FP3 Chapter 5: Vectors

Prior Knowledge Check

Before we begin, let’s review some key concepts from our previous study of vectors:

Exercise: Vector Operations

(a) Find the scalar product of the vectors $3\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$ and $4\mathbf{i} - 5\mathbf{j} + \mathbf{k}$

(b) A straight line has vector equation $\mathbf{r} = (1, 4, -2) + \lambda(2, 3, 5)$. Write down the Cartesian equation of the line.

(c) The coordinates of points P, Q and R are (1, 0, -1), (2, 4, 1) and (3, 5, 6) respectively. Find angle QPR.

Vector Operations Review

Operation	Definition
Scalar (dot) product	$\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$
Component form of scalar product	$\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$
Angle between vectors	$\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}$
Parametric equation of a line	$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$

1. Vector Product

1.1 Geometric Motivation

The scalar (dot) product of two vectors provides a scalar quantity that relates to the projection of one vector onto another. However, we often need a way to represent quantities that have both magnitude and direction that are perpendicular to both input vectors. The vector product (or cross product) fulfills this need.

Geometric Intuition for Vector Product

When two vectors $\mathbf{a}$ and $\mathbf{b}$ are not parallel, they define a plane. The vector product $\mathbf{a} \times \mathbf{b}$ gives us:

• A vector perpendicular to this plane (and thus perpendicular to both $\mathbf{a}$ and $\mathbf{b}$)
• Whose magnitude equals the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$
• Whose direction is determined by the right-hand rule

The cross product has important applications in:

• Finding areas and volumes
• Defining torque and angular momentum in physics
• Computing normal vectors to planes
• Determining perpendicular directions in 3D space

Definition (Vector Product): The vector (or cross) product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as:

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

where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, and $\hat{\mathbf{n}}$ is the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, with direction determined by the right-hand rule.

Right-Hand Rule

You can use the ‘right-hand rule’ to determine the direction of $\hat{\mathbf{n}}$:

• If $\mathbf{a}$ is your first finger and $\mathbf{b}$ is your second finger
• Then $\mathbf{a} \times \mathbf{b}$ acts in the direction of your thumb

1.2 Key Properties

• $\mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b}$ (not commutative)
• $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$ (distributive)
• $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ if and only if $\mathbf{a} = \mathbf{0}$, $\mathbf{b} = \mathbf{0}$, or $\mathbf{a}$ and $\mathbf{b}$ are parallel

Why is $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$?

When we swap the order of vectors in a cross product, the plane they define remains the same, but the direction of rotation from the first vector to the second is reversed:

This change in rotational direction causes the perpendicular vector to point in the opposite direction, resulting in the negative sign.

1.3 Unit Vector Products

Products of Unit Vectors

By the key properties of the cross product, we can obtain the following identities for unit vectors.

	$\times\ \mathbf{i}$	$\times\ \mathbf{j}$	$\times\ \mathbf{k}$
$\mathbf{i}\ \times$	$\mathbf{0}$	$\mathbf{k}$	$-\mathbf{j}$
$\mathbf{j}\ \times$	$-\mathbf{k}$	$\mathbf{0}$	$\mathbf{i}$
$\mathbf{k}\ \times$	$\mathbf{j}$	$-\mathbf{i}$	$\mathbf{0}$

1.4 Component Form and Its Derivation

The component form of the cross product can be derived from its geometric definition and the properties of unit vector products.

Starting with vectors $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$, and using the distributive property:

$$\begin{aligned} \mathbf{a} \times \mathbf{b} &= (a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}) \times (b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k})\\ &= a_1b_1(\mathbf{i} \times \mathbf{i}) + a_1b_2(\mathbf{i} \times \mathbf{j}) + a_1b_3(\mathbf{i} \times \mathbf{k})\\ &\quad + a_2b_1(\mathbf{j} \times \mathbf{i}) + a_2b_2(\mathbf{j} \times \mathbf{j}) + a_2b_3(\mathbf{j} \times \mathbf{k})\\ &\quad + a_3b_1(\mathbf{k} \times \mathbf{i}) + a_3b_2(\mathbf{k} \times \mathbf{j}) + a_3b_3(\mathbf{k} \times \mathbf{k}) \end{aligned}$$

