Vectors, lines and scalar products

What you’ll learn

• Calculate displacement, distance and unit vectors in three dimensions.
• Find and interpret vector equations of lines.
• Use the scalar product for angles and perpendicularity.

Understand the idea

A position vector locates a point relative to an origin. The displacement from A to B is the position vector of B minus that of A. Magnitude gives distance, and dividing a non-zero vector by its magnitude gives a unit vector in the same direction. Scalar multiplication changes length and may reverse direction.

A line is described by a fixed position vector plus a scalar multiple of a direction vector. Parallel lines have parallel direction vectors. To test intersection, use different parameters for the two lines, solve two coordinate equations and check the third. In three dimensions, non-parallel lines can be skew and never meet.

The scalar product is both the sum of componentwise products and the product of magnitudes times the cosine of their angle. Non-zero perpendicular vectors have scalar product zero. For the acute angle between unoriented lines, take the absolute value of the direction-vector dot product in the cosine formula.

Key formulae

$$\overrightarrow{AB}=\mathbf b-\mathbf a,\qquad |\mathbf a|=\sqrt{a_1^2+a_2^2+a_3^2}$$

$$\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}\quad(\mathbf a\ne\mathbf0)$$

$$\mathbf r=\mathbf a+\lambda\mathbf d$$

$$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta$$

Worked example

Find the line through $A=(1,2,-1)$ and $B=(3,1,3)$, and the distance $AB$.

1. The displacement vector is $B-A=(2,-1,4)$.
2. Use A as a fixed point and this displacement as the direction vector.
3. The distance is $\sqrt{2^2+(-1)^2+4^2}=\sqrt{21}$.

Answer: $\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\4\end{pmatrix}$, with $AB=\sqrt{21}$.