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M1: i,j vectors

Hi. I'm currently having a problem with ANY I,J vector question that involves interception or an entity being parallel to another entity. Could anyone outline methods to these following questions with explanations? I'd really appreciate it.

1) At time t=0 the particle P is at the point with position vector 4i, and moving with constant velocity i+j m/s. A second particle Q is at the point with position vector -3j and moving with velocity v m/s. After 8 seconds, the paths of P and Q meet. Find the speed of Q.

2)Given that c = 3i + 4j and d = i -2j, find lamba if c + lambda*d is parallel to i+j
Reply 1
Avatar for raheem94
raheem94
13
Original post by Sarabande
Hi. I'm currently having a problem with ANY I,J vector question that involves interception or an entity being parallel to another entity. Could anyone outline methods to these following questions with explanations? I'd really appreciate it.

1) At time t=0 the particle P is at the point with position vector 4i, and moving with constant velocity i+j m/s. A second particle Q is at the point with position vector -3j and moving with velocity v m/s. After 8 seconds, the paths of P and Q meet. Find the speed of Q.


For question 1,
First consider the motion of P,
We know,
r0=4i v=i+j t=8 \displaystyle r_0=4i \ v=i+j \ t=8

Use the formula, r=r0+vt \displaystyle r=r_0+vt to find an equation for r \displaystyle r

Now consider the motion of Q,
We have the information,
r0=−3j t=8 \displaystyle r_0=-3j \ t=8
Use the formula r=r0+vt \displaystyle r=r_0+vt to find another equation for 'r' in terms of 'v'.

Now equate the two 'r' equations to find 'v'.
Reply 2
Avatar for baphomet
baphomet
In Q1 it says that the paths of P and Q meet. That is not the same thing as saying P and Q actually intercept. However the question seems worded such that they do rendez-vous.

HINT: Find P's new position at t=8.
Then assume it takes Q 8s to also get to this point from its starting position. From this youcan work out Q's velocity vector and extract its speed.

If this is too vague - let me know.

Q2 HINT: parallel vectors have their i's and j's in the same ratio. Parallel to i+j means any vector whose i and j bits are equal e.g. 3i+3j or even -2i-2j
Reply 3
Avatar for ztibor
ztibor
10
Original post by Sarabande
Hi. I'm currently having a problem with ANY I,J vector question that involves interception or an entity being parallel to another entity. Could anyone outline methods to these following questions with explanations? I'd really appreciate it.

1) At time t=0 the particle P is at the point with position vector 4i, and moving with constant velocity i+j m/s. A second particle Q is at the point with position vector -3j and moving with velocity v m/s. After 8 seconds, the paths of P and Q meet. Find the speed of Q.

2)Given that c = 3i + 4j and d = i -2j, find lamba if c + lambda*d is parallel to i+j


Let the two vector and b wirh following coordinates
a⃗=a1i⃗+a2j⃗\displaystyle \vec a =a_1\vec i+a_2\vec j
b⃗=b1i⃗+b2j⃗\displaystyle \vec b=b_1 \vec i +b_2\vec j
the two vector is parallel when
a2a1=b2b1\displaystyle \frac{a_2}{a_1}=\frac{b_2}{b_1}
Reply 4
Avatar for Sarabande
Sarabande
OP
15
Original post by raheem94
For question 1,
First consider the motion of P,
We know,
r0=4i v=i+j t=8 \displaystyle r_0=4i \ v=i+j \ t=8

Use the formula, r=r0+vt \displaystyle r=r_0+vt to find an equation for r \displaystyle r

Now consider the motion of Q,
We have the information,
r0=−3j t=8 \displaystyle r_0=-3j \ t=8
Use the formula r=r0+vt \displaystyle r=r_0+vt to find another equation for 'r' in terms of 'v'.

Now equate the two 'r' equations to find 'v'.


Thank you. Using that method I was able to get the right answer.
Reply 5
Avatar for raheem94
raheem94
13
Original post by Sarabande
Thank you. Using that method I was able to get the right answer.


No problem.