Substituting the unit vector products:

$$\begin{aligned} \mathbf{a} \times \mathbf{b} &= a_1b_2\mathbf{k} - a_1b_3\mathbf{j} - a_2b_1\mathbf{k} + a_2b_3\mathbf{i} + a_3b_1\mathbf{j} - a_3b_2\mathbf{i}\\ &= (a_2b_3 - a_3b_2)\mathbf{i} + (a_3b_1 - a_1b_3)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} \end{aligned}$$

This can be written in determinant form:

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

Intuition for the Component Form

The components of $\mathbf{a} \times \mathbf{b}$ represent the areas of projections of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ onto the three coordinate planes:

• $a_2b_3 - a_3b_2$: Area of projection onto the $yz$-plane (perpendicular to $\mathbf{i}$)
• $a_3b_1 - a_1b_3$: Area of projection onto the $xz$-plane (perpendicular to $\mathbf{j}$)
• $a_1b_2 - a_2b_1$: Area of projection onto the $xy$-plane (perpendicular to $\mathbf{k}$)

This gives us another way to understand why the cross product represents area.

Example (Vector Product in Component Form): Given that $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j}$ and $\mathbf{b} = 4\mathbf{i} + \mathbf{j} - \mathbf{k}$, find $\mathbf{a} \times \mathbf{b}$:

(a) directly

(b) by a method involving a determinant

(c) Verify that $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$

Common Mistake

A common error is forgetting that the vector product is not commutative. Always remember that:

$$\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$$

This means the order matters when calculating a vector product.

2. Geometric Applications of Vector Product

2.1 Areas

Theorem (Area of a Triangle): If $\mathbf{a}$ and $\mathbf{b}$ are position vectors of points A and B relative to an origin O, then:

$$\text{Area of triangle OAB} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$$

Proof. The area of triangle OAB is $\frac{1}{2}|\mathbf{a}||\mathbf{b}|\sin\theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. But this is exactly $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$. $\square$

Key Formulas for Areas

• Area of triangle OAB = $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$
• Area of triangle ABC = $\frac{1}{2}|(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})|$
• Area of parallelogram OABC = $|\mathbf{a} \times \mathbf{b}|$
• Area of parallelogram ABCD = $|(\mathbf{b} - \mathbf{a}) \times (\mathbf{d} - \mathbf{a})|$

Example (Area of a Triangle): Find the area of triangle ABC, where the position vectors of A, B and C are $4\mathbf{i} - 2\mathbf{j} + \mathbf{k}$, $-12\mathbf{i} + 14\mathbf{j} + \mathbf{k}$ and $-4\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ respectively.

3. Scalar Triple Product

3.1 Definition and Properties

Definition (Scalar Triple Product): The scalar triple product of three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ is defined as:

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

Geometric Interpretation

The scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ represents:

• The signed volume of the parallelepiped formed by the three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$
• A positive value if $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ form a right-handed system
• A negative value if they form a left-handed system
• Zero if the three vectors are coplanar (or any vector is zero)

Key Properties of Scalar Triple Product

• Cyclic property: $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})$
• If a vector is repeated, the scalar triple product is zero: $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{p}) = 0$ for any vector $\mathbf{p}$
• In component form:

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

Example (Computing Scalar Triple Product): Calculate the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ for the vectors:

$$\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}, \quad \mathbf{c} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$$

Solution:

Method 1: First compute the cross product $\mathbf{b} \times \mathbf{c}$, then take the dot product with $\mathbf{a}$.

$$\begin{aligned} \mathbf{b} \times \mathbf{c} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix}\\ &= \mathbf{i}(4 - 1) - \mathbf{j}(2 + 3) + \mathbf{k}(-1 - 6)\\ &= 3\mathbf{i} - 5\mathbf{j} - 7\mathbf{k} \end{aligned}$$

