Figure 3.5: Plane with Normal Vector Generated by Cross Product $\mathbf{u} \times \mathbf{v}$
1. The Vector (Cross) Product
The vector cross product of two 3D vectors $\mathbf{u}$ and $\mathbf{v}$ produces a new vector that is strictly orthogonal (perpendicular) to both $\mathbf{u}$ and $\mathbf{v}$, with direction dictated by the right-hand rule:
$$\mathbf{u} \times \mathbf{v} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} \times \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} u_2 v_3 - u_3 v_2 \\ u_3 v_1 - u_1 v_3 \\ u_1 v_2 - u_2 v_1 \end{pmatrix}$$- Anti-Commutative Property: Swapping the vectors reverses the direction: $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$.
- Magnitude Formula: $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| \sin\theta$, where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$ ($0 \le \theta \le \pi$).
- Parallelism Criterion: Non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel if and only if: $$\mathbf{u} \parallel \mathbf{v} \iff \mathbf{u} \times \mathbf{v} = \mathbf{0}$$
- Geometric Area Interpretations:
• Area of Parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$: $\text{Area} = |\mathbf{u} \times \mathbf{v}|$.
• Area of Triangle with adjacent sides $\mathbf{u}$ and $\mathbf{v}$: $\text{Area} = \frac{1}{2} |\mathbf{u} \times \mathbf{v}|$.
2. Equations of a Plane in $\mathbb{R}^3$
A plane $\Pi$ in three-dimensional space is uniquely fixed by a point $A(\mathbf{a})$ lying on the plane and a non-zero normal vector $\mathbf{n} = \begin{pmatrix} A \\ B \\ C \end{pmatrix}$ perpendicular to every vector lying entirely in the plane.
Standard Representations of a Plane
• Vector Equation: For any point $R(\mathbf{r})$ in the plane, $\vec{AR} = \mathbf{r} - \mathbf{a} \perp \mathbf{n} \implies (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$: $$\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} = d$$
• Cartesian Form: Expanding the scalar product with $\mathbf{r} = (x, y, z)$ and $\mathbf{n} = (A, B, C)$: $$Ax + By + Cz = d, \quad \text{where } d = Ax_0 + By_0 + Cz_0$$
• Parametric Vector Form (2 Direction Parameters): $$\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}, \quad \lambda, \mu \in \mathbb{R}$$ where $\mathbf{u}$ and $\mathbf{v}$ are two non-parallel vectors lying parallel to the plane. The normal is $\mathbf{n} = \mathbf{u} \times \mathbf{v}$.
3. Intersections and Spatial Angles
- Line-Plane Intersection: Given line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ and plane $\mathbf{r} \cdot \mathbf{n} = d$:
$$(\mathbf{a} + \lambda\mathbf{b}) \cdot \mathbf{n} = d \implies \mathbf{a}\cdot\mathbf{n} + \lambda(\mathbf{b}\cdot\mathbf{n}) = d \implies \lambda = \frac{d - \mathbf{a}\cdot\mathbf{n}}{\mathbf{b}\cdot\mathbf{n}}$$
• If $\mathbf{b}\cdot\mathbf{n} \neq 0$: Exactly one intersection point.
• If $\mathbf{b}\cdot\mathbf{n} = 0$: The line is parallel to the plane. (If $\mathbf{a}\cdot\mathbf{n} = d$, line lies in plane; if $\mathbf{a}\cdot\mathbf{n} \neq d$, line never meets plane). - Angle $\phi$ Between a Line and a Plane: The angle between the line's direction $\mathbf{b}$ and normal $\mathbf{n}$ is the complementary angle $\frac{\pi}{2} - \phi$. Hence: $$\sin\phi = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|}$$
- Angle $\theta$ Between Two Planes $\Pi_1$ and $\Pi_2$: Exactly equal to the angle between their normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$: $$\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$$
- Line of Intersection of Two Planes: If two planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$ intersect in a line $L$, the direction vector $\mathbf{d}_L$ of the common line is orthogonal to both normals: $$\mathbf{d}_L = \mathbf{n}_1 \times \mathbf{n}_2$$
Worked Examples
Worked Example 1: Cross Product & Exact Triangle Area (Paper 1 Style · 6 marks)
The vertices of a triangle in 3D are $A(1, 2, 0)$, $B(3, 0, 1)$, and $C(2, 4, 3)$.
