Maths vectors question

The question is:

OACB is a parallelogram with OA=a and OB =b. The points M,S,N,T divide OB,BC,CA and AO in the ratio 1:4 respectively. The lines ST and MN intersect at the point D. Show that the lines MN and ST bisect one another.

(edited 6 years ago)

Original post by dont know it

The question is:

OACB is a parallelogram with OA=a and OB =b. The points M,S,N,T divide OB,BC,CA and AO in the ratio 1:4 respectively. The lines ST and MN intersect at the point D. Show that the lines MN and ST bisect one another.

Ideally, you want to show that

\overrightarrow{SD} = \frac{1}{2}\overrightarrow{ST}

and

\overrightarrow{MD} = \frac{1}{2}\overrightarrow{MN}

.

So, begin by determining what the point D is in terms of

\mathbf{a} . \mathbf{b}

.

Reply 2

dont know it

OP

9

Original post by RDKGames

Ideally, you want to show that

\overrightarrow{SD} = \frac{1}{2}\overrightarrow{ST}

and

\overrightarrow{MD} = \frac{1}{2}\overrightarrow{MN}

.

So, begin by determining what the point D is in terms of

\mathbf{a} . \mathbf{b}

.

I got SD= u(3/5a-b) and MD=y(3/5b+a). I don't really get what to do after this.

Original post by dont know it

I got SD= u(3/5a-b) and MD=y(3/5b+a). I don't really get what to do after this.

Then you can play around and choose any path you want between two known coordinates, and express it in terms of the two you got.

I.e. We know what the path

\overrightarrow{MS}

is in terms of

\mathbf{a}, \mathbf{b}

. We can also express it in terms of

\overrightarrow{SD}, \overrightarrow{MD}

. Make the two expressions equal to one another, then we must have the scalar multiples of

\mathbf{a}, \mathbf{b}

equal on both sides of the equation. This gives us two linear eqs. in

u,y

to solve simultaneously.

Reply 4