My GMA Question Thread

[alchemy]

A vectors question. Need help with (iv). Thanks.

PS: My answers to the written parts, (i) (ii) and (iii), are correct. The answer to question (iv) is

[M_BONG]

A vectors question. Need help with (iv). Thanks.

PS: My answers to the written parts, (i) (ii) and (iii), are correct. The answer to question (iv) is

The question tells you vector XC equals vector 1/2 OX
Since you already have vector OX from iii) you can find it.

Thus, 1/2 (2/3a + 1/3b)
= 1/3a + 1/6b

v) vector AC = vector AX + vector XC
Question tells you vector XA equals 1/2 BX

As XA = 1/2 BX
AX = -(1/2 x BX) - moving in opposite direction.

From part ii.  AX = - [1/2 x 2/3a-2/3b]
= -1/3a + 1/3b

Vector AC = Ax + XC  - connecting head with tail.
XC = 1/3 + 1/6b from part iv.

Therefore AC = -1/3a+1/3b+1/3a+1/6b

AC= 1/2B



               =

[alchemy]

The question tells you vector XC equals vector 1/2 OX
Since you already have vector OX from iii) you can find it.

Oh my, I really need to read the information provided... Thanks for making me realize, LOL.

[alchemy]

My answers for part (iv) and b are mixed up, for some reason. Can you help me to identify the error in my working out. The answers for all the other written parts are already correct, so they can be used to figure out (iv) and b.

My working out:
(iv) BM = BO + OM =-7j + OM
OM = AB/2 = (3j +5i)/2 = 3i/2 + 5j/2
BM = 3i/2 + (5j/2 -7j) = 3i/2 - 9j/2

Answers:
(iv) BM = 3i/2 + 5j/2
(b)  M = (-3/2, 9/2)

[LOLs99]

I think iv answer is wrong?
And OM is not equal to half AB.

That's how i will do it . (Damn, can't upload the pic cuz I did the working out on a paper)

Iv. BA= -3i-5j so BM= 0.5BA= 0.5(-3i-5j)= -1.5i-2.5j

b. Find OM. OB= 7j, BM= -1.5i-2.5j ...... Therefore, OM=OB+BM =7j+(-1.5i-2.5j)= -1.5i+4.5j. So coordinate of M =(-1.5,4.5)
It should be in exact value unless otherwise specified. But it will be messy with fraction without latex so sorry about that
 

[alchemy]

New questions here:

1) Find the point P on the line such that is parallel to the vector .

2) An aircraft is heading south with an airspeed of 600km/h. A 50km/h wind is blowing in a S30W direction. Find the direction in which the aircraft is heading and the ground speed.

3) A and B are defined by the position vectors a = 2i - 2j - k and b = 3i + 4k. Find the unit vector which bisects AOB.

You don't have to answer all questions at once. Just one answer at a time will be gladly appreciated   

[RKTR]

1) O (0,0) P(x,y)
     OP parallel to 3i+j
     gradient = 1/3

     y=1/3 x 

     Sub this into x-6y=11
       x-2x=11
        x=-11
        -11-6y=11
         -6y=22
            y=-11/3
       
       P(-11,-11/3)

2.
 using cosine rule x^2 =600^2 +50^2 -2(600)(50)cos150
                              x=643.79 km/h
using sin rule  643.79/ sin(150)  = 50/sin (a)
                         a=2.23

    so direction = S2.23W  , ground speed = 643.79 km/h 

3. find midpoint of a and b which is 5/2 i - j +3/2 k
    question says bisects AOB
     so 5/2 i -j+3/2 k is your vector
     to find unit vector,find magnitude 1st which is root 10
   so your answer = 1/(root 10) [ 5/2 i - j +3/2 k]

EDIT : answered 2nd question.. sorry for the bad diagram
EDIT : answered 3rd question

[alchemy]

Your answer to question 3 is incorrect. The answer is meant to be.
Have another try 

[alchemy]

C and D are points defined respectively by position vectors c and d. If ,
 and . Find . 