(a) Find the vectors $\vec{AB}$ and $\vec{AC}$. [2 marks]
(b) Calculate the cross product $\vec{AB} \times \vec{AC}$. [2 marks]
(c) Hence find the exact area of triangle $ABC$. [2 marks]
Official Markscheme Solution:
(a) Displacement Vectors: [2 marks]
$\vec{AB} = \begin{pmatrix} 3 - 1 \\ 0 - 2 \\ 1 - 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$ [A1].
$\vec{AC} = \begin{pmatrix} 2 - 1 \\ 4 - 2 \\ 3 - 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ [A1].
(b) Cross Product: [2 marks]
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -2 & 1 \\ 1 & 2 & 3 \end{vmatrix}$ [M1]
$= \mathbf{i}((-2)(3) - (1)(2)) - \mathbf{j}((2)(3) - (1)(1)) + \mathbf{k}((2)(2) - (-2)(1))$
$= \mathbf{i}(-6 - 2) - \mathbf{j}(6 - 1) + \mathbf{k}(4 + 2) = \begin{pmatrix} -8 \\ -5 \\ 6 \end{pmatrix}$ [A1].
(c) Triangle Area: [2 marks]
$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$ [M1]
$|\vec{AB} \times \vec{AC}| = \sqrt{(-8)^2 + (-5)^2 + 6^2} = \sqrt{64 + 25 + 36} = \sqrt{125} = 5\sqrt{5}$ [A1].
$\text{Area} = \frac{5\sqrt{5}}{2}$ square units [A1].
Worked Example 2: Equation of a Plane through Three Points (Paper 1 Style · 6 marks)
Find the Cartesian equation of the plane $\Pi$ passing through the points $A(1, 1, 1)$, $B(2, 3, 0)$, and $C(-1, 2, 4)$.
Official Markscheme Solution:
Step 1: Vectors in the Plane [2 marks]
$\vec{AB} = \begin{pmatrix} 2 - 1 \\ 3 - 1 \\ 0 - 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ [A1].
$\vec{AC} = \begin{pmatrix} -1 - 1 \\ 2 - 1 \\ 4 - 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix}$ [A1].
Step 2: Normal Vector [2 marks]
$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} (2)(3) - (-1)(1) \\ (-1)(-2) - (1)(3) \\ (1)(1) - (2)(-2) \end{pmatrix} = \begin{pmatrix} 6 + 1 \\ 2 - 3 \\ 1 + 4 \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 5 \end{pmatrix}$ [M1][A1].
Step 3: Cartesian Equation [2 marks]
Cartesian equation is $7x - y + 5z = d$ [M1].
Substitute coordinates of point $A(1, 1, 1)$: [M1]
$d = 7(1) - 1(1) + 5(1) = 7 - 1 + 5 = 11$ [A1].
Equation of the plane: $7x - y + 5z = 11$ [AG].
Check with B: $7(2) - 3 + 5(0) = 14 - 3 = 11$ (verified!).
Worked Example 3: Line Piercing a Plane (Paper 1 Style · 6 marks)
Find the coordinates of the point of intersection between the line $L: \mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$ and the plane $\Pi: 2x + y - 3z = 7$.
Official Markscheme Solution:
Write parametric equations of the line: [M1]
$x = 1 + 2\lambda, \quad y = -1 + 3\lambda, \quad z = 2 - \lambda$ [A1].
Substitute into the plane equation: [M1]
$2(1 + 2\lambda) + (-1 + 3\lambda) - 3(2 - \lambda) = 7$
$2 + 4\lambda - 1 + 3\lambda - 6 + 3\lambda = 7$ [A1]
$10\lambda - 5 = 7 \implies 10\lambda = 12 \implies \lambda = \frac{12}{10} = \frac{6}{5} = 1.2$ [A1].
Substitute $\lambda = 1.2$ back into the parametric equations: [M1]
$x = 1 + 2(1.2) = 3.4$
$y = -1 + 3(1.2) = 2.6$
$z = 2 - 1.2 = 0.8$
Point of intersection is $(3.4, 2.6, 0.8)$ [A1].