Now we compute the dot product:

$$\begin{aligned} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) &= (2\mathbf{i} - \mathbf{j} + 3\mathbf{k}) \cdot (3\mathbf{i} - 5\mathbf{j} - 7\mathbf{k})\\ &= 6 + 5 - 21 = -10 \end{aligned}$$

Method 2: Use the determinant formula directly.

$$\begin{aligned} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) &= \begin{vmatrix} 2 & -1 & 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}\\ &= 2(4 - 1) + 1(2 + 3) + 3(-1 - 6)\\ &= 6 + 5 - 21 = -10 \end{aligned}$$

Applications of Scalar Triple Product

1. Volume calculation: The absolute value gives the volume of a parallelepiped or a tetrahedron (one-sixth of the parallelepiped).
2. Testing coplanarity: Three vectors are coplanar (lie in the same plane) if and only if their scalar triple product is zero.
3. Orientation: The sign of the scalar triple product determines whether the three vectors form a right-handed or left-handed system.

Exercise: Scalar Triple Product

1. Calculate the scalar triple product $[\mathbf{a}, \mathbf{b}, \mathbf{c}]$ for: $\mathbf{a} = (3, 0, -2)$, $\mathbf{b} = (1, 2, 4)$, and $\mathbf{c} = (2, -1, 3)$

2. Determine whether the vectors $\mathbf{p} = (1, 2, 1)$, $\mathbf{q} = (3, 1, -1)$, and $\mathbf{r} = (5, 5, -1)$ are coplanar.

3. Show that $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})$ for any three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.

3.2 Volumes

Theorem (Volume of a Parallelepiped): The volume of a parallelepiped with edges $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ from a common vertex is:

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

Theorem (Volume of a Tetrahedron): The volume of a tetrahedron with vertices at the origin and at points with position vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ is:

$$\text{Volume} = \frac{1}{6}|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$$

Example (Volume of a Tetrahedron): Find the volume of the tetrahedron with vertices at A(1, 3, 0), B(3, 2, -1), C(2, -1, -1) and D(-1, 3, 6).

Solution:

$$\begin{aligned} \overrightarrow{AB} &= (2, -1, -1) \\ \overrightarrow{AC} &= (1, -4, -1) \\ \overrightarrow{AD} &= (-2, 0, 6) \end{aligned}$$

The volume is given by:

$$\begin{aligned} V &= \frac{1}{6}|\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})| \\ &= \frac{1}{6}\left|\begin{vmatrix} 2 & -1 & -1 \\ 1 & -4 & -1 \\ -2 & 0 & 6 \end{vmatrix}\right| \\ &= \frac{1}{6}|36| = 6 \end{aligned}$$

Exercise: Finding Volume

Find the volume of the tetrahedron with vertices at O(0, 0, 0), A(2, 2, 1), B(3, -1, 2), and C(1, 1, 3).

4. Vector Equations of Lines and Planes

4.1 Vector Equation of a Line

A line in three-dimensional space can be represented in several equivalent forms. We’ll start with the parametric form and then explore the vector equation.

Definition (Parametric Form of a Line): If a line passes through a point with coordinates $(x_1, y_1, z_1)$ and has direction vector $\mathbf{b} = (l, m, n)$, then any point $(x, y, z)$ on the line can be represented parametrically as:

$$\begin{aligned} x &= x_1 + \lambda l\\ y &= y_1 + \lambda m\\ z &= z_1 + \lambda n \end{aligned}$$

where $\lambda$ is a scalar parameter.

This can be equivalently written in the form:

$$\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} = \lambda$$

Definition (Vector Equation of a Line): For a line passing through point $A$ with position vector $\mathbf{a}$ and parallel to vector $\mathbf{b}$, the parametric vector form is:

$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$

where $\mathbf{r}$ is the position vector of any point on the line and $\lambda$ is a scalar parameter.

