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1. C4 vectors- coordinates of intersection
Hello there

I was just doing the Jan 06 paper when I came across the vectors question, and part (b) is the bit I'm stuck on. I understand that to find the point of intersection you equate 2 equations , usually, but I'm not really sure how to go about this one. I looked at the mark scheme but I don't understand why they multiply the l1 equation by (1,1,-1) and I'm just a bit confused

I've attached the question and the corresponding bit of the mark scheme as well, thanks in advance

2. Re: C4 vectors- coordinates of intersection
let's call P xi+yj+zk it is on l1 so you can equate it to the equation, but you also know it is perpendicular to l1 so the dot product (of l1 and OP (which is P)) will be zero. You can then use simultaneous equations
3. Re: C4 vectors- coordinates of intersection
(Original post by Silkysam)
let's call P xi+yj+zk it is on l1 so you can equate it to the equation, but you also know it is perpendicular to l1 so the dot product (of l1 and OP (which is P)) will be zero. You can then use simultaneous equations
Thanks for your help but I still don't understand- I get that the dot product will be 0 and I have to solve it etc I just don't understand why they multiply it by (1,1,-1) ? Where does that come from? If I multiply my line equation coordinates by xi yj zk surely there are too many unknowns to solve because I don't know what the lamda is yet?
4. Re: C4 vectors- coordinates of intersection
(Original post by Nic93)
Thanks for your help but I still don't understand- I get that the dot product will be 0 and I have to solve it etc I just don't understand why they multiply it by (1,1,-1) ? Where does that come from? If I multiply my line equation coordinates by xi yj zk surely there are too many unknowns to solve because I don't know what the lamda is yet?
You can cancel one unknown through simultaneous equations, but as you rightly pointed out you still have two left. In this case you can just let one of the unknown's equal a constant (any you like, but obviosuly using 1 makes the most sense), you can then work out the last unknown left.

With the first unknown found you can use this value and the simultaneous equations to work out the next one and then you both values to work out the final unknown.

Edit: Scratch that, I just worked through the question properly. It's slightly easier than I said. You know the i component of l1 equals x, j component equals y etc. you can therefore put this into the dot product equation, allowing you to work out lamda. Once you have lamda you can you the equation for l1 to work out P.

Edit2: Also I'm guessing the (1,1,-1) in the mark scheme comes from that being the direction vector of l1 (as the dot product is only interested the direction vector and not the position vector).
Last edited by Silkysam; 20-06-2012 at 22:41.
5. Re: C4 vectors- coordinates of intersection
(Original post by Silkysam)
You can cancel one unknown through simultaneous equations, but as you rightly pointed out you still have two left. In this case you can just let one of the unknown's equal a constant (any you like, but obviosuly using 1 makes the most sense), you can then work out the last unknown left.

With the first unknown found you can use this value and the simultaneous equations to work out the next one and then you both values to work out the final unknown.

Edit: Scratch that, I just worked through the question properly. It's slightly easier than I said. You know the i component of l1 equals x, j component equals y etc. you can therefore put this into the dot product equation, allowing you to work out lamda. Once you have lamda you can you the equation for l1 to work out P.

Edit2: Also I'm guessing the (1,1,-1) in the mark scheme comes from that being the direction vector of l1 (as the dot product is only interested the direction vector and not the position vector).

Hiya thanks again for the help, I've just looked around for similar type questions (Jun 06 is practically the same!!) as well, and reading your explanation, and it's a lot clearer to me now thanks I didn't realise why you had to really equate the components at first but now reading this helped me see you had to plug in your x/y/z's into the dot product equation thing, and set that equal to zero? Many thanks for your help
6. Re: C4 vectors- coordinates of intersection
Yea OP (also just P as O is the origin) is perpendicular to l1 therefore the dot product will be zero, and the dot product is the sum of each indivdual compoenent multiplied (i.e: (1)(x)+(1)(y)+(-1)(z)=0). Would you like me to post a picture of my workings?