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# *M1 -- Vector parallel to (i + j) or (i - j)* watch

1. (1) A pedestrian moves with the constant velocity [(2q^2 - 3)i + (q + 2)j] ms^-1.
Given that the velocity of the pedestrian is parallel to the vector (i - j), show that one possible valu of q is -1 and find the other possible value of q.

(2) The velocity of a car at time t seconds is given by:

v = (3t^2 - 2t + 8)i + (5t + 6)j ms^-1.

Find the values of t for which the velocity of the car is parallel to the vector (i + j).

I have noticed that in the markschemes for "parallel to (i + j)" they just make the i component = j component, and for "parallel to (i - j) they make the i component = - (j component). Can someone please explain why that is?

Also if it said the velocity was parallel to something else...e.g. (-7i + 10j)...then what is the method there? Is there a general way of doing the above questions for when the velocity vector is parallel to (xi + yj) ?

2. If it's parallel to (i-j) that means it's moving in that direction. So, it can have any magnitude, as long as its directed along the vector (i-j). So it follows that:
2q^2 - 3 = -(q+2)
and solve...

This is what I do for a general 'parallel to (xi+yj)' question:
If it's parallel, it can have a magnitude 'k' but must be directed along that vector. And let's for example take the velocity in (1).

=> |k(xi+yj)| = |[(2q^2 - 3)i + (q + 2)j|
k^2(x^2 + y^2) = (2q^2 - 3)^2 + (q+2)^2

And carry on from there to find 'k'
Then you say v = k(xi+yj), and sub in the value of k.
3. (Original post by mockel)
If it's parallel to (i-j) that means it's moving in that direction. So, it can have any magnitude, as long as its directed along the vector (i-j). So it follows that:
2q^2 - 3 = -(q+2)
and solve...

This is what I do for a general 'parallel to (xi+yj)' question:
If it's parallel, it can have a magnitude 'k' but must be directed along that vector. And let's for example take the velocity in (1).

=> |k(xi+yj)| = |[(2q^2 - 3)i + (q + 2)j|
k^2(x^2 + y^2) = (2q^2 - 3)^2 + (q+2)^2

And carry on from there to find 'k'
Then you say v = k(xi+yj), and sub in the value of k.
Thanks mate...but lol I still don't quite understand. I understood that "parallel to" means "in the direction of"...and I have used the method you've shown for a general (xi + yj) question before for questions that for example ask:

"Given that the magnitude of the velocity is 4(sqrt13) and the velocity is in the direction (2i + 3j), find the velocity in i and j form"
...in which case the velocity would be 4(2i + 3j)

But if you do the same in question 1 you end up with a messy polynomial to solve...

Hmm the main thing I don't get is why is it that if it's parallel to (i + j) you make the i component = j component? And if it's parallel to (2i + 3j) would you make (2*i component) = (3*j component)?
4. Nah, don't do it for that question. Like you said, messy polynomial.

Basically, they've already given you the velocity with i and j components. And you have the direction that its going in, in i and j components. So it's like equating coefficients. Say the velocity was k(i-j) = ki - kj. Equating coeffs:

k = 2q^2 - 3

-k = q+2
k = -(q+2)

Now, since they're both equal to 'k' we can equate them, thus:
2q^2 - 3 = -(q+2)

etc...

Say it was k(2i+3j), then you can do the same thing:

2k = 2q^2 - 3

3k = q+2

You can divide the first by 2 and the second by 3, and then equate.
5. (Original post by mockel)
Nah, don't do it for that question. Like you said, messy polynomial.

Basically, they've already given you the velocity with i and j components. And you have the direction that its going in, in i and j components. So it's like equating coefficients. Say the velocity was k(i-j) = ki - kj. Equating coeffs:

k = 2q^2 - 3

-k = q+2
k = -(q+2)

Now, since they're both equal to 'k' we can equate them, thus:
2q^2 - 3 = -(q+2)

etc...

Say it was k(2i+3j), then you can do the same thing:

2k = 2q^2 - 3

3k = q+2

You can divide the first by 2 and the second by 3, and then equate.
Ahhhhh there now I get it...lol ....cheers mate...really appreciate it

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