Quick questions on Vector and Cartesian equations of lines in QCE Specialist Mathematics Unit 3

A line through a point with position vector $\mathbf{a}$ and direction $\mathbf{d}$ has vector equation

Eliminating the parameter $t$ (when each $d_i \neq 0$) gives the symmetric Cartesian form

A common starting point is two given points $A$ and $B$ rather than a point and a direction. Take the direction to be the displacement $\vec{AB} = \mathbf{b} - \mathbf{a}$ and the base point to be either $A$ or $B$; both choices describe the same line. This is why a line has many equivalent vector equations: any point on the line and any nonzero scalar multiple of the direction will do. When checking whether a given point lies on a line, substitute its coordinates into the parametric equations and confirm the same value of $t$ satisfies all three.

The angle between two lines is the angle between their direction vectors, found with the dot product:

To find how far a point $P$ is from a line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$, take the displacement $\vec{AP}$ from the base point, then use the cross product: the perpendicular distance is $\dfrac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|}$. This works because $|\vec{AP} \times \mathbf{d}| = |\vec{AP}||\mathbf{d}|\sin\theta$ is the area of the parallelogram, and dividing by the base length $|\mathbf{d}|$ leaves the height, which is exactly the perpendicular distance.