Worked Example 4: Angle Between a Line and a Plane (Paper 2 Style · 7 marks)
Consider the line $L: \mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$ and the plane $\Pi: x + 2y - z = 4$.
(a) Find the point of intersection of $L$ and $\Pi$. [3 marks]
(b) Find the acute angle between the line $L$ and the plane $\Pi$ in degrees. [4 marks]
Official Markscheme Solution:
(a) Point of Intersection: [3 marks]
Substitute $x = 2 + \lambda, y = 3 - \lambda, z = -1 + 2\lambda$ into plane equation: [M1]
$(2 + \lambda) + 2(3 - \lambda) - (-1 + 2\lambda) = 4$
$2 + \lambda + 6 - 2\lambda + 1 - 2\lambda = 4 \implies 9 - 3\lambda = 4 \implies 3\lambda = 5 \implies \lambda = \frac{5}{3}$ [A1].
Coordinates: $x = 2 + 5/3 = 11/3, y = 3 - 5/3 = 4/3, z = -1 + 10/3 = 7/3$.
Point of intersection is $\left(\frac{11}{3}, \frac{4}{3}, \frac{7}{3}\right)$ [A1].
(b) Angle Between Line and Plane: [4 marks]
Direction vector of line: $\mathbf{b} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$; Normal to plane: $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ [M1].
• Scalar product: $\mathbf{b}\cdot\mathbf{n} = 1(1) + (-1)(2) + 2(-1) = 1 - 2 - 2 = -3$ [A1].
• Magnitudes: $|\mathbf{b}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$, $|\mathbf{n}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$ [A1].
Use the sine formula for line-plane angle $\phi$: [M1]
$\sin\phi = \frac{|\mathbf{b}\cdot\mathbf{n}|}{|\mathbf{b}||\mathbf{n}|} = \frac{|-3|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ [A1].
$\phi = \arcsin(0.5) = 30^\circ$ (or $\frac{\pi}{6}$ radians) [A1].
Worked Example 5: Line of Intersection of Two Planes (Paper 2 Style · 8 marks)
Two planes are given by $\Pi_1: x + y + z = 3$ and $\Pi_2: 2x - y + 3z = 4$.
(a) Find the angle between the two planes in degrees. [3 marks]
(b) Find a vector equation of the line of intersection of $\Pi_1$ and $\Pi_2$. [5 marks]
Official Markscheme Solution:
(a) Angle Between Planes: [3 marks]
Normals are $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ [A1].
$\mathbf{n}_1 \cdot \mathbf{n}_2 = 1(2) + 1(-1) + 1(3) = 2 - 1 + 3 = 4$ [A1].
$|\mathbf{n}_1| = \sqrt{1 + 1 + 1} = \sqrt{3}$, $|\mathbf{n}_2| = \sqrt{4 + 1 + 9} = \sqrt{14}$ [A1].
$\cos\theta = \frac{4}{\sqrt{3}\sqrt{14}} = \frac{4}{\sqrt{42}} \approx 0.6172 \implies \theta = \arccos(0.6172) \approx 51.9^\circ$ [A1].
(b) Line of Intersection: [5 marks]
The direction of the common line is orthogonal to both normals: [M1]
$\mathbf{d}_L = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 1(3) - 1(-1) \\ 1(2) - 1(3) \\ 1(-1) - 1(2) \end{pmatrix} = \begin{pmatrix} 3 + 1 \\ 2 - 3 \\ -1 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}$ [A1].
Find a common point: let $z = 0$ [M1]:
$x + y = 3$
$2x - y = 4$
Add equations: $3x = 7 \implies x = \frac{7}{3}$ [A1].
$y = 3 - \frac{7}{3} = \frac{2}{3}$.
Common point is $\mathbf{a} = \begin{pmatrix} 7/3 \\ 2/3 \\ 0 \end{pmatrix}$ [A1].
Vector equation: $\mathbf{r} = \begin{pmatrix} 7/3 \\ 2/3 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}$ [AG].
Practice Exercises (10 Tiered Questions)
Work through these 10 examination-calibrated exercises ranging from core foundational skills to advanced Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.
Find $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$.
