Vectors in 3D

Vectors in 3D

This chapter extends the study of vectors from A Level Mathematics into three dimensions, Introducing the vector (cross) product, equations of planes, and the scalar triple product. These Tools are essential for geometry, mechanics, and physics at university level.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	3D vectors, scalar product, vector product, planes, scalar triple product
Edexcel	FP1	3D vectors, scalar product, vector product, lines and planes
OCR (A)	Paper 1	3D vectors, scalar product, vector product, planes
CIE	P1	3D vectors, scalar product, vector product, lines, planes, intersections

:::info All boards cover 3D vectors, the scalar product, and the vector product. AQA includes the Scalar triple product for volumes. CIE places particular emphasis on intersections of lines and Planes. :::

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1. Review of A Level Vectors

A vector has magnitude and direction. In 2D, $\mathbf{a} = \begin{pmatrix}a_1\\a_2\end{pmatrix}$ has Magnitude $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}$.

Key results from A Level:

• Scalar (dot) product: $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$
• Perpendicularity: $\mathbf{a} \perp \mathbf{b} \iff \mathbf{a}\cdot\mathbf{b} = 0$
• Vector equation of a line: $\mathbf{r} = \mathbf{a} + t\mathbf{d}$

This chapter extends all of these ideas into three dimensions and introduces new operations.

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2. Vectors in 3D

2.1 Notation

A vector in 3D is written as a column vector or in component form:

$\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$

Where \mathbf{i}$$\mathbf{j}$$\mathbf{k} are unit vectors along the $x$-, $y$-, $z$-axes.

2.2 Position vectors and displacement

Definition. The position vector of a point $P(x,y,z)$ relative to origin $O$ is

$\overrightarrow{OP} = \begin{pmatrix}x\\y\\z\end{pmatrix}$

The displacement from $A$ to $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.

2.3 Magnitude

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

2.4 Direction cosines

Definition. The direction cosines of $\mathbf{a}$ are

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

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

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

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3. Scalar (Dot) Product

3.1 Definition

Definition. The scalar (dot) product of $\mathbf{a}$ and $\mathbf{b}$ in 3D is

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

3.2 Geometric interpretation

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

Where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

3.3 Key properties

• $\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2$
• $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$ (commutative)
• $\mathbf{a}\cdot(\mathbf{b} + \mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}$ (distributive)
• $\mathbf{a}\cdot\mathbf{b} = 0 \iff \mathbf{a} \perp \mathbf{b}$ (when neither is zero)

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4. Vector (Cross) Product

4.1 Definition

Definition. The vector (cross) 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

$\boxed{\mathbf{a}\times\mathbf{b} = \begin{pmatrix}a_2b_3 - a_3b_2\\a_3b_1 - a_1b_3\\a_1b_2 - a_2b_1\end{pmatrix}}$

4.2 Determinant form

The cross product can be computed using a symbolic determinant:

$\mathbf{a}\times\mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}$

4.3 Geometric interpretation

Theorem. $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$Where $\theta$ is The angle between $\mathbf{a}$ and $\mathbf{b}$.

Proof that the cross product magnitude equals the area of the parallelogram

Consider the parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$.

$|\mathbf{a}\times\mathbf{b}|^2 = (a_2b_3 - a_3b_2)^2 + (a_3b_1 - a_1b_3)^2 + (a_1b_2 - a_2b_1)^2$

Expanding and collecting terms:

$|\mathbf{a}\times\mathbf{b}|^2 = (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1+a_2b_2+a_3b_3)^2$

$= |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a}\cdot\mathbf{b})^2$

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

$|\mathbf{a}\times\mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2(1 - \cos^2\theta) = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta$

$|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$

The area of the parallelogram is base $\times$ height $= |\mathbf{a}| \times |\mathbf{b}|\sin\theta$Which equals $|\mathbf{a}\times\mathbf{b}|$. $\square$

The cross product $\mathbf{a}\times\mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$And its direction is given by the right-hand rule.

4.4 Properties

• $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$ (anti-commutative)
• $\mathbf{a}\times\mathbf{a} = \mathbf{0}$
• $\mathbf{a}\times\mathbf{b} = \mathbf{0} \iff \mathbf{a}$ and $\mathbf{b}$ are parallel (or one is zero)
• $\mathbf{a}\times(\mathbf{b}+\mathbf{c}) = \mathbf{a}\times\mathbf{b} + \mathbf{a}\times\mathbf{c}$ (distributive)
• $\mathbf{i}\times\mathbf{j} = \mathbf{k}$, $\mathbf{j}\times\mathbf{k} = \mathbf{i}$ $\mathbf{k}\times\mathbf{i} = \mathbf{j}$

:::caution warning $\mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a}$. The cross product is Only defined in 3D. :::

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5. Equation of a Line in 3D

5.1 Vector form

Definition. The vector equation of a line through point $A$ (position vector $\mathbf{a}$) in Direction $\mathbf{d}$ is

$\boxed{\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}, \quad \lambda \in \mathbb{R}}$

5.2 Cartesian form

If $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{d} = \begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}$The parametric equations are:

$x = a_1 + \lambda d_1, \quad y = a_2 + \lambda d_2, \quad z = a_3 + \lambda d_3$

When all $d_i \neq 0$The Cartesian (symmetric) form is:

$\boxed{\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}}$

5.3 Intersection of two lines in 3D

Given $\mathbf{r}_1 = \mathbf{a}_1 + \lambda\mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{a}_2 + \mu\mathbf{d}_2$:

1. Equate components to get three equations in $\lambda$ and $\mu$.
2. Solve two equations for $\lambda$ and $\mu$.
3. Check the third equation:

• If consistent: the lines intersect.
• If inconsistent: the lines are skew (not parallel, not intersecting).

