The line r=(2i+3j-k) +t(-i+3j+k) passes through A (where t=1) and B (where t=5). Find the position vector of point C where AC:BA=-1:4
Did you not even try to do this yourself? What are A and B? BA is the distance between them. AC is the signed between this new point, C, and A. I would just write it as CA: BA= 1:4 C is 4 times as far from A as B is. Rather than calculating the direct line distance from A to B or A to C, by "similar triangles" it is sufficient to find C such that the difference of each coordinate satisfies that condition.
Actually I did try, and I asked because I couldn't go any further. FYI, I did eventually get the answer.
A: <1,6,0> B: <-3,18,4> BA= <4, -12,-4> AC:BA= -1:4 CA:BA=1:4 Dividing BA by 4, <1, -3, -1> CA=<1,6,0> - OC
Therefore, OC=9j+k
PS, please don't feel obliged to reply if it's such trouble