Show Full Markscheme
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} (0)(1) - (2)(3) \\ (2)(0) - (1)(1) \\ (1)(3) - (0)(0) \end{pmatrix} = \begin{pmatrix} -6 \\ -1 \\ 3 \end{pmatrix}$ [M1][A1].
Find the Cartesian equation of the plane passing through $P(2, -3, 1)$ with normal vector $\mathbf{n} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}$.
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Cartesian equation form: $4x + y - 2z = d$ [M1].
Substitute $P(2, -3, 1)$: $d = 4(2) + 1(-3) - 2(1) = 8 - 3 - 2 = 3$ [A1].
Equation: $4x + y - 2z = 3$ [A1].
Find the area of the parallelogram defined by adjacent vectors $\mathbf{u} = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{v} = 2\mathbf{j} + 3\mathbf{k}$.
Show Full Markscheme
$\mathbf{u} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix}$ [M1].
$\mathbf{u} \times \mathbf{v} = \begin{pmatrix} (2)(3) - (0)(2) \\ (0)(0) - (1)(3) \\ (1)(2) - (2)(0) \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}$ [A1].
$\text{Area} = |\mathbf{u} \times \mathbf{v}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$ square units [A1].
Explain why planes $\Pi_1: 2x - y + 2z = 4$ and $\Pi_2: 4x - 2y + 4z = 17$ are parallel, and find the shortest perpendicular distance between them.
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Normals are $\mathbf{n}_1 = (2, -1, 2)$ and $\mathbf{n}_2 = (4, -2, 4) = 2\mathbf{n}_1$ [M1].
Since the normal vectors are scalar multiples, the planes are parallel [A1].
Rewrite $\Pi_2$ by dividing by 2: $2x - y + 2z = 8.5$ [M1].
Pick point on $\Pi_1$: let $y = 0, z = 0 \implies 2x = 4 \implies P(2, 0, 0)$ [M1].
Perpendicular distance from $P(2, 0, 0)$ to $\Pi_2$:
$D = \frac{|2(2) - 0 + 2(0) - 8.5|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|4 - 8.5|}{3} = \frac{4.5}{3} = 1.5$ units [A1].
Find the acute angle between the planes $x + y = 1$ and $y + z = 2$ in radians.
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Normals: $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ [M1].
$\mathbf{n}_1 \cdot \mathbf{n}_2 = 1(0) + 1(1) + 0(1) = 1$ [A1].
$|\mathbf{n}_1| = \sqrt{2}, \quad |\mathbf{n}_2| = \sqrt{2}$ [A1].
$\cos\theta = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \implies \theta = \frac{\pi}{3}$ radians (or $60^\circ$) [A1].
Find the Cartesian equation of the plane passing through $A(3, 4, -1)$ that is parallel to the plane $5x - 2y + 4z = 9$.
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Parallel plane has the identical normal vector: $\mathbf{n} = \begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix}$ [M1].
Equation: $5x - 2y + 4z = d$.
Substitute $A(3, 4, -1)$: $d = 5(3) - 2(4) + 4(-1) = 15 - 8 - 4 = 3$ [A1].
Cartesian equation: $5x - 2y + 4z = 3$ [A1].
Find the Cartesian equation of the plane containing the intersecting lines $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$.
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Normal to plane is $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$: [M1]
$\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (0)(2) - (-1)(1) \\ (-1)(0) - (1)(2) \\ (1)(1) - (0)(0) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ [A1].
Cartesian equation: $x - 2y + z = d$ [M1].
Substitute common point $(1, 2, 3)$: $d = 1 - 2(2) + 3 = 1 - 4 + 3 = 0$ [A1].
Equation: $x - 2y + z = 0$ (plane passes through origin!) [A1].
Find the coordinates of the foot of the perpendicular from point $P(1, 2, 3)$ to the plane $\Pi: 2x - y + z = 5$, and calculate the shortest distance.
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The line perpendicular to $\Pi$ passing through $P(1, 2, 3)$ has direction $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$: [M1]
$\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ [A1].
Substitute into plane: $2(1 + 2t) - (2 - t) + (3 + t) = 5$
$2 + 4t - 2 + t + 3 + t = 5 \implies 6t + 3 = 5 \implies 6t = 2 \implies t = \frac{1}{3}$ [A1].