4. If $\mathbf{d}_1 \times \mathbf{d}_2 = \mathbf{0}$: the lines are parallel.

Example. Find the intersection of $\mathbf{r}_1 = \begin{pmatrix}1\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}$ And $\mathbf{r}_2 = \begin{pmatrix}3\\1\\4\end{pmatrix} + \mu\begin{pmatrix}2\\1\\-1\end{pmatrix}$.

Equating: 1+\lambda = 3+2\mu$$2-\lambda = 1+\mu$$2\lambda = 4-\mu.

From equation 2: $\lambda = 1 - \mu$. From equation 1: $1+(1-\mu) = 3+2\mu \implies 2-\mu = 3+2\mu \implies -3\mu = 1 \implies \mu = -1/3$ $\lambda = 4/3$.

Check equation 3: $2(4/3) = 4-(-1/3) \implies 8/3 = 13/3$. Not consistent — the lines are skew.

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6. Equation of a Plane

6.1 Vector form

Definition. The equation of a plane with normal vector $\mathbf{n}$ passing through point $A$ (position vector $\mathbf{a}$) is

$\boxed{\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}}$

This works because every point $P$ on the plane satisfies $\overrightarrow{AP} \perp \mathbf{n}$ I.e., $(\mathbf{r} - \mathbf{a})\cdot\mathbf{n} = 0$.

6.2 Cartesian form

If $\mathbf{n} = \begin{pmatrix}a\\b\\c\end{pmatrix}$ and $\mathbf{a}\cdot\mathbf{n} = d$:

$\boxed{ax + by + cz = d}$

6.3 Finding the normal to a plane

Given three points $A$, $B$, $C$ on the plane, the normal is

$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$

Example. Find the equation of the plane through A(1,0,2)$$B(3,1,0)$$C(0,2,1).

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

$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}(1)(-1)-(-2)(2)\\(-2)(-1)-(2)(-1)\\(2)(2)-(1)(-1)\end{pmatrix} = \begin{pmatrix}3\\4\\5\end{pmatrix}$.

$\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} = 1(3)+0(4)+2(5) = 13$.

$\boxed{3x + 4y + 5z = 13}$

6.4 Angle between two planes

The angle between two planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$ is

$\boxed{\cos\theta = \frac◆LB◆|\mathbf{n}_1\cdot\mathbf{n}_2|◆RB◆◆LB◆|\mathbf{n}_1||\mathbf{n}_2|◆RB◆}$

The acute angle is found by taking the absolute value.

6.5 Angle between a line and a plane

The angle $\phi$ between a line with direction $\mathbf{d}$ and a plane with normal $\mathbf{n}$ Satisfies:

$\sin\phi = \frac◆LB◆|\mathbf{d}\cdot\mathbf{n}|◆RB◆◆LB◆|\mathbf{d}||\mathbf{n}|◆RB◆$

Equivalently, if $\alpha$ is the angle between $\mathbf{d}$ and $\mathbf{n}$Then $\phi = 90° - \alpha$.

6.6 Line of intersection of two planes

To find the line of intersection of $a_1 x + b_1 y + c_1 z = d_1$ and $a_2 x + b_2 y + c_2 z = d_2$:

1. The direction is $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$.
2. Find a point satisfying both equations (set one variable to zero and solve).

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7. Distance from a Point to a Plane

7.1 Formula

Theorem. The perpendicular distance from point $P$ with position vector $\mathbf{p}$ to the Plane $\mathbf{r}\cdot\mathbf{n} = d$ is

$\boxed{D = \frac◆LB◆|\mathbf{p}\cdot\mathbf{n} - d|◆RB◆◆LB◆|\mathbf{n}|◆RB◆}$

In Cartesian form, for plane $ax + by + cz = d$ and point $(x_0, y_0, z_0)$:

$\boxed{D = \frac◆LB◆|ax_0 + by_0 + cz_0 - d|◆RB◆◆LB◆\sqrt{a^2+b^2+c^2}◆RB◆}$

Proof of the distance formula

Let the plane have equation $\mathbf{r}\cdot\hat{\mathbf{n}} = p$ where $\hat{\mathbf{n}}$ is a unit Normal and $p$ is the perpendicular distance from the origin to the plane.

For any point $P$ with position vector $\mathbf{p}$The distance from $P$ to the plane is the Magnitude of the projection of $\mathbf{p}$ onto $\hat{\mathbf{n}}$Minus $p$:

$D = |\mathbf{p}\cdot\hat{\mathbf{n}} - p|$

If the plane is given as $\mathbf{r}\cdot\mathbf{n} = d$ (where $\mathbf{n}$ is not necessarily a Unit vector), then $\hat{\mathbf{n}} = \mathbf{n}/|\mathbf{n}|$ and $p = d/|\mathbf{n}|$:

$D = \left|\mathbf{p}\cdot\frac◆LB◆\mathbf{n}◆RB◆◆LB◆|\mathbf{n}|◆RB◆ - \frac◆LB◆d◆RB◆◆LB◆|\mathbf{n}|◆RB◆\right| = \frac◆LB◆|\mathbf{p}\cdot\mathbf{n} - d|◆RB◆◆LB◆|\mathbf{n}|◆RB◆$

$\square$

Example. Find the distance from $P(1, 2, 3)$ to the plane $2x - y + 2z = 5$.

$D = \dfrac◆LB◆|2(1)-1(2)+2(3)-5|◆RB◆◆LB◆\sqrt{4+1+4}◆RB◆ = \dfrac◆LB◆|2-2+6-5|◆RB◆◆LB◆3◆RB◆ = \dfrac{1}{3}$.

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8. Scalar Triple Product

8.1 Definition

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

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

In component form:

$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix}a_1 & a_2 & a_3\\b_1 & b_2 & b_3\\c_1 & c_2 & c_3\end{vmatrix}$

8.2 Geometric interpretation

Proof that the scalar triple product equals the volume of the parallelepiped

The parallelepiped with edges \mathbf{a}$$\mathbf{b}$$\mathbf{c} has base area $|\mathbf{b}\times\mathbf{c}|$ (from Section 4.3).