Do you understand the 2nd question? or do you still need any explanation?
Reply 6
Avatar for Sarabande
Sarabande
OP
15
Original post by ztibor
Let the two vector and b wirh following coordinates
a⃗=a1i⃗+a2j⃗\displaystyle \vec a =a_1\vec i+a_2\vec j
b⃗=b1i⃗+b2j⃗\displaystyle \vec b=b_1 \vec i +b_2\vec j
the two vector is parallel when
a2a1=b2b1\displaystyle \frac{a_2}{a_1}=\frac{b_2}{b_1}


I'm not trying to prove they're parallel, I'm trying to find an unknown when 2 vectors are parallel lol.
Reply 7
Avatar for Sarabande
Sarabande
OP
15
Original post by raheem94
No problem.

Do you understand the 2nd question? or do you still need any explanation?


Yes please
Reply 8
Avatar for raheem94
raheem94
13
Original post by Sarabande
Yes please


c=3i+4j and d=i−2j hence c+λd=3i+4j+λ(1−2j)=3i+4j+λi−2λj=(3+λ)i+(4−2λ)j \displaystyle c=3i+4j \text{ and } d=i-2j \text{ hence } c+\lambda d = 3i+4j+\lambda(1-2j)=3i+4j+\lambda i -2\lambda j = (3+\lambda)i + (4-2\lambda)j

As c+λd \displaystyle c+\lambda d is parallel to i+j, hence its a multiple of it.

e.g. 3i+3j \displaystyle 3i+3j is paralled to i+j because 3i+3j=3(i+j) \displaystyle 3i+3j=3(i+j) so it is a multiple of it.

So now lets say that (3+λ)i+(4−2λ)j=k(i+j) \displaystyle (3+\lambda)i + (4-2\lambda)j =k(i+j) where 'k' is a constant. As they are multiples hence adding the constant will let us make them equal.

(3+λ)i+(4−2λ)j=k(i+j)(3+λ)i+(4−2λ)j=ki+kj \displaystyle (3+\lambda)i + (4-2\lambda)j =k(i+j) \\ (3+\lambda)i + (4-2\lambda)j = ki + kj

Now we get two equations,
3+λ=k and 4−2λ=k \displaystyle 3+\lambda = k \text{ and } 4-2\lambda =k

Solve the equations simultaneously to get the value of λ \displaystyle \lambda

Do you get it?
Reply 9
Avatar for Sarabande
Sarabande
OP
15
Original post by raheem94
c=3i+4j and d=i−2j hence c+λd=3i+4j+λ(1−2j)=3i+4j+λi−2λj=(3+λ)i+(4−2λ)j \displaystyle c=3i+4j \text{ and } d=i-2j \text{ hence } c+\lambda d = 3i+4j+\lambda(1-2j)=3i+4j+\lambda i -2\lambda j = (3+\lambda)i + (4-2\lambda)j

As c+λd \displaystyle c+\lambda d is parallel to i+j, hence its a multiple of it.

e.g. 3i+3j \displaystyle 3i+3j is paralled to i+j because 3i+3j=3(i+j) \displaystyle 3i+3j=3(i+j) so it is a multiple of it.

So now lets say that (3+λ)i+(4−2λ)j=k(i+j) \displaystyle (3+\lambda)i + (4-2\lambda)j =k(i+j) where 'k' is a constant. As they are multiples hence adding the constant will let us make them equal.

(3+λ)i+(4−2λ)j=k(i+j)(3+λ)i+(4−2λ)j=ki+kj \displaystyle (3+\lambda)i + (4-2\lambda)j =k(i+j) \\ (3+\lambda)i + (4-2\lambda)j = ki + kj

Now we get two equations,
3+λ=k and 4−2λ=k \displaystyle 3+\lambda = k \text{ and } 4-2\lambda =k

Solve the equations simultaneously to get the value of λ \displaystyle \lambda

Do you get it?


I understand it now. Thanks so much.
Reply 10
Avatar for raheem94
raheem94
13
Original post by Sarabande
I understand it now. Thanks so much.


Actually there is an alternative technique to this question as well, that's the one which ztibor did.
How do you write a new position vector for p

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