Quick questions on Vector geometry applications explained: H2 Further Mathematics

The foot $F$ of the perpendicular from a point $P$ to a line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ is the point on the line closest to $P$. Parametrise $F = \mathbf{a} + \lambda\mathbf{d}$, then impose

The foot $F$ of the perpendicular from $P$ to a plane lies along the normal $\mathbf{n}$ through $P$. Write the line $\mathbf{r} = \mathbf{p} + t\mathbf{n}$ and substitute into the plane equation to find $t$, then $F$. Equivalently, step from $P$ along $\mathbf{n}$ by the signed distance.

The reflection $P'$ of $P$ in a line or plane is on the far side of the mirror, the same distance away. Once the foot $F$ is known, the foot is the midpoint of $P$ and $P'$, so

Position vectors turn geometry into algebra. Useful tools: the midpoint of $A$ and $B$ is $\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$; the point dividing $AB$ in ratio $m : n$ is $\dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$ (the ratio theorem); two segments are parallel when their vectors are scalar multiples, and three points are collinear when two of the joining vectors are parallel. Showing such relations proves results like "the diagonals of a parallelogram bisect each other".

The defining condition is $\overrightarrow{PF}\cdot\mathbf{d} = 0$; guessing $F$ without it gives the wrong point.

The point dividing $AB$ in ratio $m : n$ weights $\mathbf{b}$ by $m$ and $\mathbf{a}$ by $n$ (the opposite of the naive guess); check with an endpoint.

State the condition that determines the foot of the perpendicular from $P$ to a line with direction $\mathbf{d}$. [1 mark]

If $F$ is the foot of the perpendicular from $P$ to a plane, write the reflection $P'$. [1 mark]

Write the position vector of the midpoint of points with position vectors $\mathbf{a}$ and $\mathbf{b}$. [1 mark]