M1 -- Vector parallel to (i + j) or (i - j)

(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) ?

Thx in advance!

Reply 1

m:)ckel

14

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.

Reply 2

devesh254

OP

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)?

Reply 3

m:)ckel

14

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.

Reply 4

devesh254

OP

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.