The height is the component of $\mathbf{a}$ perpendicular to the base, which is the projection of $\mathbf{a}$ onto the direction of $\mathbf{b}\times\mathbf{c}$:

$\mathrm{height} = |\mathbf{a}|\cos\phi = \frac◆LB◆\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})◆RB◆◆LB◆|\mathbf{b}\times\mathbf{c}|◆RB◆$

Where $\phi$ is the angle between $\mathbf{a}$ and $\mathbf{b}\times\mathbf{c}$.

$V = \mathrm{base} \times \mathrm{height} = |\mathbf{b}\times\mathbf{c}| \cdot \frac◆LB◆\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})◆RB◆◆LB◆|\mathbf{b}\times\mathbf{c}|◆RB◆ = \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$

Taking the absolute value to get a positive volume:

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

$\square$

8.3 Properties

• $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \mathbf{b}\cdot(\mathbf{c}\times\mathbf{a}) = \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})$ (cyclic permutation)
• $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = -\mathbf{a}\cdot(\mathbf{c}\times\mathbf{b})$ (swapping two vectors changes sign)
• $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = 0$ if and only if \mathbf{a}$$\mathbf{b}$$\mathbf{c} are coplanar

:::info info C$$D are coplanar, then $\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = 0$.

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9. Distance Between Two Skew Lines

The shortest distance between two skew lines $\mathbf{r}_1 = \mathbf{a}_1 + \lambda\mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{a}_2 + \mu\mathbf{d}_2$ is

$\boxed{D = \frac◆LB◆|(\mathbf{a}_2 - \mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2)|◆RB◆◆LB◆|\mathbf{d}_1\times\mathbf{d}_2|◆RB◆}$

Intuition. The shortest distance is measured along the common perpendicular. The direction of The common perpendicular is $\mathbf{d}_1 \times \mathbf{d}_2$. The formula projects the vector Between any point on each line onto this perpendicular direction.

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10. Key Results Summary

| Quantity | Formula | | --------------------- | ------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------- | -------- | -------------------------------- | ----------- | | Dot product | $\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = | \mathbf{a} | | \mathbf{b} | \cos\theta$ | | Cross product | $\mathbf{a}\times\mathbf{b} = \begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$ | | Line | $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ | | Plane (vector) | $\mathbf{r}\cdot\mathbf{n} = d$ | | Plane (Cartesian) | $ax + by + cz = d$ | | Point-plane distance | $D = \dfrac◆LB◆ | \mathbf{p}\cdot\mathbf{n} - d | ◆RB◆◆LB◆ | \mathbf{n} | ◆RB◆$ | | Parallelepiped volume | $V = | \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) |$ | | Skew line distance | $D = \dfrac◆LB◆ | (\mathbf{a}\_2-\mathbf{a}\_1)\cdot(\mathbf{d}\_1\times\mathbf{d}\_2) | ◆RB◆◆LB◆ | \mathbf{d}\_1\times\mathbf{d}\_2 | ◆RB◆$ |

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Problems

Problem 1

Given $\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\4\\-2\end{pmatrix}$Find $\mathbf{a}\times\mathbf{b}$ and verify That it is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$.

Solution 1

$\mathbf{a}\times\mathbf{b} = \begin{pmatrix}(-1)(-2)-(3)(4)\\(3)(1)-(2)(-2)\\(2)(4)-(-1)(1)\end{pmatrix} = \begin{pmatrix}2-12\\3+4\\8+1\end{pmatrix} = \begin{pmatrix}-10\\7\\9\end{pmatrix}$

Verify: $\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b}) = 2(-10)+(-1)(7)+3(9) = -20-7+27 = 0$. ✓

$\mathbf{b}\cdot(\mathbf{a}\times\mathbf{b}) = 1(-10)+4(7)+(-2)(9) = -10+28-18 = 0$. ✓

If you get this wrong, revise: Vector Cross Product — Section 4.

Problem 2

Find the equation of the plane through A(1, 2, 0)$$B(0, 1, 3)$$C(2, -1, 1).

Solution 2

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

$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}(-1)(1)-(3)(-3)\\(3)(1)-(-1)(1)\\(-1)(-3)-(-1)(1)\end{pmatrix} = \begin{pmatrix}8\\4\\4\end{pmatrix}$

Simplify: $\mathbf{n} = \begin{pmatrix}2\\1\\1\end{pmatrix}$.

$\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n} = 2+2+0 = 4$.

$\boxed{2x + y + z = 4}$

If you get this wrong, revise: Equation of a Plane — Section 6.

Problem 3

Find the distance from the point $P(3, 1, -2)$ to the plane $x + 2y - 2z = 6$.

Solution 3

$D = \dfrac◆LB◆|3 + 2(1) - 2(-2) - 6|◆RB◆◆LB◆\sqrt{1+4+4}◆RB◆ = \dfrac◆LB◆|3+2+4-6|◆RB◆◆LB◆3◆RB◆ = \dfrac{3}{3} = 1$.

$\boxed{D = 1}$.

If you get this wrong, revise: Distance from Point to Plane — Section 7.

Problem 4

Find the shortest distance between the skew lines $\mathbf{r}_1 = \begin{pmatrix}0\\1\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\0\\2\end{pmatrix}$ And $\mathbf{r}_2 = \begin{pmatrix}1\\0\\2\end{pmatrix} + \mu\begin{pmatrix}0\\1\\-1\end{pmatrix}$.

Solution 4

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

$\mathbf{d}_1\times\mathbf{d}_2 = \begin{pmatrix}(0)(-1)-(2)(1)\\(2)(0)-(1)(-1)\\(1)(1)-(0)(0)\end{pmatrix} = \begin{pmatrix}-2\\1\\1\end{pmatrix}$

$|\mathbf{d}_1\times\mathbf{d}_2| = \sqrt{4+1+1} = \sqrt{6}$.

