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C4 vectors

Can someone help me with this question?
All I get for OC is c, and I can't seem to find pq in terms of c so I'm stuck. Can someone please help? Thank you :smile:
IMG_20181004_145640-compressed.jpg.jpegIMG_20181004_145620-compressed.jpg.jpeg
Reply 1
I thought the old spec died off with the 2018 a levels,shouldn't you be doing pure maths instead of the old C1,C2 etc?
Reply 2
Original post by Danielnovels25
Can someone help me with this question?
All I get for OC is c, and I can't seem to find pq in terms of c so I'm stuck. Can someone please help? Thank you :smile:
IMG_20181004_145640-compressed.jpg.jpegIMG_20181004_145620-compressed.jpg.jpeg


OB and AC are parallel and same length - its a prallelogram
Reply 3
I thought the old spec died off with the final exams in 2018,shouldn't youn be doing pure maths instead?
Original post by n dog
I thought the old spec died off with the final exams in 2018,shouldn't youn be doing pure maths instead?


I think we're the last batch (or one of the last) to do this particular syllabus.
Original post by mqb2766
OB and AC are parallel and same length - its a prallelogram


Oh right... And I'm an A level student🤦🏽*♀️
Thank you :smile:
Reply 6
Original post by Danielnovels25
Oh right... And I'm an A level student🤦🏽*♀️
Thank you :smile:


No problem, the parallelogram info probably helps with the 2nd part as well
Original post by mqb2766
No problem, the parallelogram info probably helps with the 2nd part as well


But look at this example from the book. Is it a mistake? IMG_20181004_151928-compressed.jpg.jpeg
Reply 8
Original post by Danielnovels25
Can someone help me with this question?
All I get for OC is c, and I can't seem to find pq in terms of c so I'm stuck. Can someone please help? Thank you :smile:
IMG_20181004_145640-compressed.jpg.jpegIMG_20181004_145620-compressed.jpg.jpeg

It easy.
BA+OB=OA \vec{BA} + \vec {OB} = \vec{OA}

AB=(ba)AQ=2(ba)3 \vec{AB} = (b-a) \rightarrow \vec{AQ} =\frac {2(b- a)}{3}

Similarly OP+PA=OA \vec{OP} + \vec{PA} =\vec {OA}
giving AP=ab3 \vec{AP} = a - \frac {b}{3}

AQ+PQ=AP\vec{AQ} + \vec{PQ} = \vec{AP}

giving PQ=1(b+a)3 \vec{PQ} = \frac{-1(b+a)}{3}
(edited 5 years ago)
Original post by r_gup
It easy.
BA+OB=OA \vec{BA} + \vec {OB} = \vec{OA}

AB=(ba)AQ=2(ba)3 \vec{AB} = (b-a) \rightarrow \vec{AQ} =\frac {2(b- a)}{3}

Similarly OP+PA=OA \vec{OP} + \vec{PA} =\vec {OA}
giving AP=ab3 \vec{AP} = a - \frac {b}{3}

AQ+PQ=AP\vec{AQ} + \vec{PQ} = \vec{AP}

giving PQ=1(b+a)3 \vec{PQ} = \frac{-1(b+a)}{3}


Thanks. :smile:
I was a bit confused because I had forgotten a parallelogram has 2 parallel and equal sides.
Reply 10
Original post by Danielnovels25
Thanks. :smile:
I was a bit confused because I had forgotten a parallelogram has 2 parallel and equal sides.


Just make sure you do not draw a trapezium next time. I got a bit confused.

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