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# C4 - Seeing if vectors intersect, why do we only see if the two K values are equal? watch

1. When they have solved it for S and t, they then sub those values into both k values (the z directions) to see if they are equal. Obviously if they are equal then they intersect, and if not then they don't. But how come you can tell by checking only if the Z values/ K values are equal on both sides? How come you don't have to check it for the X values, the Y values and then the Z values?

For example, in this question, the i/x values are equal (both equal 8) but they don't intersect since the z values are different.

On a sidenote too, should I be using ijk or xyz in the above terminology? What's the difference?
2. (Original post by vector12)

When they have solved it for S and t, they then sub those values into both k values (the z directions) to see if they are equal. Obviously if they are equal then they intersect, and if not then they don't. But how come you can tell by checking only if the Z values/ K values are equal on both sides? How come you don't have to check it for the X values, the Y values and then the Z values?
The two equations they used to find s and t, come from equating the first pair of coordinates, and from equating the second pair of coordinates. Any solution will automatically satisfy those first two coordinates, so we only need to check the third.

On a sidenote too, should I be using ijk or xyz in the above terminology? What's the difference?
i,j,k are vectors. x,y,z refer to directions. What you have in the parentheses are coordinates. For the top number, I'd say the first coordinate, by preference. x-coordinate is probably acceptable.
3. (Original post by ghostwalker)
The two equations they used to find s and t, come from equating the first pair of coordinates, and from equating the second pair of coordinates. Any solution will automatically satisfy those first two coordinates, so we only need to check the third.

i,j,k are vectors. x,y,z refer to directions. What you have in the parentheses are coordinates. For the top number, I'd say the first coordinate, by preference. x-coordinate is probably acceptable.
Oh okay, so if you used the x coordinate and the z coordinate to solve for s and t, you would then need to use the y coordinate to check if they're equal? Is that right?

EDIT: I've just tried solving for s and t using the x and z components and I've ended up with:

6+27t = 9+6s
-6-2t=4+6s

Which gave me 12+29t = 5, and so 29t=-7. But I already know from the mark scheme that the z components aren't equal, so would I have been wrong to try and solve it using the x and z components rather than the x and y components?
4. (Original post by vector12)
Oh okay, so if you used the x coordinate and the z coordinate to solve for s and t, you would then need to use the y coordinate to check if they're equal? Is that right?
Yep - you got it.

And sorry to be a pain, but I have just said i, j, k rather than x,y,z again there?
If you're asking "should you have said ...", then no. x,y,z is fine.
5. (Original post by vector12)

EDIT: I've just tried solving for s and t using the x and z components and I've ended up with:

6+27t = 9+6s
-6-2t=2+6s

Which gave me 12+29t = 7, and so 29t=7. But I already know from the mark scheme that the z components aren't equal, so would I have been wrong to try and solve it using the x and z components rather than the x and y components?
It is NOT incorrect to use x and z.

Saying "the z components aren't equal" needs qualifying - below:

If you choose x,y, then where the lines have the same x,y coordinates, they will have different z coordinates.

If you choose x,z, then where the lines have the same x,z coordinates, they will have different y coordinates.

I would choose any pair of coordinates to start that look as if they might give an easy couple of equations - i.e. go for the ones with the smaller numbers.
Edit: That said, in all likelihood, most markschemes / solutions will probably just use x,y. Anything else would be listed as an alternative solution. Fine in an exam, but possibly not present in "worked solutions".

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