A-Level Further Mathematics: Further Vectors

Moving from two-dimensional vectors to three dimensions unlocks a powerful mathematical language for describing the space we live in. This toolkit is indispensable for fields from engineering and computer graphics to physics and robotics. Mastering 3D vector geometry allows you to precisely model lines and planes, calculate angles and intersections, and solve complex spatial problems involving distance and volume.

Vector Equations of Lines and Planes in Three Dimensions

In three dimensions, a line is defined by a point it passes through and its direction. The vector equation of a line is given by $\mathbf{r}=\mathbf{a}+\lambda \mathbf{d}$, where $\mathbf{r}$ is the position vector of any point on the line, $\mathbf{a}$ is the position vector of a known point, $\mathbf{d}$ is the direction vector, and $\lambda$ is a scalar parameter. You can also express this in Cartesian form by equating the $x$, $y$, and $z$ components from the vector form.

A plane, however, is defined by a point and a direction perpendicular to it. The vector equation of a plane can be expressed as $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$, where $\mathbf{n}$ is a vector normal (perpendicular) to the plane and $\mathbf{a}$ is the position vector of a point in the plane. Expanding this dot product leads to the Cartesian equation of a plane in the form $ax+by+cz=d$, where $(a,b,c)$ are the components of the normal vector $\mathbf{n}$.

Example: Find the vector equation of the plane containing the points $A(1,0,2)$, $B(3,1,-1)$, and $C(4,2,0)$. First, find two vectors in the plane: $\overrightarrow{AB}=(2,1,-3)$ and $\overrightarrow{AC}=(3,2,-2)$. The normal vector $\mathbf{n}$ is the cross product of these: $\mathbf{n}=\overrightarrow{AB}\times \overrightarrow{AC}$. Using point $A$, the equation is $(\mathbf{r}-(1,0,2))\cdot \mathbf{n}=0$.

The Cross Product and Its Geometric Meaning

The cross product (or vector product) of two vectors $\mathbf{a}$ and $\mathbf{b}$ is a vector defined as $\mathbf{a}\times \mathbf{b}=|\mathbf{a}||\mathbf{b}|\sin \theta \hat{\mathbf{n}}$, where $\theta$ is the angle between them and $\hat{\mathbf{n}}$ is a unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, following the right-hand rule. In component form, if $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ and $\mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$, then: $$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\mathbf{i}-(a_1{b}_3-a_3{b}_1)\mathbf{j}+(a_1{b}_2-a_2{b}_1)\mathbf{k}.$$

Its magnitude has a crucial geometric interpretation: $|\mathbf{a}\times \mathbf{b}|$ gives the area of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$. Consequently, the area of the corresponding triangle is $\frac{1}{2}|\mathbf{a}\times \mathbf{b}|$. This makes the cross product essential for calculating areas in 3D space without needing a base and perpendicular height.

The Scalar Triple Product for Volume

The scalar triple product combines the dot and cross product: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$. It yields a scalar value. Its absolute value, $|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$, gives the volume of the parallelepiped formed by the three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ emanating from the same point.

A parallelepiped is a 3D shape like a skewed box. Since a tetrahedron (triangular pyramid) with the same vectors as three edges from a vertex has one-sixth of this volume, the volume of the tetrahedron is given by $\frac{1}{6}|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$. You can compute the scalar triple product efficiently using a determinant: $$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|.$$ If this determinant is zero, the three vectors are coplanar—a key test for geometric relationships.

Finding Intersections and Angles

To find the angle between two lines, you use the direction vectors ${\mathbf{d}}_1$ and ${\mathbf{d}}_2$ with the formula $\cos \theta =\frac{|{\mathbf{d}}_1\cdot {\mathbf{d}}_2|}{|{\mathbf{d}}_1||{\mathbf{d}}_2|}$. For the angle between two planes, use their normal vectors ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$ in the same formula. The angle between a line and a plane is found using the line's direction vector $\mathbf{d}$ and the plane's normal $\mathbf{n}$: $\sin \theta =\frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.

