tutorinseoul

[Group5][Vector]Exam Question Practice

Hi everyone, hope you will enjoy this question! 

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Posted (edited)

I really like this question, but my god it's the hardest Vectors question I've ever attempted doing since I mostly stick to Section A Maths HL questions only when practicing HL.

Also, would it be possible for you to PM me the answers? Also wondering if all the questions you've posted so far are all Paper 1 questions, or if they require calculator, you'll point it out.

Thanks and keep up with the questions! Doing my best to do them all! 

Edited by IB`ez

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Heya, this is SL and paper 2 :)! share your answer and i will check them!

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Posted (edited)

8. a) i. AB = AO + OB = (3, 1, 7) 

       ii. |AB|= (9+1+49)^0.5= (59)^0.5

   b) OC OA= AC 

      (x+1)= -1

      Cx=-2

      (y-0)= 1

      Cy=1

      (z-4)= -1

      Cz=3 

   c) i. angle ADB= 180-theta 

      ii. Area of triangle ABD= 1/2 (BD) (AD) (sin180-theta) 

  d) ( (1/2) (BD) (AD) (sin180-theta) ) / (1/2) (AD) (DC) (sintheta) =3

    (BD) (sin180-theta) / (DC) (sintheta) =3

   BD/DC=3

   BD=3DC

   DC=BD/3

Since BC= BD + DC

         BC= BD + (1/3) BD

              = (4/3) BD

Therefore, BD/BC=3/4

e) 4BD=3BC 

   4x-16=3(6)

  4x=24

  Dx=10

   4y-4=3(0)

   Dy=1

   4z-12=3(0)

   Dz=3 

* I'm not sure whether these are correct * 

 

Edited by IBstudent21

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On 3/27/2017 at 2:01 PM, IBstudent21 said:

8. a) i. AB = AO + OB = (3, 1, 7) 

If AB = AO + OB, what is AO and how do you add that to OB?

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13 hours ago, kw0573 said:

If AB = AO + OB, what is AO and how do you add that to OB?

I just saw the mistake I made!! :o 

I saw it as OA, and therefore added the vector OA to OB.

AO is still easy to find though, so you just flip the signs, since OA= -AO

Therefore, AO becomes (1,0,-4)

So, AB= (5,1,-1)

|AB| changes as well, I'll fix both in my other post later. 

Thanks for checking it for me! :D 

I hope something like that doesn't happen in either of my papers. I keep making these stupid mistakes everywhere.  

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On 3/27/2017 at 2:01 PM, IBstudent21 said:

e) 4BD=3BC 

   4x-16=3(6)

  4x=24

  Dx=10

Similarly, you if you write OB = (4, 1, 3) beside OC = (-2, 1, 3). It should be clear that D, on line segment BC, has a x coordinate between -2 and 4. 

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49 minutes ago, kw0573 said:

Similarly, you if you write OB = (4, 1, 3) beside OC = (-2, 1, 3). It should be clear that D, on line segment BC, has a x coordinate between -2 and 4. 

Oh yeah, you're right. How do I find it though, if it isn't the midpoint? 

Is the method I used wrong? :P 

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Posted (edited)

7 minutes ago, IBstudent21 said:

Oh yeah, you're right. How do I find it though, if it isn't the midpoint? 

Is the method I used wrong? :P 

3BCx is not 18. 

Edited by kw0573

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