Note that this can also be expressed using the cross product:

$$(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$$

Geometric Interpretation of Line Equations

• $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$: Any point on the line can be reached by starting at point $A$ and moving some distance in the direction of $\mathbf{b}$
• $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$: The vector from any point $A$ on the line to any other point on the line is parallel to $\mathbf{b}$ (so their cross product is zero)
• $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} = \lambda$: The components of the displacement from $(x_1,y_1,z_1)$ are proportional to the components of the direction vector $(l,m,n)$

Example (Finding the Equation of a Line): Find the vector equation of the line through the points (1, 2, -1) and (3, -2, 2) in the form $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$.

4.2 Vector Equation of a Plane

Definition (Vector Equation of a Plane): A plane can be represented in several forms:

• Scalar product form: $\mathbf{r} \cdot \mathbf{n} = p$ where $\mathbf{n}$ is normal to the plane and $p$ is a scalar.
• Point-normal form: $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$ where $\mathbf{a}$ is the position vector of a point on the plane.
• Parametric form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ where $\mathbf{b}$ and $\mathbf{c}$ are non-parallel vectors in the plane, and $\lambda$ and $\mu$ are scalar parameters.
• Cartesian form: $ax + by + cz + d = 0$

Geometric Interpretation of Plane Equations

• $\mathbf{r} \cdot \mathbf{n} = p$: The projection of any position vector $\mathbf{r}$ onto the normal $\mathbf{n}$ equals $p$. The scalar $p$ is the length of the perpendicular from the origin to the plane (when $\mathbf{n}$ is a unit vector).
• $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$: The vector from any point $A$ (with position vector $\mathbf{a}$) to any other point on the plane (with position vector $\mathbf{r}$) is perpendicular to $\mathbf{n}$ (i.e., lies in the plane).
• $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$: Any point on the plane can be reached by starting at point $A$ and moving along linear combinations of vectors $\mathbf{b}$ and $\mathbf{c}$ that lie in the plane.

Example (Converting Between Plane Representations): Find different forms of the equation of a plane passing through the point $P(2,1,3)$ with normal vector $\mathbf{n} = (3,2,-1)$.

Solution:

(a) Point-normal form:

The point-normal form is:

$$(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$$

where $\mathbf{a} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ and normal vector $\mathbf{n} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}$.

(b) Scalar product form:

Expanding the point-normal form:

$$\begin{aligned} (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} &= 0\\ \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \end{aligned}$$

Computing the constant term:

$$\mathbf{a} \cdot \mathbf{n} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 6 + 2 - 3 = 5$$

Therefore, the scalar product form is $\mathbf{r} \cdot \mathbf{n} = 5$ where $\mathbf{n} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}$.

(c) Cartesian form:

Setting $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, we expand the scalar product form:

$$3x + 2y - z = 5$$

(d) Parametric form:

To find this form, we need to find two non-parallel vectors that lie in the plane. Since these vectors must be perpendicular to the normal vector $\mathbf{n} = (3,2,-1)$, we can choose $\mathbf{b} = (-2,3,0)$ and $\mathbf{c} = (1,0,3)$.

Verification:

$$\begin{aligned} \mathbf{n} \cdot \mathbf{b} &= 3(-2) + 2(3) + (-1)(0) = 0\\ \mathbf{n} \cdot \mathbf{c} &= 3(1) + 2(0) + (-1)(3) = 0 \end{aligned}$$

The parametric form is:

$$\mathbf{r} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) + \lambda(-2\mathbf{i} + 3\mathbf{j}) + \mu(\mathbf{i} + 3\mathbf{k})$$

Exercise: Converting Between Plane Representations

1. Find the point-normal form, scalar product form, and Cartesian form of the plane passing through the points $A(1,0,2)$, $B(3,1,0)$, and $C(2,2,1)$.

2. A plane has Cartesian equation $2x - 3y + z = 6$. Find: (a) The normal vector to the plane (b) A point on the plane (c) The scalar product form of the plane (d) The point-normal form of the plane (e) A parametric representation of the plane

3. A plane has parametric representation $\mathbf{r} = (1,2,3) + \lambda(2,0,1) + \mu(0,1,-1)$. Find the normal vector to this plane and then express the plane in Cartesian form.