Finding the intersection of two lines involves equating their vector equations $\mathbf{a}+\lambda \mathbf{d}$ and $\mathbf{b}+\mu \mathbf{m}$ and solving the three simultaneous equations for $\lambda$ and $\mu$. If a unique solution exists, the lines intersect. If the direction vectors are parallel but the lines are distinct, they are parallel. If they are not parallel and do not intersect, they are skew lines. The intersection of a line and a plane is found by substituting the line's equation into the plane's equation to solve for the parameter.

Calculating Shortest Distances

Shortest distance problems are a key application of these vector tools.

• Shortest distance from a point to a line: Given point $P$ with position vector $\mathbf{p}$ and line $\mathbf{r}=\mathbf{a}+\lambda \mathbf{d}$, the distance is:

$$D=\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}.$$ This formula comes from the area of the parallelogram formed by vectors $(\mathbf{p}-\mathbf{a})$ and $\mathbf{d}$.

• Shortest distance from a point to a plane: For point $P(\mathbf{p})$ and plane $(\mathbf{r}-\mathbf{a})\cdot \hat{\mathbf{n}}=0$ (where $\hat{\mathbf{n}}$ is a unit normal), the distance is the absolute value of the perpendicular component: $D=|(\mathbf{p}-\mathbf{a})\cdot \hat{\mathbf{n}}|$. For a plane in Cartesian form $ax+by+cz=d$, the distance from point $(x_0,y_0,z_0)$ is:

$$D=\frac{|a{x}_0+b{y}_0+c{z}_0-d|}{\sqrt{a^2+b^2+c^2}}.$$

• Shortest distance between two skew lines: For lines $\mathbf{r}=\mathbf{a}+\lambda \mathbf{d}$ and $\mathbf{r}=\mathbf{b}+\mu \mathbf{m}$, the connecting vector perpendicular to both lines is $\mathbf{d}\times \mathbf{m}$. The shortest distance is the projection of the vector between points $(\mathbf{a}-\mathbf{b})$ onto this common perpendicular:

$$D=\frac{|(\mathbf{a}-\mathbf{b})\cdot (\mathbf{d}\times \mathbf{m})|}{|\mathbf{d}\times \mathbf{m}|}.$$ The numerator is a scalar triple product, highlighting the interconnectedness of these concepts.

Common Pitfalls

1. Misidentifying Direction and Normal Vectors: A frequent error is using a direction vector where a normal vector is required, or vice-versa. Remember: lines use direction vectors $\mathbf{d}$; planes use normal vectors $\mathbf{n}$. When finding the angle between a line and a plane, using $\cos$ instead of $\sin$ is a direct result of this confusion.

1. Errors in the Cross Product Calculation: The determinant calculation for $\mathbf{a}\times \mathbf{b}$ is prone to sign errors, especially with the middle ($\mathbf{j}$) component which is subtracted. Always compute methodically and verify that your result is perpendicular to both original vectors by checking the dot product is zero.

1. Misinterpreting the Scalar Triple Product: Forgetting to take the absolute value for volume will give a signed volume, which can be negative depending on the orientation of the vectors. In geometric contexts, volume is always positive. Also, confusing the formula for a parallelepiped ($|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$) with that for a tetrahedron ($\frac{1}{6}$ of that value) will lead to an answer that is off by a factor of six.

1. Assuming Lines Intersect in 3D: In 2D, non-parallel lines always intersect. In 3D, this is not true—they can be skew. When solving for an intersection, always check your values of $\lambda$ and $\mu$ satisfy all three component equations. If they don't, the lines are skew.

Summary

• Lines in 3D are described by $\mathbf{r}=\mathbf{a}+\lambda \mathbf{d}$, while planes are described by $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ or $ax+by+cz=d$.
• The cross product $\mathbf{a}\times \mathbf{b}$ yields a vector perpendicular to both, with a magnitude equal to the area of the corresponding parallelogram.
• The scalar triple product, $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, gives the signed volume of the parallelepiped defined by the three vectors; its absolute value is used for geometric volume.
• Angles are found using dot products (cosine) of direction vectors (for lines) or normal vectors (for planes), and sine for the angle between a line and a plane.
• Shortest distances can be calculated using derived vector formulas: for point-line (using cross product magnitude), point-plane (using dot product with a unit normal), and between skew lines (using a combination of cross and dot products in a scalar triple product).