Represent three-dimensional vectors in component form, compute the scalar product and magnitude, and find vector equations of lines and the equation of a sphere

What this dot point is asking

NESA wants you to extend two-dimensional vectors into three dimensions. You must write vectors in component form using $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$, compute magnitude and the scalar (dot) product, find the angle between vectors, write vector equations of lines, decide whether two lines intersect or are skew, and derive the equation of a sphere. Almost every formula is the two-dimensional one with a third term bolted on; the genuinely new ideas are the third axis itself, the possibility of skew lines, and the sphere.

Vectors in component form

A point in space is described by an ordered triple $(x, y, z)$ relative to three mutually perpendicular axes. The position vector of that point is

where $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are the unit vectors along the $x$, $y$ and $z$ axes. To reach the point you step $x$ along $\mathbf{i}$, then $y$ along $\mathbf{j}$, then $z$ along $\mathbf{k}$; the three steps are independent because the axes are perpendicular.

Addition, subtraction and scalar multiplication act component by component, exactly as in two dimensions. The only discipline the third dimension demands is never to silently drop the $\mathbf{k}$ term: a vector you began in two-dimensional habit will be wrong by a whole coordinate.

Magnitude as a space diagonal

The magnitude (length) of $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ is

This is Pythagoras applied twice. The vector is the long diagonal of a rectangular box with edge lengths $a_1$, $a_2$ and $a_3$: the floor diagonal has length $\sqrt{a_1^2 + a_2^2}$, and the space diagonal then has length $\sqrt{(a_1^2 + a_2^2) + a_3^2}$, which is the formula above.

The unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$, and the distance between two points $A$ and $B$ is $|\overrightarrow{AB}| = |\mathbf{b} - \mathbf{a}|$, the magnitude of the displacement.

The scalar product in three dimensions

The scalar (dot) product is

where $\theta$ is the angle between the vectors. The component formula and the geometric formula are two faces of the same number: the first lets you compute it from coordinates, the second lets you extract the angle. Rearranging gives

Two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b} = 0$, because $\cos 90^\circ = 0$. The scalar product is commutative ($\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$), distributes over addition, and satisfies $\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2$, which is the bridge between lengths and the dot product used throughout vector proofs.

Vector equation of a line

A line through the point with position vector $\mathbf{a}$ and parallel to the direction vector $\mathbf{d}\,(\ne\mathbf{0})$ consists of all points

The recipe is the same as in two dimensions, just with three-component vectors. Build it up one piece at a time.

Stage 1, a point and a direction. Fix a point $A$ on the line with position vector $\mathbf{a}$, and a direction $\mathbf{d}$ pointing along the line. Adding one copy of $\mathbf{d}$ to $\mathbf{a}$ lands on a second point of the line.

Stage 2, the whole line. Letting $\lambda$ range over every real number sweeps out the entire line. The position vector $\mathbf{r}$ of a general point $P$ is $\mathbf{a}$ (to get onto the line at $A$) plus $\lambda$ copies of $\mathbf{d}$ (to slide along to $P$). At $\lambda = 0$ you are at $A$; positive $\lambda$ moves one way, negative the other.

In components, $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ reads

so $x = a_1 + \lambda d_1$, $y = a_2 + \lambda d_2$, $z = a_3 + \lambda d_3$. To find the line through two points $A$ and $B$, take $\mathbf{a} = \overrightarrow{OA}$ and direction $\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. The point $\mathbf{a}$ may be any point on the line and $\mathbf{d}$ any non-zero multiple of the direction, so a line has infinitely many vector equations.

To test whether a point lies on the line, solve for a single $\lambda$ that reproduces every component. Set up one equation per coordinate, solve the first for $\lambda$, and check that the same $\lambda$ satisfies the other two. If any coordinate disagrees, the point is off the line, exactly as in the worked exam question above.

Intersecting versus skew lines

Two lines meet where their position vectors coincide. Use different parameters for the two lines and set them equal:

That is one equation per coordinate, so three equations in only two unknowns $\lambda$ and $\mu$. Solve two of them, then check the third: this is where three dimensions differs sharply from two. In the plane, two non-parallel lines always meet. In space they need not.

If $\lambda$ and $\mu$ from the first two equations also satisfy the third, the lines intersect and substituting gives the meeting point. If the directions are parallel ($\mathbf{d}_1 = k\mathbf{d}_2$), the lines are parallel; test a point of one against the other to decide whether they are coincident or distinct. If the directions are not parallel and the third equation fails, the lines are skew: they are neither parallel nor intersecting, an arrangement impossible in a plane.

In the picture the two skew lines appear to cross, but $L_2$ passes in front of $L_1$ at different depths (the break in $L_1$ shows this), so there is no common point. Algebraically, that is precisely the case where the first two equations give a $(\lambda, \mu)$ pair that the third equation rejects.

The equation of a sphere

A sphere is the set of points at fixed distance $R$ (the radius) from a centre $C$ with position vector $\mathbf{c}$:

In Cartesian coordinates with $\mathbf{c} = (a, b, c)$ this is

This is the natural 3D analogue of the circle equation, with the third squared term appearing for the extra coordinate. An expanded form is converted back to centre-radius form by completing the square in each of $x$, $y$ and $z$; the constant left over equals $R^2$.

How exam questions ask about 3D vectors and lines

• "Find the magnitude of" or "the distance between $A$ and $B$": apply $\sqrt{a_1^2 + a_2^2 + a_3^2}$ (to $\mathbf{b} - \mathbf{a}$ for a distance).
• "Find the angle between $\mathbf{u}$ and $\mathbf{v}$": compute $\mathbf{u}\cdot\mathbf{v}$ and both magnitudes, then $\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$.
• "Show that $\mathbf{u}$ and $\mathbf{v}$ are perpendicular": show $\mathbf{u}\cdot\mathbf{v} = 0$; no need to find the angle.
• "Find a vector equation of the line through $A$ and $B$": take $\mathbf{d} = \mathbf{b} - \mathbf{a}$, then state $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$.
• "Does $C$ lie on the line?": solve one coordinate for $\lambda$ and check it reproduces the other two.
• "Do the lines intersect, or are they skew?": set the forms equal with different parameters, solve two coordinates, and test the third; if it fails and the directions are not parallel, say "skew".
• "Find the equation of the sphere with centre $C$ and radius $R$" or "... and write down its centre and radius": state $|\mathbf{r} - \mathbf{c}| = R$, or complete the square to read off $\mathbf{c}$ and $R$.