5. Geometric Problems in 3D Space

5.1 Intersections

Methods for Finding Intersections

1. Line-Line Intersection:

   • Set up parametric equations for both lines
   • Equate components to find parameter values
   • Check if parameter values give same point on both lines

2. Line-Plane Intersection:

   • Substitute line equation into plane equation
   • Solve for the parameter value
   • Substitute back to find intersection point

3. Plane-Plane Intersection:

   • Direction of intersection line is $\mathbf{n}_1 \times \mathbf{n}_2$
   • Find a point on both planes by solving simultaneous equations
   • Write line equation using point and direction

Example (Line-Plane Intersection): Find the point of intersection between the line $\mathbf{r} = (2, 1, 3) + t(1, 2, -1)$ and the plane $2x - y + 3z = 8$.

Example (Plane-Plane Intersection): Find the line of intersection between the planes $2x + y - z = 3$ and $x - y + 2z = 4$.

Exercise: Intersections

1. Find the point of intersection between the line $\mathbf{r} = (1, 0, 2) + t(2, 1, 3)$ and the plane $3x - 2y + z = 5$.

2. Determine whether the lines $\mathbf{r} = (1, 2, 3) + s(2, 3, 1)$ and $\mathbf{r} = (2, 1, 4) + t(4, 6, 2)$ intersect. If they do, find the point of intersection.

3. Find the line of intersection between the planes $x + 2y + z = 4$ and $2x - y + 3z = 5$.

5.2 Angles

Angle Formulas

1. Angle Between Two Vectors:

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

2. Angle Between Line and Plane:

$$\sin\theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$$

where $\mathbf{b}$ is the direction of the line and $\mathbf{n}$ is normal to the plane.

3. Angle Between Two Planes:

$$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$$

where $\mathbf{n}_1$ and $\mathbf{n}_2$ are normals to the planes.

Geometric Interpretation of Angles

1. Angle Between Line and Plane:

For a line with direction vector $\mathbf{b}$ and a plane with normal vector $\mathbf{n}$, the angle $\theta$ between the line and the plane satisfies:

$$\sin\theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$$

• The angle $\alpha$ between the direction vector $\mathbf{b}$ of the line and the normal vector $\mathbf{n}$ of the plane satisfies $\cos\alpha = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$.
• From the diagram, we can see that $\theta$ and $\alpha$ are complementary angles ($\theta = \frac{\pi}{2} - \alpha$), so $\sin\theta = \cos\alpha$.

2. Angle Between Two Planes:

For planes with normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$, the angle is:

$$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$$

• The acute angle $\theta$ between the normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ satisfies $\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$.
• From the diagram, we can see that $\theta$ is indeed the acute angle between the two planes.

Example (Angle Between Line and Plane): Find the acute angle between the line $\mathbf{r} = 2\mathbf{i} + \mathbf{j} - 5\mathbf{k} + \lambda(3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k})$ and the plane with equation $\mathbf{r}\cdot(2\mathbf{i} - 2\mathbf{j} - \mathbf{k}) = 2$.

5.3 Distances

Distance Formulas

1. Distance from Point to Plane:

$$\text{Distance} = \frac{|(\mathbf{p}-\mathbf{r}) \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{|\mathbf{p} \cdot \mathbf{n} - a|}{|\mathbf{n}|}$$

where plane equation is $\mathbf{r}\cdot\mathbf{n} = a$.

2. Distance from Point to Line:

$$\text{Distance} = |\mathbf{p} - \mathbf{a}| \sin\theta = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}$$

where line equation is $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$.

3. Distance Between Skew Lines:

$$\text{Distance} = \frac{|(\mathbf{a} - \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{d})|}{|\mathbf{b} \times \mathbf{d}|}$$

where line equations are $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ and $\mathbf{r} = \mathbf{c} + \mu\mathbf{d}$.