$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2) = 1(-2)+(-1)(1)+3(1) = -2-1+3 = 0$.

$D = \dfrac◆LB◆0◆RB◆◆LB◆\sqrt{6}◆RB◆ = 0$. The lines actually intersect (not skew).

If you get this wrong, revise: Distance Between Skew Lines — Section 9.

Problem 5

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

Solution 5

$\mathbf{b}\times\mathbf{c} = \begin{pmatrix}(1)(1)-(-1)(-1)\\(-1)(2)-(3)(1)\\(3)(-1)-(1)(2)\end{pmatrix} = \begin{pmatrix}0\\-5\\-5\end{pmatrix}$

$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = 1(0)+0(-5)+2(-5) = -10$.

$V = |-10| = \boxed{10}$.

If you get this wrong, revise: Scalar Triple Product — Section 8.

Problem 6

Find the angle between the planes $2x - y + z = 3$ and $x + y + 2z = 1$.

Solution 6

\mathbf{n}_1 = \begin{pmatrix}2\\-1\\1\end{pmatrix}$$|\mathbf{n}_1| = \sqrt{6}. \mathbf{n}_2 = \begin{pmatrix}1\\1\\2\end{pmatrix}$$|\mathbf{n}_2| = \sqrt{6}.

$\cos\theta = \dfrac◆LB◆|2-1+2|◆RB◆◆LB◆\sqrt{6}\cdot\sqrt{6}◆RB◆ = \dfrac{3}{6} = \dfrac{1}{2}$.

$\theta = \boxed{60°}$.

If you get this wrong, revise: Angle Between Two Planes — Section 6.4.

Problem 7

Show that the points A(1, 2, 3)$$B(3, 1, 2)$$C(2, 3, 1)$$D(0, 4, 4) are coplanar.

Solution 7

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

$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = \begin{vmatrix}2&-1&-1\\1&1&-2\\-1&2&1\end{vmatrix}$

$= 2(1\cdot 1-(-2)\cdot 2) - (-1)(1\cdot 1-(-2)(-1)) + (-1)(1\cdot 2-1\cdot(-1))$

$= 2(1+4) + 1(1-2) - 1(2+1) = 10 - 1 - 3 = 6 \neq 0$.

Wait — $6 \neq 0$So the points are not coplanar. Let me verify.

Actually, let me recompute $\overrightarrow{AC}\times\overrightarrow{AD}$:

$= \begin{pmatrix}(1)(1)-(-2)(2)\\(-2)(-1)-(1)(1)\\(1)(2)-(1)(-1)\end{pmatrix} = \begin{pmatrix}5\\1\\3\end{pmatrix}$

$\overrightarrow{AB}\cdot\begin{pmatrix}5\\1\\3\end{pmatrix} = 10-1-3 = 6 \neq 0$.

The points are not coplanar.

If you get this wrong, revise: Scalar Triple Product — Section 8.

Problem 8

Find the line of intersection of the planes $x + y + z = 6$ and $2x - y + z = 3$.

Solution 8

$\mathbf{n}_1 = \begin{pmatrix}1\\1\\1\end{pmatrix}$ $\mathbf{n}_2 = \begin{pmatrix}2\\-1\\1\end{pmatrix}$.

Direction: $\mathbf{d} = \mathbf{n}_1\times\mathbf{n}_2 = \begin{pmatrix}(1)(1)-(1)(-1)\\(1)(2)-(1)(1)\\(1)(-1)-(1)(2)\end{pmatrix} = \begin{pmatrix}2\\1\\-3\end{pmatrix}$.

Set $z = 0$: $x + y = 6$ and $2x - y = 3$. Adding: 3x = 9 \implies x = 3$$y = 3.

Point: $(3, 3, 0)$.

$\boxed{\mathbf{r} = \begin{pmatrix}3\\3\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-3\end{pmatrix}}$

If you get this wrong, revise: Line of Intersection — Section 6.6.

Problem 9

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

Solution 9

$\mathbf{d} = \begin{pmatrix}3\\1\\-1\end{pmatrix}$ $\mathbf{n} = \begin{pmatrix}2\\-1\\2\end{pmatrix}$.

$\sin\phi = \dfrac◆LB◆|\mathbf{d}\cdot\mathbf{n}|◆RB◆◆LB◆|\mathbf{d}||\mathbf{n}|◆RB◆ = \dfrac◆LB◆|6-1-2|◆RB◆◆LB◆\sqrt{11}\sqrt{9}◆RB◆ = \dfrac◆LB◆3◆RB◆◆LB◆3\sqrt{11}◆RB◆ = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{11}◆RB◆$.

$\phi = \arcsin\!\left(\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{11}◆RB◆\right) \approx \boxed{17.6°}$.

If you get this wrong, revise: Angle Between Line and Plane — Section 6.5.

Problem 10

Find the shortest distance between the skew lines $\mathbf{r}_1 = \begin{pmatrix}1\\0\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\3\end{pmatrix}$ And $\mathbf{r}_2 = \begin{pmatrix}0\\1\\0\end{pmatrix} + \mu\begin{pmatrix}2\\3\\4\end{pmatrix}$.

Solution 10

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

$\mathbf{d}_1\times\mathbf{d}_2 = \begin{pmatrix}(2)(4)-(3)(3)\\(3)(2)-(1)(4)\\(1)(3)-(2)(2)\end{pmatrix} = \begin{pmatrix}-1\\2\\-1\end{pmatrix}$

$|\mathbf{d}_1\times\mathbf{d}_2| = \sqrt{1+4+1} = \sqrt{6}$.

$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2) = (-1)(-1)+1(2)+0(-1) = 1+2 = 3$.

