# [Group5][Vector]Exam Question Practice

Hi everyone, hope you will enjoy this question!

2 people like this

##### Share on other sites

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

##### Share on other sites

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

##### Share on other sites

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

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

##### Share on other sites
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?

##### Share on other sites
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!!

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!

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

##### Share on other sites
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.

##### Share on other sites
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?

##### Share on other sites

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?

3BCx is not 18.

Edited by kw0573

##### Share on other sites
16 minutes ago, kw0573 said:

3BCx is not 18.

Is it -18?