

Many geometry problems can be solved elegantly using vectors. The general strategy is:

1. Assign position vectors to key points.
2. Express the relevant geometric conditions in vector form (parallelism via scalar multiples, perpendicularity via dot products, midpoints via averages).
3. Compute and simplify algebraically.

Example. In triangle $ABC$Let $M$ be the midpoint of $AB$. Prove that $\overrightarrow{CM} = \dfrac{1}{2}\overrightarrow{CA} + \dfrac{1}{2}\overrightarrow{CB}$.

$\mathbf{m} = \dfrac{\mathbf{a} + \mathbf{b}}{2}$ (midpoint formula).

$\overrightarrow{CM} = \mathbf{m} - \mathbf{c} = \dfrac{\mathbf{a} + \mathbf{b}}{2} - \mathbf{c} = \dfrac{\mathbf{a} - \mathbf{c}}{2} + \dfrac{\mathbf{b} - \mathbf{c}}{2} = \dfrac{1}{2}\overrightarrow{CA} + \dfrac{1}{2}\overrightarrow{CB}$. $\blacksquare$

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Problem 1

Given $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$Find $\mathbf{a} + \mathbf{b}$$\mathbf{a} - \mathbf{b}$$|\mathbf{a}|$And a unit vector in the direction of $\mathbf{a}$.

Solution 1

$\mathbf{a} + \mathbf{b} = 4\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} = \begin{pmatrix}4\\2\\-2\end{pmatrix}$.

$\mathbf{a} - \mathbf{b} = 2\mathbf{i} - 6\mathbf{j} + 4\mathbf{k} = \begin{pmatrix}2\\-6\\4\end{pmatrix}$.

$|\mathbf{a}| = \sqrt{9+4+1} = \sqrt{14}$.

$\hat{\mathbf{a}} = \dfrac{1}{\sqrt{14}}\begin{pmatrix}3\\-2\\1\end{pmatrix}$.

If you get this wrong, revise: Magnitude, Unit Vectors — Section 2.

Problem 2

Find the angle between $\mathbf{a} = \begin{pmatrix}2\\1\\-1\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\-3\\2\end{pmatrix}$.

Solution 2

$\mathbf{a}\cdot\mathbf{b} = 2-3-2 = -3$. $|\mathbf{a}| = \sqrt{4+1+1} = \sqrt{6}$$|\mathbf{b}| = \sqrt{1+9+4} = \sqrt{14}$.

$\cos\theta = \dfrac{-3}{\sqrt{6}\sqrt{14}} = \dfrac{-3}{\sqrt{84}} = \dfrac{-3}{2\sqrt{21}} = \dfrac{-\sqrt{21}}{14}$.

$\theta = \arccos\!\left(\dfrac{-\sqrt{21}}{14}\right) \approx 109.1^\circ$.

If you get this wrong, revise: The Scalar (Dot) Product — Section 4.

Problem 3

Find the vector equation of the line through $A(1, 2, -1)$ and $B(3, 0, 4)$.

Solution 3

Direction: $\overrightarrow{AB} = \begin{pmatrix}3-1\\0-2\\4-(-1)\end{pmatrix} = \begin{pmatrix}2\\-2\\5\end{pmatrix}$.

$\mathbf{r} = \begin{pmatrix}1\\2\\-1\end{pmatrix} + t\begin{pmatrix}2\\-2\\5\end{pmatrix}$

If you get this wrong, revise: Vector Equation of a Line — Section 5.

Problem 4

Show that the lines $\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix}3\\1\\1\end{pmatrix} + s\begin{pmatrix}1\\-1\\1\end{pmatrix}$ intersect, and find the point of intersection.

Solution 4

Equating components: $1+2t = 3+s$$t = 1-s$$2-t = 1+s$.

From $t = 1-s$ and $2-t = 1+s$: $2-(1-s) = 1+s \implies 1+s = 1+s$ ✓ (consistent).

Check first: 1+2(1-s) = 3+s \implies 3-2s = 3+s \implies s = 0$$t = 1.

Point: $\begin{pmatrix}1+2\\0+1\\2-1\end{pmatrix} = \begin{pmatrix}3\\1\\1\end{pmatrix}$.

If you get this wrong, revise: Intersection of Lines — Section 6.

Problem 5

Find $\lambda$ such that $\begin{pmatrix}\lambda\\3\\-1\end{pmatrix}$ is perpendicular to $\begin{pmatrix}2\\\lambda\\4\end{pmatrix}$.