$D = \dfrac◆LB◆3◆RB◆◆LB◆\sqrt{6}◆RB◆ = \dfrac◆LB◆3\sqrt{6}◆RB◆◆LB◆6◆RB◆ = \boxed{\dfrac◆LB◆\sqrt{6}◆RB◆◆LB◆2◆RB◆}$.

If you get this wrong, revise: Distance Between Skew Lines — Section 9.

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11. Advanced Worked Examples

Example 11.1: Reflection in a plane

Problem. Find the reflection of the point $P(1, 2, 3)$ in the plane $x + y + z = 6$.

Solution. The reflected point $P'$ satisfies:

$P' = P - 2D\hat{\mathbf{n}}$

Where $D = \dfrac◆LB◆1+2+3-6◆RB◆◆LB◆\sqrt{3}◆RB◆ = 0$ and $\hat{\mathbf{n}} = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆(1, 1, 1)$.

Since $D = 0$The point $P$ lies on the plane, so its reflection is itself: $P' = (1, 2, 3)$.

Let me use a point not on the plane. The reflection of $Q(0, 0, 0)$:

$D = \frac◆LB◆0 + 0 + 0 - 6◆RB◆◆LB◆\sqrt{3}◆RB◆ = -2\sqrt{3}$

$Q' = (0, 0, 0) - 2(-2\sqrt{3})\frac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆(1, 1, 1) = (0, 0, 0) + 4(1, 1, 1) = (4, 4, 4)$

Check: the midpoint of $Q$ and $Q'$ is $(2, 2, 2)$Which satisfies $2+2+2 = 6$. Correct.

Example 11.2: Angle between a line and a plane

Problem. Find the acute angle between the line $\mathbf{r} = (1, -1, 2) + \lambda(3, 0, -1)$ and The plane $x - 2y + 2z = 5$.

Solution. \mathbf{d} = (3, 0, -1)$$\mathbf{n} = (1, -2, 2).

$\sin\phi = \frac◆LB◆|\mathbf{d}\cdot\mathbf{n}|◆RB◆◆LB◆|\mathbf{d}||\mathbf{n}|◆RB◆ = \frac◆LB◆|3 + 0 - 2|◆RB◆◆LB◆\sqrt{10}\sqrt{9}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆3\sqrt{10}◆RB◆$

$\phi = \arcsin\!\left(\frac◆LB◆1◆RB◆◆LB◆3\sqrt{10}◆RB◆\right) \approx 6.1^\circ$

Example 11.3: Volume of a tetrahedron using the scalar triple product

Problem. Find the volume of the tetrahedron with vertices O(0,0,0)$$A(1,0,0)$$B(0,2,0) $C(0,0,3)$.

Solution. \overrightarrow{OA} = (1,0,0)$$\overrightarrow{OB} = (0,2,0) $\overrightarrow{OC} = (0,0,3)$.

$V = |\overrightarrow{OA}\cdot(\overrightarrow{OB}\times\overrightarrow{OC})| = \left|\begin{vmatrix}1&0&0\\0&2&0\\0&0&3\end{vmatrix}\right| = |6| = 6$

This equals $\frac{1}{6} \times 1 \times 2 \times 3 = 1$Confirming the standard formula.

Example 11.4: Shortest distance using calculus

Problem. Find the shortest distance between the lines $\mathbf{r}_1 = (1, 0, 0) + \lambda(1, 1, 0)$ and $\mathbf{r}_2 = (0, 1, 0) + \mu(0, 1, 1)$.

Solution. \mathbf{d}_1 = (1, 1, 0)$$\mathbf{d}_2 = (0, 1, 1).

These are not parallel (not scalar multiples), so the lines are either intersecting or skew.

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

Check intersection: $1+\lambda = 0$ and $1+\lambda = 1+\mu$ and $\lambda = \mu$.

From the first: $\lambda = -1$. From the second: $0 = 1 + (-1) = 0$. From the third: $-1 = -1$. Consistent! The lines intersect, so the shortest distance is $0$.

Example 11.5: Finding the equation of a plane from three points

Problem. Find the equation of the plane through P(1, 1, 0)$$Q(2, 0, 1)$$R(0, 1, 1).

Solution. \overrightarrow{PQ} = (1, -1, 1)$$\overrightarrow{PR} = (-1, 0, 1).

$\mathbf{n} = \overrightarrow{PQ}\times\overrightarrow{PR} = \begin{pmatrix}(-1)(1) - (1)(0)\\(1)(-1) - (1)(1)\\(1)(0) - (-1)(-1)\end{pmatrix} = \begin{pmatrix}-1\\-2\\-1\end{pmatrix}$

$\mathbf{r}\cdot\mathbf{n} = (1)(-1) + (1)(-2) + (0)(-1) = -3$.

Equation: $-x - 2y - z = -3$I.e., $\boxed{x + 2y + z = 3}$.

Example 11.6: Verifying coplanarity

Problem. Determine whether the points A(1, 0, 1)$$B(2, 1, 3)$$C(3, 1, 4)$$D(0, -1, -1) Are coplanar.

Solution. \overrightarrow{AB} = (1, 1, 2)$$\overrightarrow{AC} = (2, 1, 3) $\overrightarrow{AD} = (-1, -1, -2)$.

$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}(1)(3)-(2)(1)\\(2)(2)-(1)(3)\\(1)(1)-(1)(2)\end{pmatrix} = \begin{pmatrix}1\\1\\-1\end{pmatrix}$

$(\overrightarrow{AB}\times\overrightarrow{AC})\cdot\overrightarrow{AD} = 1(-1) + 1(-1) + (-1)(-2) = -1 - 1 + 2 = 0$

Since the scalar triple product is zero, the four points are coplanar. $\blacksquare$

Example 11.7: Projection of a vector onto a plane

Problem. Find the projection of the vector $\mathbf{a} = (2, 1, -1)$ onto the plane $x + y + z = 1$.

Solution. The unit normal is $\hat{\mathbf{n}} = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆(1, 1, 1)$.

