Tricky questions

[luffy]

Here's another:

If the position vectors of A, B and C are , and respectively, determine whether the lines and intersect each other.

The easiest thing is to construct vector paths along these two lines. (i.e. if a plane flies with some vector velocity from some initial position vector, blah blah)




If at any point OA(t) = BC(s) for some arbitrary values of t and s, then the two paths do cross. We thus have a system of 3 equations (equating x y and z) and 2 variables (s and t):


Solving the first two simultaneously, we get s=0.5, t=0
Substituting into the last eqn, we see that this does not satisfy the final relationship. Thus the two lines do not cross.

For 3 dimensional vectors, how do we check if the vectors 'paths' cross? (i.e. they don't necessarily intersect)

[Mao]

@luffy, exactly as I did.

Let and be two vector paths. We seek the point of intersection, i.e. when the coordinates are the same, though the time doesn't have to be the same.

If such a point of intersection exists, let this point be , and reaches at time . reaches at time . Then, we're saying .

If this point of intersection exists, then there must be a set of and at which is true. This may sound tricky to find, but is a set of 3 simultaneous equations. You would solve the first two equations to find out possible value(s) of and , then substitute into the third equation to see if they fit.

[funkyducky]

Mao, would you say the second question is outside of the spesh syllabus? I know how to solve this stuff from what I've learnt in unimaths, but other spesh kids wouldn't have an easy time of it...

[luffy]

@luffy, exactly as I did.

Let and be two vector paths. We seek the point of intersection, i.e. when the coordinates are the same, though the time doesn't have to be the same.

If such a point of intersection exists, let this point be , and reaches at time . reaches at time . Then, we're saying .

If this point of intersection exists, then there must be a set of and at which is true. This may sound tricky to find, but is a set of 3 simultaneous equations. You would solve the first two equations to find out possible value(s) of and , then substitute into the third equation to see if they fit.

Fair enough. That seems really long-winded.

Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?

[dc302]

@luffy, exactly as I did.

Let and be two vector paths. We seek the point of intersection, i.e. when the coordinates are the same, though the time doesn't have to be the same.

If such a point of intersection exists, let this point be , and reaches at time . reaches at time . Then, we're saying .

If this point of intersection exists, then there must be a set of and at which is true. This may sound tricky to find, but is a set of 3 simultaneous equations. You would solve the first two equations to find out possible value(s) of and , then substitute into the third equation to see if they fit.

Fair enough. That seems really long-winded.

Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?

If you just want the intersection of the paths, regardless of time, then yes that should also work.

[Mao]

@luffy, exactly as I did.

Let and be two vector paths. We seek the point of intersection, i.e. when the coordinates are the same, though the time doesn't have to be the same.

If such a point of intersection exists, let this point be , and reaches at time . reaches at time . Then, we're saying .

If this point of intersection exists, then there must be a set of and at which is true. This may sound tricky to find, but is a set of 3 simultaneous equations. You would solve the first two equations to find out possible value(s) of and , then substitute into the third equation to see if they fit.

Fair enough. That seems really long-winded.

Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?

That is exactly what I'm saying. =S awks.
I guess I should explain myself clearer next time.

[Mao]

Mao, would you say the second question is outside of the spesh syllabus? I know how to solve this stuff from what I've learnt in unimaths, but other spesh kids wouldn't have an easy time of it...

Which is the second question?

The first locus question is a bit too hard for specialist level imo. The second and third locus questions aren't too difficult (though it relies on the result of the first question).

The vector question isn't above the syllabus, I have encountered them in my time in specialist.  Though this type of questions would be considered very difficult at this level.