Vector and Cartesian equations of linesAQA A-Level Further Maths: Revision notes

A line in 3D is fixed by a point on it and a direction. If $A$ has position vector $\mathbf a$ and $\mathbf d$ is a direction vector, every point on the line has position vector $\mathbf r=\mathbf a+\lambda\mathbf d,$ where the parameter $\lambda$ takes every real value. Each value of $\lambda$ gives one point. In components, $\mathbf r=(a_1+\lambda d_1,\;a_2+\lambda d_2,\;a_3+\lambda d_3)$. The equation is not unique: any point on the line can be the base point and any non-zero multiple of $\mathbf d$ can be the direction. To test whether a point lies on the line, equate one component to find $\lambda$, then check the other two give the same $\lambda$.

For points $A$ and $B$ with position vectors $\mathbf a$ and $\mathbf b$, the direction is $\overrightarrow{AB}=\mathbf b-\mathbf a$ (end minus start), so $\mathbf r=\mathbf a+\lambda(\mathbf b-\mathbf a).$ Example: $A(1,2,-1)$, $B(3,-2,3)$ give $\overrightarrow{AB}=(2,-4,4)$ and $\mathbf r=(1,2,-1)+\lambda(2,-4,4)$. This can be simplified to direction $(1,-2,2)$. At $\lambda=0$ you are at $A$ and at $\lambda=1$ you are at $B$.

Write the vector equation as $x=a_1+\lambda d_1$, $y=a_2+\lambda d_2$, $z=a_3+\lambda d_3$ and make $\lambda$ the subject of each: $\frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}\;(=\lambda).$ So $\mathbf r=(4,-3,1)+\mu(2,-1,3)$ becomes $\frac{x-4}{2}=\frac{y+3}{-1}=\frac{z-1}{3}$. In reverse, read the point from the numerators (change the signs: $y+3$ means $-3$) and the direction from the denominators. If a direction component is $0$, you cannot divide by it. For direction $(0,2,5)$ through $(3,1,-2)$ write $x=3,\;\frac{y-1}{2}=\frac{z+2}{5}$.

Two lines are parallel if their direction vectors are multiples of each other. Parallel lines are the same line only if a point on one also lies on the other. Example: $\mathbf r=(2,-1,5)+\lambda(2,1,-1)$ and $\mathbf r=(1,4,-2)+\mu(4,2,-2)$ have directions $(2,1,-1)$ and $2(2,1,-1)$, so are parallel. Testing $(1,4,-2)$ in the first line: $\lambda=-\frac12$ from $x$, but then $y=-\frac32\ne4$. So they are parallel and distinct. A line parallel to a given line through a new point uses the same direction vector with the new base point.