The projection of $\mathbf{a}$ onto the normal direction:

$\text{proj}_{\mathbf{n}}\,\mathbf{a} = (\mathbf{a}\cdot\hat{\mathbf{n}})\hat{\mathbf{n}} = \frac{2+1-1}{3}(1,1,1) = \frac{2}{3}(1,1,1)$

The projection onto the plane (i.e., the component parallel to the plane):

$\mathbf{a}_{\parallel} = \mathbf{a} - \text{proj}_{\mathbf{n}}\,\mathbf{a} = (2, 1, -1) - \frac{2}{3}(1, 1, 1) = \left(\frac{4}{3}, \frac{1}{3}, -\frac{5}{3}\right)$

Example 11.8: Vector product proof of the sine rule

Problem. Using the vector product, prove that $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$.

Solution. See Section 4.3 of this document. The proof uses the identity:

$|\mathbf{a}\times\mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a}\cdot\mathbf{b})^2$

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

$= |\mathbf{a}|^2|\mathbf{b}|^2(1 - \cos^2\theta) = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta$

Taking square roots gives the result.

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12. Connections to Other Topics

12.1 Vectors and matrices

The cross product $\mathbf{a}\times\mathbf{b}$ can be computed as a $3 \times 3$ determinant. See Matrices.

12.2 Vectors and mechanics

Vector methods are essential in mechanics for resolving forces, moments, and angular momentum. The Moment of a force $\mathbf{F}$ about point $O$ is $\mathbf{r}\times\mathbf{F}$. See Projectile Motion.

12.3 Planes and coordinate geometry

The equation of a plane and the distance formula connect vectors to 3D coordinate geometry. See Polar Coordinates.

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13. Additional Exam-Style Questions

Question 11

The lines $L_1$ and $L_2$ are given by:

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

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

Find the shortest distance between $L_1$ and $L_2$.

Solution

\mathbf{d}_1 = (2,-1,1)$$\mathbf{d}_2 = (1,-1,-1).

$\mathbf{d}_1\times\mathbf{d}_2 = \begin{pmatrix}(-1)(-1)-(1)(-1)\\(1)(1)-(2)(-1)\\(2)(-1)-(-1)(1)\end{pmatrix} = \begin{pmatrix}2\\3\\-3\end{pmatrix}$

$|\mathbf{d}_1\times\mathbf{d}_2| = \sqrt{4+9+9} = \sqrt{22}$.

$\mathbf{a}_2 - \mathbf{a}_1 = (2, -1, 2)$.

$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{d}_1\times\mathbf{d}_2) = 4 - 3 - 6 = -5$.

$D = \dfrac◆LB◆|-5|◆RB◆◆LB◆\sqrt{22}◆RB◆ = \dfrac◆LB◆5◆RB◆◆LB◆\sqrt{22}◆RB◆ = \dfrac◆LB◆5\sqrt{22}◆RB◆◆LB◆22◆RB◆$.

Question 12

Find the volume of the parallelepiped with edges \mathbf{a} = (2, 0, 1)$$\mathbf{b} = (1, 3, 0) $\mathbf{c} = (0, -1, 2)$.

Solution

$V = |\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})| = \left|\det\begin{pmatrix}2&0&1\\1&3&0\\0&-1&2\end{pmatrix}\right|$

$= |2(6-0) - 0 + 1(-1-0)| = |12 - 1| = 11$.

Question 13

Prove that the points equidistant from two fixed points lie on the perpendicular bisector plane Of the segment joining them.

Solution

Let the fixed points be $A$ and $B$ with position vectors $\mathbf{a}$ and $\mathbf{b}$. A point $P$ Is equidistant from $A$ and $B$ when:

$|\mathbf{p} - \mathbf{a}| = |\mathbf{p} - \mathbf{b}|$

Squaring: $(\mathbf{p}-\mathbf{a})\cdot(\mathbf{p}-\mathbf{a}) = (\mathbf{p}-\mathbf{b})\cdot(\mathbf{p}-\mathbf{b})$

$|\mathbf{p}|^2 - 2\mathbf{a}\cdot\mathbf{p} + |\mathbf{a}|^2 = |\mathbf{p}|^2 - 2\mathbf{b}\cdot\mathbf{p} + |\mathbf{b}|^2$

$2(\mathbf{b} - \mathbf{a})\cdot\mathbf{p} = |\mathbf{b}|^2 - |\mathbf{a}|^2$

This is the equation of a plane with normal $\mathbf{b} - \mathbf{a}$ (perpendicular to $AB$), which Passes through the midpoint $\dfrac◆LB◆\mathbf{a}+\mathbf{b}◆RB◆◆LB◆2◆RB◆$. This is the Perpendicular bisector. $\blacksquare$

Question 14

Find the equation of the plane containing the line $L: \mathbf{r} = (1, 0, 2) + \lambda(1, 2, -1)$ And the point $P(3, 1, 4)$.

Solution

The direction of $L$ is $\mathbf{d} = (1, 2, -1)$. Two vectors in the plane are $\overrightarrow{PQ} = (1, 0, 2) - (3, 1, 4) = (-2, -1, -2)$ (wait, $Q$ should be on $L$Not $P$).

Actually, the point on $L$ at $\lambda = 0$ is $(1, 0, 2)$. Vectors in the plane: $\overrightarrow{PQ} = (1-3, 0-1, 2-4) = (-2, -1, -2)$ and $\mathbf{d} = (1, 2, -1)$.

$\mathbf{n} = (-2,-1,-2)\times(1,2,-1) = \begin{pmatrix}(-1)(-1)-(-2)(2)\\(-2)(1)-(-2)(-1)\\(-2)(2)-(-1)(1)\end{pmatrix} = \begin{pmatrix}1+4\\-2-2\\-4+1\end{pmatrix} = \begin{pmatrix}5\\-4\\-3\end{pmatrix}$

$\mathbf{r}\cdot\mathbf{n} = (1)(5) + (0)(-4) + (2)(-3) = 5 - 6 = -1$.

