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# Urgent mechanics question !!!! Watch

1. Question 6 B

How do you work it out ?

The examiner's report states :

the first mark was for adding the two vectors together but many students then stated that this sum was equal to (i – 2j) rather than a multiple of it and were unable to make any progress.

I am more interested in the THEORY behind the part in bold. What is the concept ? Why do we equate the multiple of (i-2j) ?
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3. (Original post by Ari Ben Canaan)
Question 6 B

How do you work it out ?

The examiner's report states :

the first mark was for adding the two vectors together but many students then stated that this sum was equal to (i – 2j) rather than a multiple of it and were unable to make any progress.

I am more interested in the THEORY behind the part in bold. What is the concept ? Why do we equate the multiple of (i-2j) ?
R=(4i-5j)+(pi+qj)=(4+p)i + (-5+q)j

So this R is parallel to (i-2j), you equate R to multiple of (i-2j) because the vector R can be (i-2j) or (2i-4j) or (10i-20j) etc. R is parallel to i-2j not equals to i-2j

So
4. (Original post by cazzy-joe)
R=(4i-5j)+(pi+qj)=(4+p)i + (-5+q)j

So this R is parallel to (i-2j), you equate R to multiple of (i-2j) because the vector R can be (i-2j) or (2i-4j) or (10i-20j) etc. R is parallel to i-2j not equals to i-2j

So

Kon Kun Krab

EDIT : A little birdie whispered into my ear that its actually Kob Kun Krab :P

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