Foot of perpendicular $F$:
$x = 1 + 2(1/3) = \frac{5}{3}, \quad y = 2 - 1/3 = \frac{5}{3}, \quad z = 3 + 1/3 = \frac{10}{3}$ [A1].
Coordinates: $F\left(\frac{5}{3}, \frac{5}{3}, \frac{10}{3}\right)$.
Shortest distance: $D = |t||\mathbf{n}| = \frac{1}{3}\sqrt{2^2 + (-1)^2 + 1^2} = \frac{\sqrt{6}}{3} \approx 0.816$ units [A1].
The volume of a tetrahedron with vertices at $O(0, 0, 0)$, $A(\mathbf{a})$, $B(\mathbf{b})$, and $C(\mathbf{c})$ is given by $V = \frac{1}{6} |(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}|$. Find the volume if $A(2, 0, 0)$, $B(0, 3, 0)$, and $C(1, 1, 4)$.
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$\mathbf{a} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ 0 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}$ [M1].
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix}$ [A1].
Scalar triple product: $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = 0(1) + 0(1) + 6(4) = 24$ [A1].
$\text{Volume} = \frac{1}{6} |24| = 4$ cubic units [A1].
Prove algebraically that $|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2$ (Lagrange's Identity).
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Let $\theta$ be the angle between $\mathbf{u}$ and $\mathbf{v}$ [M1].
Recall $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| \sin\theta$ and $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta$ [M1].
Square both quantities:
$|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 \sin^2\theta$ [A1]
$(\mathbf{u} \cdot \mathbf{v})^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 \cos^2\theta$ [A1].
Add the two squared quantities: [M1]
$|\mathbf{u} \times \mathbf{v}|^2 + (\mathbf{u} \cdot \mathbf{v})^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 (\sin^2\theta + \cos^2\theta)$
Since $\sin^2\theta + \cos^2\theta = 1$:
$|\mathbf{u} \times \mathbf{v}|^2 + (\mathbf{u} \cdot \mathbf{v})^2 = |\mathbf{u}|^2 |\mathbf{v}|^2$ [A1].
Subtract $(\mathbf{u}\cdot\mathbf{v})^2$: $|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - (\mathbf{u} \cdot \mathbf{v})^2$ [AG].
IB Formula Booklet References (Topic 3)
Syllabus 3.13 & 3.14: Vector Cross Product & Planes
The official IB DP Mathematics AA Formula Booklet provides:
- Vector product: $\mathbf{v} \times \mathbf{w} = \begin{pmatrix} v_2 w_3 - v_3 w_2 \\ v_3 w_1 - v_1 w_3 \\ v_1 w_2 - v_2 w_1 \end{pmatrix}$
- Magnitude of cross product: $|\mathbf{v} \times \mathbf{w}| = |\mathbf{v}||\mathbf{w}|\sin\theta$
- Area of parallelogram: $A = |\mathbf{v} \times \mathbf{w}|$
- Vector equation of a plane: $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$
- Cartesian equation of a plane: $ax + by + cz = d$
Key Formulas & Identities
1. Angle Formulas for Lines and Planes
- Two Planes $\Pi_1, \Pi_2$: $\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
- Line $L$ and Plane $\Pi$: $\sin\phi = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$ (sine, NOT cosine!)
- Two Lines $L_1, L_2$: $\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$
2. Line of Intersection of Two Planes
- Direction of line: $\mathbf{d}_L = \mathbf{n}_1 \times \mathbf{n}_2$
- Common point: set $z = 0$ (or another coordinate) and solve $2 \times 2$ linear system.
Common Exam Pitfalls
Pitfall 1: Line-Plane Angle vs. Plane-Plane Angle
The angle between a line and a plane uses SINE ($\sin\phi = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$) because the normal vector $\mathbf{n}$ is perpendicular to the plane surface. If you use cosine, you calculate the angle between the line and the normal, which is the complementary angle $\frac{\pi}{2} - \phi$!
Pitfall 2: Anti-Commutativity of the Cross Product
$\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u})$. Reversing the vectors flips the sign of every single component. While either normal is valid for a plane equation (since $Ax+By+Cz=d$ is equivalent to $-Ax-By-Cz=-d$), sign slips during the cross product determinant will ruin the plane equation.