Geometric Interpretation of Distances

1. Point to Plane:

The shortest distance from a point to a plane is along the perpendicular to the plane. The formula $\frac{|(\mathbf{p}-\mathbf{r}) \cdot \mathbf{n}|}{|\mathbf{n}|}$ calculates this perpendicular distance by:

• Picking a point on the plane $\mathbf{r}$
• Projecting the position vector $\mathbf{p}-\mathbf{r}$ onto the unit normal direction $\mathbf{n}$
• Computing the magnitude of the projection

2. Point to Line:

The formula $\frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}$ calculates the distance by:

• Picking a point on the line $\mathbf{a}$
• Computing the magnitude of the cross product of $\mathbf{p} - \mathbf{a}$ and the direction vector $\mathbf{b}$
• This can be visualized as the quantity $|\mathbf{p} - \mathbf{a}||\mathbf{b}|\sin\theta$ where $\theta$ is the angle between $\mathbf{p} - \mathbf{a}$ and $\mathbf{b}$
• Dividing by $|\mathbf{b}|$ gives the distance

3. Skew Lines:

For skew lines (lines in 3D space that neither intersect nor are parallel), the formula $\frac{|(\mathbf{a} - \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{d})|}{|\mathbf{b} \times \mathbf{d}|}$ calculates the shortest distance between the two lines by:

• Picking a point on each line $\mathbf{a}$ and $\mathbf{c}$
• Computing the cross product of the direction vectors $\mathbf{b}$ and $\mathbf{d}$
• $\mathbf{b} \times \mathbf{d}$ gives a vector perpendicular to both direction vectors
• Projecting the displacement vector $\mathbf{a} - \mathbf{c}$ onto the perpendicular direction $\mathbf{b} \times \mathbf{d}$
• Computing the magnitude of the projection

Example (Distance Between Skew Lines): Find the shortest distance between the two skew lines with equations $\mathbf{r} = \mathbf{i} + \lambda(\mathbf{j} + \mathbf{k})$ and $\mathbf{r} = -\mathbf{i} + 3\mathbf{j} - \mathbf{k} + \mu(2\mathbf{i} - \mathbf{j} - \mathbf{k})$, where $\lambda$ and $\mu$ are scalars.

Example (Distance from Point to Plane): Find the distance from the point $P(3, 1, -2)$ to the plane with equation $2x - y + 2z = 5$.

Exercise: Finding Distances

1. Find the distance from the point $P(2, 3, 4)$ to the line with equation $\mathbf{r} = (1, 0, 1) + t(2, 1, 1)$.

2. Find the shortest distance between the skew lines $\mathbf{r} = (1, 2, 0) + s(1, 0, 1)$ and $\mathbf{r} = (0, 0, 3) + t(0, 1, 1)$.

6. Summary and Key Formulas

Key Definitions and Formulas

• Vector Product: $\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\, \hat{\mathbf{n}}$

• Component Form: $\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2)\mathbf{i} + (a_3b_1 - a_1b_3)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}$

• Areas:

  • Triangle OAB: $\frac{1}{2}|\mathbf{a} \times \mathbf{b}|$
  • Triangle ABC: $\frac{1}{2}|(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})|$
  • Parallelogram OABC: $|\mathbf{a} \times \mathbf{b}|$

• Scalar Triple Product: $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$

• Volumes:

  • Parallelepiped: $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$
  • Tetrahedron: $\frac{1}{6}|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$

• Line Equations:

  • Vector form: $(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$
  • Parametric form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$
  • Cartesian form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$

• Plane Equations:

  • Scalar product form: $\mathbf{r} \cdot \mathbf{n} = p$
  • Parametric form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$
  • Cartesian form: $ax + by + cz = d$

• Angles:

  • Between line and plane: $\sin\theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$
  • Between two planes: $\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$

• Distances:

  • Point to plane: $\frac{|\mathbf{p} \cdot \mathbf{n} - d|}{|\mathbf{n}|}$
  • Point to line: $\frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|}$
  • Between skew lines: $\frac{|(\mathbf{a} - \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{d})|}{|\mathbf{b} \times \mathbf{d}|}$

