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C4 vectors questions

where did (1, 2, 1) come from on part b)ii? I used (2, 4, 2) and still got the right answer.
Reply 1
Is my diagram ok for this question and would I use the dot product of AB and AD to find the coordinates of D (part c)?
(edited 6 years ago)
They're basically the same thing when considering angles, so it doesn't matter. In the end, 1(4) - 1(4) = 0 is the same as 2(4) - 2(4) = 0.
Original post by kiiten
where did (1, 2, 1) come from on part b)ii? I used (2, 4, 2) and still got the right answer.


That's because (2, 4, 2) is a multiple of (1, 2, 1) - both have the equivalent direction just different sizes, and it's only the direction that you care about.
Original post by kiiten
where did (1, 2, 1) come from on part b)ii? I used (2, 4, 2) and still got the right answer.


Basically, since this is a movement vector, as long as the proportion is the same, you should be fine :smile:
Reply 5
The mark scheme is just using 1/2 the vector you did but it doesn't make a difference to the angle between them. You can make the vector as long as you want and you'll see that in the calculation it cancels out o give the right answer.
Original post by kiiten
Is my diagram ok for this question and would I use the dot product of AB and AD to find the coordinates of D (part c)?


No the diagram doesn't seem correct.

For D you would use the fact that ODCD=0\displaystyle \overrightarrow{OD} \cdot \overrightarrow{CD} = 0
(edited 6 years ago)
Reply 7
Original post by kiiten
Is my diagram ok for this question and would I use the dot product of AB and AD to find the coordinates of D (part c)?

You need to work on your diagram. In your diagram angle ODC is not 90 and OC2OB\overrightarrow{OC} \neq 2 \overrightarrow{OB}
Reply 8
Original post by RDKGames
That's because (2, 4, 2) is a multiple of (1, 2, 1) - both have the equivalent direction just different sizes, and it's only the direction that you care about.



So would i still get a mark for my answer?

Have they just simplified the vector like with a fraction?
Original post by kiiten
So would i still get a mark for my answer?

Have they just simplified the vector like with a fraction?


Yes you would. They divided by 2.
Reply 10
Original post by RDKGames
No the diagram doesn't seem correct.

For D you would use the fact that ODCD=0\displaystyle \overrightarrow{OD} \cdot \overrightarrow{CD} = 0


Original post by notnek
You need to work on your diagram. In your diagram angle ODC is not 90 and OC2OB\overrightarrow{OC} \neq 2 \overrightarrow{OB}


Im not sure what you mean OC = 2OB ?

zz.png
Reply 11
Original post by kiiten
Im not sure what you mean OC = 2OB ?

zz.png

That's better but I feel you would be better off without drawing an axis - it is restricting your diagrams. Also you are not told that angle OAB is 90 so it's best not to assume it.

I'm going to post a diagram I've made again:



The key things we know for sure are that the line ll is parallel to AB, angle ODC is 90 and OC = 2OB so this is all shown on the diagram.

SInce OD\overrightarrow{OD} and AB\overrightarrow{AB} are perpendicular, it is true that ODAB=0\overrightarrow{OD}\cdot \overrightarrow{AB}=0.

You already have AB\overrightarrow{AB} and you know that OD\overrightarrow{OD} is the position vector of a point that lies on ll. Any ideas how to carry on?
(edited 6 years ago)
Reply 12
Attachment not found
Original post by notnek
That's better but I feel you would be better off without drawing an axis - it is restricting your diagrams. Also you are not told that angle OAB is 90 so it's best not to assume it.

I'm going to post a diagram I've made again:



The key things we know for sure are that the line ll is parallel to AB, angle ODC is 90 and OC = 2OB so this is all shown on the diagram.

SInce OD\overrightarrow{OD} and AB\overrightarrow{AB} are perpendicular, it is true that ODAB=0\overrightarrow{OD}\cdot \overrightarrow{AB}=0.

You already have AB\overrightarrow{AB} and you know that OD\overrightarrow{OD} is the position vector of a point that lies on ll. Any ideas how to carry on?


I cant see your diagram right now but would you do the dot product of AB and OD = 0 ?
(edited 6 years ago)
Reply 13
Original post by notnek
That's better but I feel you would be better off without drawing an axis - it is restricting your diagrams. Also you are not told that angle OAB is 90 so it's best not to assume it.

I'm going to post a diagram I've made again:



The key things we know for sure are that the line ll is parallel to AB, angle ODC is 90 and OC = 2OB so this is all shown on the diagram.

SInce OD\overrightarrow{OD} and AB\overrightarrow{AB} are perpendicular, it is true that ODAB=0\overrightarrow{OD}\cdot \overrightarrow{AB}=0.

You already have AB\overrightarrow{AB} and you know that OD\overrightarrow{OD} is the position vector of a point that lies on ll. Any ideas how to carry on?


I know it doesnt say but if OD and AB are perpendicular then isnt angle OAB = 90?
Reply 14
Original post by kiiten
I know it doesnt say but if OD and AB are perpendicular then isnt angle OAB = 90?

Why do you think that?
Reply 15
Original post by kiiten
Attachment not found


I cant see your diagram right now but would you do the dot product of AB and OD = 0 ?

That seems okay at first glance - did you get the right answer?

And do you understand the working?

I'm on my phone so can't properly help today but I can tomorrow.
Reply 16
Original post by notnek
Why do you think that?


If 2 lines are perpendicular doesnt that mean the angle between is 90?

Original post by notnek
That seems okay at first glance - did you get the right answer?

And do you understand the working?

I'm on my phone so can't properly help today but I can tomorrow.


Oh thats ok, i appreciate your help :smile:

Yeah i got it right but i would have initially used the dot product of OD and DC (not AB) because like you said, it mentions that ODC = 90 so they are definitely perpendicular.
Reply 17
Original post by kiiten
If 2 lines are perpendicular doesnt that mean the angle between is 90?



Oh thats ok, i appreciate your help :smile:

Yeah i got it right but i would have initially used the dot product of OD and DC (not AB) because like you said, it mentions that ODC = 90 so they are definitely perpendicular.

You said OAB is a right angle. This would mean that OA and AB are perpendicular but you are not told this.

I'm on my phone so someone will hopefully correct me if I'm talking rubbish :smile:
Reply 18
Original post by notnek
You said OAB is a right angle. This would mean that OA and AB are perpendicular but you are not told this.

I'm on my phone so someone will hopefully correct me if I'm talking rubbish :smile:


Ohh sorry yes i see what you mean because you dont know if A extends to the left of O (like in your diagram). Sorry if that doesnt make sense :redface:

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