Vector question

...

A point on the line AP will have position vector $k(\beta b + \gamma c )$ for some k.

If you want it to be on BC as well, you need the sum of the two multipliers to be 1.

Now try and get an equation involving k, and hence find its value and hence the position vector of D.

For part ii): I did the following: AP=
BC= $b +\mu(c-b)$. Intersect when: $(1-\mu)b+ \mu c=t \beta b+t \gamma c$. Solving gives $t=\frac{1}{\beta + \gamma}$. So the point of intersection is at: $\frac{\beta}{\beta + \gamma} + \frac{\gamma}{\beta + \gamma}$. But then how do i calcualte the ratio between the two lines...

Well what are BD and DC?

PS: You need a space after "\beta", etc. in your LaTex.

so BD would be: $b+ (\frac{\beta b}{\beta + \gamma}+ \frac{\gamma c}{\beta + \gamma} - b)$

DC is: $\frac{\beta b}{\beta + \gamma} + \frac{\gamma c}{\beta + \gamma} + (c-\frac{\beta b}{\beta + \gamma} + \frac{\gamma c}{\beta + \gamma})$.

which once simplified I'm assuming will cancel down to just $\frac{\gamma c}{\beta b}$ ?.

Was my part i) of the question correct, and if so, what is the significance of t+s=1...? Sorry for all the questions, just trying to fully understand.

BD is AD - AB, i.e. the position vector for D minus the position vector for B.

You seem to have an extra "b+" at the front.

Similar problem with DC.



Don't think so. See what you get.



Just noticed the image for part i). The locus is a straight line from B to C extending either side.

BD is AD - AB, i.e. the position vector for D minus the position vector for B.

You seem to have an extra "b+" at the front.

Similar problem with DC.



Don't think so. See what you get.



Just noticed the image for part i). The locus is a straight line from B to C extending either side.

Yeah, way I drew my diagram on paper was dodgy. I'll retry that now. Can you explain why the locus is a straight line from B to C... I thought it would just pass through the line segment BC. I'm confused :/.

x= sa+tb

But s+t=1, so t=1-s

Thus x=sa+(1-s)b)

= b +(a-b)s

Which is he equation of a line. If s=0 we have b, if s=1 we have a, and thus it's a line through A and B.