Pitfall 3: Triangle Area Factor
Remember that $|\mathbf{u} \times \mathbf{v}|$ gives the area of a parallelogram. For a triangle with vertices $A, B, C$, you must multiply by $\frac{1}{2}$: $\text{Area} = \frac{1}{2}|\vec{AB} \times \vec{AC}|$.
Topic Diagnostic Test
Complete this timed diagnostic test to verify your conceptual command of cross products, planes, and spatial intersections.
Section A: Diagnostic Multiple Choice Questions
What is $\mathbf{i} \times \mathbf{j}$?
What is a normal vector to the plane $3x - 2y + 4z = 11$?
If $|\mathbf{u} \times \mathbf{v}| = 10$, what is the area of the triangle spanned by $\mathbf{u}$ and $\mathbf{v}$?
Which formula correctly determines the angle $\phi$ between a line with direction $\mathbf{b}$ and a plane with normal $\mathbf{n}$?
Section B: Examination-Style Free Response Questions
Points $A(1, 0, -1)$, $B(2, 1, 1)$, and $C(1, 3, 2)$ lie in a plane $\Pi$.
(a) Find the vectors $\vec{AB}$ and $\vec{AC}$. [2 marks]
(b) Find a normal vector to the plane $\Pi$. [3 marks]
(c) Hence write down the Cartesian equation of the plane $\Pi$. [2 marks]
Show Full Markscheme
(a) Vectors: [2 marks]
$\vec{AB} = \begin{pmatrix} 2 - 1 \\ 1 - 0 \\ 1 - (-1) \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ [A1].
$\vec{AC} = \begin{pmatrix} 1 - 1 \\ 3 - 0 \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ 3 \end{pmatrix}$ [A1].
(b) Normal Vector: [3 marks]
$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2 \\ 0 & 3 & 3 \end{vmatrix}$ [M1]
$= \mathbf{i}(3 - 6) - \mathbf{j}(3 - 0) + \mathbf{k}(3 - 0) = \begin{pmatrix} -3 \\ -3 \\ 3 \end{pmatrix} = -3\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ [A1].
A simplified normal is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ [A1].
(c) Cartesian Equation: [2 marks]
$x + y - z = d$ [M1].
Substitute $A(1, 0, -1)$: $d = 1 + 0 - (-1) = 2$ [A1].
Cartesian equation: $x + y - z = 2$ [AG].
The plane $\Pi$ has Cartesian equation $2x - 3y + z = 7$. The line $L$ passes through $P(4, 1, -2)$ and is perpendicular to $\Pi$.
(a) Write down a vector equation for line $L$. [2 marks]
(b) Find the coordinates of the point of intersection between line $L$ and plane $\Pi$. [3 marks]
(c) Hence find the shortest distance from point $P$ to the plane $\Pi$. [3 marks]
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(a) Line Equation: [2 marks]
Since $L$ is perpendicular to $\Pi$, its direction vector is the normal to $\Pi$: $\mathbf{d} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ [M1].
$\mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ [A1].
(b) Point of Intersection: [3 marks]
Substitute $x = 4 + 2\lambda, y = 1 - 3\lambda, z = -2 + \lambda$ into plane equation: [M1]
$2(4 + 2\lambda) - 3(1 - 3\lambda) + (-2 + \lambda) = 7$
$8 + 4\lambda - 3 + 9\lambda - 2 + \lambda = 7 \implies 14\lambda + 3 = 7 \implies 14\lambda = 4 \implies \lambda = \frac{2}{7}$ [A1].
Coordinates of foot $F$:
$x = 4 + 4/7 = 32/7, \quad y = 1 - 6/7 = 1/7, \quad z = -2 + 2/7 = -12/7$ [A1].
Intersection point is $F\left(\frac{32}{7}, \frac{1}{7}, -\frac{12}{7}\right)$.
(c) Shortest Distance: [3 marks]
Distance $= |\vec{PF}| = |\lambda \mathbf{n}| = \frac{2}{7}\sqrt{2^2 + (-3)^2 + 1^2} = \frac{2}{7}\sqrt{4 + 9 + 1} = \frac{2\sqrt{14}}{7} \approx 1.07$ units [M1][A1][A1].