Equation: $5x - 4y - 3z = -1$.

Verify: $(3)(-4) - 4(1) - 3(4) = -15 - 4 - 12 = -31 \neq -1$. Let me recheck.

Using point $P(3, 1, 4)$: $5(3) - 4(1) - 3(4) = 15 - 4 - 12 = -1$. Correct.

Question 15

Prove that $\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}\times(\mathbf{c}\times\mathbf{a}) = \mathbf{c}\times(\mathbf{a}\times\mathbf{b})$.

Solution

This is the cyclic permutation property of the scalar triple product. In determinant form:

$\det\begin{pmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{pmatrix} = \det\begin{pmatrix}b_1&b_2&b_3\\c_1&c_2&c_3\\a_1&a_2&a_3\end{pmatrix} = \det\begin{pmatrix}c_1&c_2&c_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{pmatrix}$

Each equality follows from the fact that swapping two rows of a determinant changes its sign, and Two swaps return to the original sign. $\blacksquare$

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14. Connections to Other Topics

14.1 Vectors and matrices

The cross product and scalar triple product can be expressed as determinants. See Matrices.

14.2 Vectors and mechanics

Vector methods are essential for resolving forces, computing moments ($\mathbf{M} = \mathbf{r} \times \mathbf{F}$), and angular momentum. See Projectile Motion.

14.3 Planes and coordinate geometry

The equation of a plane connects vectors to 3D geometry. See Polar Coordinates for parametric Representations of curves.

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15. Key Results Summary

| Result | Formula | | ------------------------ | ----------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------- | -------- | -------------------------------- | ---------------------------------- | ------------- | ----- | | Scalar product | $\mathbf{a}\cdot\mathbf{b} = | \mathbf{a} | | \mathbf{b} | \cos\theta = a_1b_1+a_2b_2+a_3b_3$ | | Vector product | $\mathbf{a}\times\mathbf{b} = \begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}$ | | Scalar triple product | $[\mathbf{a},\mathbf{b},\mathbf{c}] = \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \det(\mathbf{a}\;\mathbf{b}\;\mathbf{c})$ | | Distance: point to plane | $D = \dfrac◆LB◆ | \mathbf{a}\cdot\mathbf{n} - d | ◆RB◆◆LB◆ | \mathbf{n} | ◆RB◆$ | | Distance: skew lines | $D = \dfrac◆LB◆ | (\mathbf{a}\_2-\mathbf{a}\_1)\cdot(\mathbf{d}\_1\times\mathbf{d}\_2) | ◆RB◆◆LB◆ | \mathbf{d}\_1\times\mathbf{d}\_2 | ◆RB◆$ | | Angle: line to plane | $\sin\phi = \dfrac◆LB◆ | \mathbf{d}\cdot\mathbf{n} | ◆RB◆◆LB◆ | \mathbf{d} | | \mathbf{n} | ◆RB◆$ | | Angle: two planes | $\cos\theta = \dfrac◆LB◆ | \mathbf{n}\_1\cdot\mathbf{n}\_2 | ◆RB◆◆LB◆ | \mathbf{n}\_1 | | \mathbf{n}\_2 | ◆RB◆$ | | Volume of tetrahedron | $V = \dfrac{1}{6} | [\mathbf{a},\mathbf{b},\mathbf{c}] |$ | | Reflection in plane | $P' = P - 2D\hat{\mathbf{n}}$ where $D = \dfrac◆LB◆P\cdot\mathbf{n}-d◆RB◆◆LB◆ | \mathbf{n} | ◆RB◆$ |

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16. Further Exam-Style Questions

Question 16

Find the equation of the plane that passes through the points A(1,0,0)$$B(0,1,0)$$C(0,0,1) and Verify that $D(1/3, 1/3, 1/3)$ lies on it.

Solution

\overrightarrow{AB} = (-1,1,0)$$\overrightarrow{AC} = (-1,0,1).

$\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = (1,1,1)$.

$\mathbf{r}\cdot\mathbf{n} = (1)(1)+(0)(1)+(0)(1) = 1$.

Equation: $\boxed{x+y+z = 1}$.

Check $D$: $\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3} = 1$. ✓

Question 17

Prove that $\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$ (the vector triple product expansion).

Solution

Let \mathbf{a} = (a_1,a_2,a_3)$$\mathbf{b} = (b_1,b_2,b_3)$$\mathbf{c} = (c_1,c_2,c_3).

$\mathbf{b}\times\mathbf{c} = (b_2c_3-b_3c_2,\; b_3c_1-b_1c_3,\; b_1c_2-b_2c_1)$.

The first component of $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$:

$a_2(b_1c_2-b_2c_1) - a_3(b_3c_1-b_1c_3) = a_2b_1c_2-a_2b_2c_1-a_3b_3c_1+a_3b_1c_3$

$= b_1(a_2c_2+a_3c_3) - c_1(a_2b_2+a_3b_3)$

$= b_1(\mathbf{a}\cdot\mathbf{c}-a_1c_1) - c_1(\mathbf{a}\cdot\mathbf{b}-a_1b_1)$

$= b_1(\mathbf{a}\cdot\mathbf{c}) - c_1(\mathbf{a}\cdot\mathbf{b}) - a_1b_1c_1+a_1c_1b_1 = b_1(\mathbf{a}\cdot\mathbf{c})-c_1(\mathbf{a}\cdot\mathbf{b})$.

Similarly for the other two components. $\blacksquare$

Question 18

Two lines are given by $\mathbf{r}_1 = (0,1,0)+\lambda(1,0,-1)$ and $\mathbf{r}_2 = (0,0,1)+\mu(0,1,1)$. Determine whether they intersect, are parallel, or are skew.