Common Mistakes to Avoid

1. Forgetting that the vector product is not commutative: $\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}$
2. Calculating $\mathbf{a} \times \mathbf{a}$ and expecting a non-zero result (it’s always zero)
3. Using the wrong formula for the angle between a line and a plane (it’s $\sin\theta$, not $\cos\theta$)
4. Not normalizing the normal vector when finding the perpendicular distance from the origin to a plane
5. Forgetting to take the absolute value in distance formulas
6. Confusing the parametric representation of a line ($\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$) with the vector equation ($(\mathbf{r} - \mathbf{a}) \times \mathbf{b} = \mathbf{0}$)

7. Practice Problems

Exercise 1

The points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively, relative to a fixed origin $O$.

It is given that: $\mathbf{a} = \mathbf{i} + \mathbf{j}$, $\quad \mathbf{b} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}$, $\quad \mathbf{c} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$

Calculate: (a) $\mathbf{b} \times \mathbf{c}$ (3 marks) (b) $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ (2 marks) (c) the area of triangle $OBC$ (2 marks) (d) the volume of the tetrahedron $OABC$ (1 mark)

Exercise 2

The position vectors of the points $A$, $B$ and $C$ relative to an origin $O$ are $\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}$, $7\mathbf{i} - 3\mathbf{k}$ and $4\mathbf{i} + 4\mathbf{j}$ respectively.

Find: (a) $\overrightarrow{AC} \times \overrightarrow{BC}$ (4 marks) (b) the area of triangle $ABC$ (2 marks) (c) an equation of the plane $ABC$ in the form $\mathbf{r} \cdot \mathbf{n} = p$ (2 marks)

Exercise 3

The plane $P$ has equation

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

(a) Find a vector perpendicular to the plane $P$. (2 marks)

The line $l$ passes through the point $A(1, 3, 3)$ and meets $P$ at $(3, 1, 2)$.

The acute angle between the plane $P$ and the line $l$ is $\alpha$.

(b) Find $\alpha$ to the nearest degree. (4 marks) (c) Find the perpendicular distance from $A$ to the plane $P$. (4 marks)

Exercise 4

The lines $l_1$ and $l_2$ have equations

$$\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 3 \\ 4 \end{pmatrix} \quad \text{and} \quad \mathbf{r} = \begin{pmatrix} a \\ -4 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix}$$

If the lines $l_1$ and $l_2$ intersect, find: (a) the value of $a$, (4 marks) (b) an equation for the plane containing the lines $l_1$ and $l_2$, giving your answer in the form $ax + by + cz + d = 0$, where $a$, $b$, $c$ and $d$ are constants. (4 marks)

For other values of $a$, the lines $l_1$ and $l_2$ do not intersect and are skew lines.

Given that $a = 2$,

(c) find the shortest distance between the lines $l_1$ and $l_2$. (3 marks)

Exercise 5

The plane $\Pi$ has vector equation

$$\mathbf{r} = 3\mathbf{i} + \mathbf{k} + \lambda(-4\mathbf{i} + \mathbf{j}) + \mu(6\mathbf{i} - 2\mathbf{j} + \mathbf{k})$$

(a) Find an equation of $\Pi$ in the form $\mathbf{r}\cdot\mathbf{n} = p$, where $\mathbf{n}$ is a vector perpendicular to $\Pi$ and $p$ is a constant. (5 marks)

The point $P$ has coordinates $(6, 13, 5)$. The line $l$ passes through $P$ and is perpendicular to $\Pi$. The line $l$ intersects $\Pi$ at the point $N$.

(b) Show that the coordinates of $N$ are $(3, 1, -1)$. (4 marks)

The point $R$ lies on $\Pi$ and has coordinates $(1, 0, 2)$.

(c) Find the perpendicular distance from $N$ to the line $PR$. Give your answer to 3 significant figures. (5 marks)