Solution 5

Perpendicular $\iff$ dot product $= 0$:

$2\lambda + 3\lambda - 4 = 0 \implies 5\lambda = 4 \implies \lambda = \frac{4}{5}$

If you get this wrong, revise: Perpendicularity test — Section 4.3.

Problem 6

Find the distance from $P(1, 2, 3)$ to the line $\mathbf{r} = \begin{pmatrix}0\\1\\-1\end{pmatrix} + t\begin{pmatrix}1\\1\\1\end{pmatrix}$.

Solution 6

$\overrightarrow{PQ} = \begin{pmatrix}t\\1+t\\-1+t\end{pmatrix} - \begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}t-1\\t-1\\t-4\end{pmatrix}$.

$\overrightarrow{PQ} \cdot \begin{pmatrix}1\\1\\1\end{pmatrix} = 0$: $(t-1)+(t-1)+(t-4) = 0 \implies 3t = 6 \implies t = 2$.

$Q = \begin{pmatrix}2\\3\\1\end{pmatrix}$ $\overrightarrow{PQ} = \begin{pmatrix}1\\1\\-2\end{pmatrix}$ $|\overrightarrow{PQ}| = \sqrt{1+1+4} = \sqrt{6}$.

If you get this wrong, revise: Distance from a Point to a Line — Section 8.

Problem 7

Prove that the direction cosines satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

Solution 7

For $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ with $|\mathbf{a}| = m$:

\cos\alpha = a_1/m$$\cos\beta = a_2/m$$\cos\gamma = a_3/m.

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \frac{a_1^2+a_2^2+a_3^2}{m^2} = \frac{m^2}{m^2} = 1 \quad \blacksquare$

If you get this wrong, revise: Direction Cosines — Section 2.3.

Problem 8

Points $A$$B$$C$ have position vectors $\mathbf{a} = \begin{pmatrix}1\\-1\\2\end{pmatrix}$$\mathbf{b} = \begin{pmatrix}3\\1\\0\end{pmatrix}$$\mathbf{c} = \begin{pmatrix}4\\0\\3\end{pmatrix}$. Determine whether $\triangle ABC$ is right-angled.

Solution 8

$\overrightarrow{AB} = \begin{pmatrix}2\\2\\-2\end{pmatrix}$$\overrightarrow{AC} = \begin{pmatrix}3\\1\\1\end{pmatrix}$$\overrightarrow{BC} = \begin{pmatrix}1\\-1\\3\end{pmatrix}$.

$\overrightarrow{AB}\cdot\overrightarrow{AC} = 6+2-2 = 6 \neq 0$. $\overrightarrow{AB}\cdot\overrightarrow{BC} = 2-2-6 = -6 \neq 0$. $\overrightarrow{AC}\cdot\overrightarrow{BC} = 3-1+3 = 5 \neq 0$.

No pair is perpendicular, so $\triangle ABC$ is not right-angled.

If you get this wrong, revise: Perpendicularity test — Section 4.3.

Problem 9

Show that the lines $\mathbf{r} = \begin{pmatrix}0\\0\\1\end{pmatrix} + t\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + s\begin{pmatrix}1\\0\\1\end{pmatrix}$ intersect, and find the point of intersection.

Solution 9

Equating: $t = s$$t = 1$$1 = s$.

From $t = 1$ and $t = s$: $s = 1$. Check third: $1 = s = 1$ ✓.

Wait — all three are consistent! Let me re-check. Line 1: $(t, t, 1)$. Line 2: $(s, 1, s)$.

t = s$$t = 1$$1 = s. So $t = s = 1$. Point: $(1, 1, 1)$.

Actually the lines intersect at $(1,1,1)$They are not skew.

If you get this wrong, revise: Skew Lines — Section 6.2.

Problem 10

Given $\mathbf{a} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 3\mathbf{j}$Find the vector projection of $\mathbf{b}$ onto $\mathbf{a}$.

Solution 10

The projection of $\mathbf{b}$ onto $\mathbf{a}$ is $\mathrm{proj}_{\mathbf{a}}\mathbf{b} = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|^2}\,\mathbf{a}$.

$\mathbf{a}\cdot\mathbf{b} = 2-3 = -1$. $|\mathbf{a}|^2 = 4+1 = 5$.

$\mathrm{proj}_{\mathbf{a}}\mathbf{b} = \frac{-1}{5}(2\mathbf{i}+\mathbf{j}) = -\frac{2}{5}\mathbf{i} - \frac{1}{5}\mathbf{j}$

If you get this wrong, revise: Geometric Interpretation — Section 4.2.

