Porisma

Worked solution

Locate internal and external division points in coordinates

Given the line segment ABAB, where A(3,1,2)A(3,1,2) and B(2,3,5)B(2,3,5), find the coordinates of PP on (i) ABAB with AP:PB=2:3AP:PB=2:3; (ii) ABAB produced with AP:PB=2:1AP:PB=2:1.

CJC · Vectors Basic · Example 6

(i) Internal point

Answer

P=(13/5,9/5,16/5)P=(13/5,9/5,16/5).

Weight the two positions

OP→=3OA→+2OB→3+2=15[3(312)+2(235)]=15(13916).\begin{aligned}\overrightarrow{OP}&=\frac{3\overrightarrow{OA}+2\overrightarrow{OB}}{3+2}\\&=\frac15\left[3\begin{pmatrix}3\\1\\2\end{pmatrix}+2\begin{pmatrix}2\\3\\5\end{pmatrix}\right]=\frac15\begin{pmatrix}13\\9\\16\end{pmatrix}.\end{aligned}

P lies between A and B. AP:PB=2:3 gives weights 3 for A and 2 for B.

How did this part go?

(ii) Beyond B

Answer

P=(1,5,8)P=(1,5,8).

Recognise the midpoint B

OB→=12(OA→+OP→)\overrightarrow{OB}=\frac12(\overrightarrow{OA}+\overrightarrow{OP})

The order is A,B,P. AP=2PB implies AB=PB, so B is the midpoint of AP.

Rearrange before substituting

OP→=2OB→−OA→=2(235)−(312)=(158)\overrightarrow{OP}=2\overrightarrow{OB}-\overrightarrow{OA}=2\begin{pmatrix}2\\3\\5\end{pmatrix}-\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}1\\5\\8\end{pmatrix}

This external point uses a negative coefficient for A, rather than positive internal weights.

How did this part go?

Solution notes

Both Student and Teacher copies print the complete working and the two ratio diagrams on page 14.