Solution

\mathbf{d}_1 = (1,0,-1)$$\mathbf{d}_2 = (0,1,1). Not parallel (not scalar multiples).

For intersection: \lambda = 0$$1 = \mu$$-\lambda = 1+\mu.

From $\lambda = 0$: $1 = \mu$ and $0 = 1+1 = 2$. Contradiction.

The lines are skew.

Distance: $\mathbf{d}_1\times\mathbf{d}_2 = (0\cdot1-(-1)\cdot1,\; (-1)\cdot0-1\cdot1,\; 1\cdot1-0\cdot0) = (1,-1,1)$.

$|\mathbf{d}_1\times\mathbf{d}_2| = \sqrt{3}$.

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

$D = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{3}◆RB◆ = \boxed{\dfrac◆LB◆2\sqrt{3}◆RB◆◆LB◆3◆RB◆}$.

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17. Advanced Topics

17.1 Vector equations of planes

The equation of a plane can be written in three equivalent forms:

• Scalar product: $\mathbf{r}\cdot\mathbf{n} = d$
• Cartesian: $ax+by+cz = d$
• Parametric: $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$

17.2 The shortest distance between two skew lines — alternative derivation

The shortest distance between skew lines equals the perpendicular distance from any point on one Line to the parallel plane containing the other line.

17.3 Triple vector product identity

The BAC-CAB rule: $\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$.

This identity is extensively used in mechanics (e.g., angular momentum, moments).

17.4 Applications in geometry

• Coplanarity test: $[\mathbf{a},\mathbf{b},\mathbf{c}] = 0 \iff$ the three vectors are coplanar.
• Volume of parallelepiped: $V = |[\mathbf{a},\mathbf{b},\mathbf{c}]|$.
• Volume of tetrahedron: $V = \dfrac{1}{6}|[\mathbf{a},\mathbf{b},\mathbf{c}]|$.

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18. Further Exam-Style Questions

Question 19

Find the angle between the planes $2x - y + z = 3$ and $x + y + 2z = 1$.

Solution

\mathbf{n}_1 = (2,-1,1)$$\mathbf{n}_2 = (1,1,2).

$\cos\theta = \dfrac◆LB◆|2-1+2|◆RB◆◆LB◆\sqrt{6}\sqrt{6}◆RB◆ = \dfrac{3}{6} = \dfrac{1}{2}$.

$\boxed{\theta = 60°}$

Question 20

Prove that the line of intersection of the planes $x+y+z=1$ and $2x-y+z=3$ is parallel to the Vector $(2, 1, -3)$.

Solution

The direction of the line of intersection is $\mathbf{n}_1 \times \mathbf{n}_2 = (1,1,1) \times (2,-1,1)$.

$= \begin{pmatrix}(1)(1)-(1)(-1)\\(1)(2)-(1)(1)\\(1)(-1)-(1)(2)\end{pmatrix} = (2, 1, -3)$.

Since the cross product gives $(2,1,-3)$The line is parallel to this vector. $\blacksquare$

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19. Advanced Topics in 3D Vectors

19.1 Direction cosines

If $\mathbf{a}$ makes angles $\alpha, \beta, \gamma$ with the coordinate axes, then:

$\cos\alpha = \dfrac◆LB◆a_1◆RB◆◆LB◆|\mathbf{a}|◆RB◆$ $\cos\beta = \dfrac◆LB◆a_2◆RB◆◆LB◆|\mathbf{a}|◆RB◆$ $\cos\gamma = \dfrac◆LB◆a_3◆RB◆◆LB◆|\mathbf{a}|◆RB◆$

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

19.2 Triple vector product (BAC-CAB rule)

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

Note: $(\mathbf{a}\times\mathbf{b})\times\mathbf{c} = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{a}(\mathbf{b}\cdot\mathbf{c})$ (different!)

19.3 Shortest distance from a point to a line

The shortest distance from point $P$ (position vector $\mathbf{p}$) to the line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ is:

$D = \frac◆LB◆|(\mathbf{p}-\mathbf{a})\times\mathbf{d}|◆RB◆◆LB◆|\mathbf{d}|◆RB◆$

19.4 Vector planes — parametric form

A plane through point $\mathbf{a}$ spanned by vectors $\mathbf{b}$ and $\mathbf{c}$:

$\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$

The normal is $\mathbf{n} = \mathbf{b}\times\mathbf{c}$.

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20. Further Exam-Style Questions

Question 21

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

Solution

$\mathbf{p}-\mathbf{a} = (1,2,3)-(0,1,-1) = (1,1,4)$. $\mathbf{d} = (1,1,0)$.

$(\mathbf{p}-\mathbf{a})\times\mathbf{d} = \begin{pmatrix}1\cdot0-4\cdot1\\4\cdot1-1\cdot0\\1\cdot1-1\cdot1\end{pmatrix} = (-4, 4, 0)$.

$D = \dfrac◆LB◆|(-4,4,0)|◆RB◆◆LB◆|(1,1,0)|◆RB◆ = \dfrac◆LB◆\sqrt{32}◆RB◆◆LB◆\sqrt{2}◆RB◆ = \dfrac◆LB◆4\sqrt{2}◆RB◆◆LB◆\sqrt{2}◆RB◆ = \boxed{4}$.

Question 22

Prove that the lines $\mathbf{r}_1 = (1,0,0)+\lambda(1,1,1)$ and $\mathbf{r}_2 = (0,1,0)+\mu(1,-1,0)$ intersect and find the point of intersection.

Solution

1+\lambda = \mu$$\lambda = 1-\mu$$\lambda = 0.

From $\lambda = 0$: $\mu = 1$. Check: $1 = 1$ ✓, $0 = 0$ ✓, $0 = 0$ ✓.

The lines intersect at $\boxed{(1,0,0)}$.

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. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

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

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

6. Incorrectly applying integration by parts by choosing $u$ and $\frac{dv}{dx}$ the wrong way around.

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