Problem 11

Find the angle between the line $\mathbf{r} = \begin{pmatrix}1\\2\\-1\end{pmatrix} + t\begin{pmatrix}3\\-1\\2\end{pmatrix}$ and the plane $2x - y + z = 5$.

Solution 11

The normal to the plane is $\mathbf{n} = \begin{pmatrix}2\\-1\\1\end{pmatrix}$And the direction of the line is $\mathbf{d} = \begin{pmatrix}3\\-1\\2\end{pmatrix}$.

The angle between the line and the plane equals $90^\circ$ minus the angle between $\mathbf{d}$ and $\mathbf{n}$.

$\cos\phi = \dfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|} = \dfrac{6+1+2}{\sqrt{14}\sqrt{6}} = \dfrac{9}{2\sqrt{21}}$.

Angle between line and normal: $\phi = \arccos\!\left(\dfrac{9}{2\sqrt{21}}\right)$.

Angle between line and plane: $90° - \phi$.

If you get this wrong, revise: Angle Between Two Vectors — Section 7.

Problem 12

Find the direction cosines of the vector $\mathbf{v} = \begin{pmatrix}1\\-2\\2\end{pmatrix}$And verify that they satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

Solution 12

$|\mathbf{v}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

\cos\alpha = \dfrac{1}{3}$$\quad \cos\beta = \dfrac{-2}{3}$$\quad \cos\gamma = \dfrac{2}{3}.

Check: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = \dfrac{1}{9} + \dfrac{4}{9} + \dfrac{4}{9} = \dfrac{9}{9} = 1$. ✓

The angles are $\alpha = \arccos(1/3) \approx 70.5^\circ$ \beta = \arccos(-2/3) \approx 131.8^\circ$$\gamma = \arccos(2/3) \approx 48.2^\circ.

If you get this wrong, revise: Direction Cosines — Section 2.3.

Problem 13

Find the vector equation of the line through $A(2, -3, 1)$ that is parallel to the line $\mathbf{r} = \begin{pmatrix}0\\1\\-2\end{pmatrix} + t\begin{pmatrix}4\\-1\\3\end{pmatrix}$.

Solution 13

Since the line is parallel, it has the same direction vector $\begin{pmatrix}4\\-1\\3\end{pmatrix}$.

Using point $A(2, -3, 1)$:

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

If you get this wrong, revise: Vector Equation of a Line in 3D — Section 5.4.

Problem 14

Find the volume of the parallelepiped with edges $\mathbf{a} = \begin{pmatrix}1\\0\\2\end{pmatrix}$$\mathbf{b} = \begin{pmatrix}3\\1\\-1\end{pmatrix}$$\mathbf{c} = \begin{pmatrix}-1\\2\\1\end{pmatrix}$.

Solution 14

$[\mathbf{a},\, \mathbf{b},\, \mathbf{c}] = \begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & -1 \\ -1 & 2 & 1 \end{vmatrix}$

$= 1\begin{vmatrix}1 & -1\\2 & 1\end{vmatrix} - 0\begin{vmatrix}3 & -1\\-1 & 1\end{vmatrix} + 2\begin{vmatrix}3 & 1\\-1 & 2\end{vmatrix}$

$= 1(1+2) + 0 + 2(6+1) = 3 + 14 = 17$.

Volume $= |17| = 17$ cubic units.

Volume of the tetrahedron $= \dfrac{17}{6}$ cubic units.

If you get this wrong, revise: Scalar Triple Product — Section 9.

Problem 15

Determine whether the points $P(1, 2, 3)$$Q(4, 5, 6)$$R(7, 8, 9)$ are collinear. If they are, find the ratio $PQ : QR$.

Solution 15

$\overrightarrow{PQ} = \begin{pmatrix}3\\3\\3\end{pmatrix}$ $\overrightarrow{QR} = \begin{pmatrix}3\\3\\3\end{pmatrix}$.

Since $\overrightarrow{PQ} = \overrightarrow{QR}$The points are collinear. The ratio is $PQ : QR = 1 : 1$.

If you get this wrong, revise: Proving Collinear Points — Section 10.1.

Problem 16

Show that the lines $\mathbf{r}_1 = \begin{pmatrix}1\\0\\0\end{pmatrix} + t\begin{pmatrix}1\\2\\3\end{pmatrix}$ and $\mathbf{r}_2 = \begin{pmatrix}0\\1\\-1\end{pmatrix} + s\begin{pmatrix}2\\-1\\1\end{pmatrix}$ are skew.

Solution 16

Equating components: $1+t = 2s$$2t = 1-s$$3t = -1+s$.

From equation 2: $s = 1 - 2t$. Substitute into equation 1: $1+t = 2(1-2t) = 2-4t \implies 5t = 1 \implies t = 1/5$.

Then $s = 1 - 2/5 = 3/5$.

Check equation 3: $3(1/5) = 3/5$ and $-1 + 3/5 = -2/5$.

$3/5 \neq -2/5$So the third equation is inconsistent. The lines are skew. $\blacksquare$

If you get this wrong, revise: Skew Lines — Section 6.2.

Problem 17

Find the shortest distance from the point $P(3, -1, 2)$ to the line $\mathbf{r} = \begin{pmatrix}1\\2\\-1\end{pmatrix} + t\begin{pmatrix}1\\0\\3\end{pmatrix}$.

Solution 17

Let $\mathbf{a} = \begin{pmatrix}1\\2\\-1\end{pmatrix}$$\mathbf{d} = \begin{pmatrix}1\\0\\3\end{pmatrix}$ $\mathbf{p} = \begin{pmatrix}3\\-1\\2\end{pmatrix}$.

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

$\overrightarrow{PQ} = \mathbf{a} + t\mathbf{d} - \mathbf{p} = \begin{pmatrix}t-2\\t+2\\-1+3t-2\end{pmatrix} = \begin{pmatrix}t-2\\t+2\\3t-3\end{pmatrix}$.

Set $\overrightarrow{PQ}\cdot\mathbf{d} = 0$: $(t-2)(1) + (t+2)(0) + (3t-3)(3) = 0$

$t - 2 + 9t - 9 = 0 \implies 10t = 11 \implies t = 11/10$.

$Q = \begin{pmatrix}1+11/10\\2\\-1+33/10\end{pmatrix} = \begin{pmatrix}21/10\\2\\23/10\end{pmatrix}$.

$\overrightarrow{PQ} = \begin{pmatrix}21/10-3\\2-(-1)\\23/10-2\end{pmatrix} = \begin{pmatrix}-9/10\\3\\3/10\end{pmatrix}$.

$d = \sqrt{81/100 + 9 + 9/100} = \sqrt{81/100 + 900/100 + 9/100} = \sqrt{990/100} = \dfrac{3\sqrt{110}}{10}$.

If you get this wrong, revise: Distance from a Point to a Line — Section 8.

Problem 18

Points $A$$B$$C$$D$ have position vectors $\mathbf{a} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$\mathbf{b} = \begin{pmatrix}4\\1\\-2\end{pmatrix}$$\mathbf{c} = \begin{pmatrix}6\\3\\1\end{pmatrix}$$\mathbf{d} = \begin{pmatrix}2\\2\\3\end{pmatrix}$. Show that $ABCD$ is a parallelogram, and determine whether it is a rectangle.

Solution 18

$\overrightarrow{AB} = \begin{pmatrix}4\\1\\-2\end{pmatrix}$ $\overrightarrow{DC} = \mathbf{c} - \mathbf{d} = \begin{pmatrix}4\\1\\-2\end{pmatrix}$.

Since \overrightarrow{AB} = \overrightarrow{DC}$$ABCD is a parallelogram. ✓

Check for rectangle: $\overrightarrow{AB}\cdot\overrightarrow{AD} = \begin{pmatrix}4\\1\\-2\end{pmatrix}\cdot\begin{pmatrix}2\\2\\3\end{pmatrix} = 8 + 2 - 6 = 4 \neq 0$.

The adjacent sides are not perpendicular, so $ABCD$ is not a rectangle.

If you get this wrong, revise: Proving Points Form a Parallelogram — Section 10.3.

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## Common Pitfalls

1. Confusing position vectors with direction vectors. Position vectors point from the origin.

2. Forgetting that the scalar product gives a scalar, not a vector.

3. Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.

4. Losing marks by not showing sufficient working. Always write out each step, especially in proof questions.

5. Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.

6. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

• Coordinates and Geometry — The distance formula, perpendicular lines, and circle equations are expressed using vector notation.
• Proof — Vector methods provide elegant alternative proofs for geometric results such as collinearity and perpendicularity.
• Trigonometry — The angle between two vectors uses the cosine rule and direction cosines from trigonometry.
• Mechanics — Velocity and acceleration vectors in mechanics are particular applications of the vector concept.

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.