[RKTR]

c.d =lcl x ldl x cos @


Using cosine rule
CD^2 = 25 + 49 -2(4)
CD^2 = 66
CD=root 66

Is it right? I think I forget vectors already

[alchemy]

c.d =lcl x ldl x cos @


Using cosine rule
CD^2 = 25 + 49 -2(4)
CD^2 = 66
CD=root 66

Is it right? I think I forget vectors already

Yes, that's correct. Thank you.
Is anyone able to explain question 3? Here it is again:
3) A and B are defined by the position vectors a = 2i - 2j - k and b = 3i + 4k. Find the unit vector which bisects AOB.

4) Simplify cos(u-v) sin v + sin(u-v) cos v

[alchemy]

Uhh, anybody able to give me some assistance on the last question I posted up?

Anyways, attached is another question in which my working/answers to parts (I) and (ii) are correct. I need help on (b)(i).

Cheers,

[brightsky]

1. A and B are defined by the position vectors a = 2i - 2j - k and b = 3i + 4k. Find the unit vector which bisects AOB.

In general, how would you go about finding the unit vector which bisects two vectors a and b? There are infinitely many vectors which bisect a and b. We need only to find one of the vectors (which would give us the desired direction) and then adjust the magnitude to form the unit vector.

Now. We know that the perpendicular bisector of the base of an isosceles triangle passes through the apex. It also happens to bisect the angle of the apex. So we want to create an isosceles triangle. How do we do that? We know we can adjust the magnitude of a and b to our heart's content. So why not adjust the magnitude to 1? So we have (a hat) and (b hat). (Sorry, I ceebs with latex, but those are unit vectors.) Then if we put these new unit vectors tail to tail, and then connect their endpoints, we have an isosceles triangle. We know, from the above, that the perpendicular bisector of the line connecting the endpoints will bisect the angle AOB. So find that. The perpendicular bisector is 1/2 (a hat + b hat).

Good. We have the desired direction now. All we have to do is adjust the magnitude to 1 by dividing the vector by its original magnitude. Then we are done.

2. Simplify cos(u-v) sin v + sin(u-v) cos (v).

Recall the compound angle formula for sine. sin(a+b) = sin(a)cos(b) + cos(a)sin(b). Bear any resemblance to the expression above? In this case, we have a = v and b = u-v, because sin(v + u - v) = sin(v)cos(u-v) + cos(v) sin(u-v). Therefore the simplified expression is sin(v + u - v) which is equal to sin(u).

3. Given that OPQ is a straight line, find i) the value of k, ii) the ratio OP/PQ.

i)

Have you heard of the term collinear? Basically points A, B, and C are collinear if they lie in a straight line. So in this case, we are given that O, P and Q are collinear. What does this mean. Well it means that OQ = q*OP (should have vector symbols on top), for some constant q, or in other words that OQ and OP are parallel.

So far so good. Now, what do we know? We know that:

OP = OA + 2/3*AB = a + 2/3*(b-a) = 1/3*a + 2/3*b
OQ = OC + 6/7*CB = k*a + 6/7*(b-k*a) = 1/7 k*a + 6/7*b

So we have:

OQ = q*OP
1/7*k*a + 6/7*b = 1/3*q*a + 2/3*q*b

Now we can equate coefficients (because a and b are linearly independent, i.e. they are not parallel):

1/7 k = 1/3 q....(1)
6/7 = 2/3 q....(2)

From (2), we get q = 9/7. Sub into (1), we get k = 3. Yay.

ii)

So we know now that:

OQ = q*OP = 9/7*OP

Now we know that OQ = OP + PQ. So:

OP + PQ = 9/7*OP
PQ = 2/7*OP

These are vectors PQ and OP but we can glean from this that the magnitude of PQ is 2/7 times the magnitude of OP. So:

PQ = 2/7*OP (magnitudes now not vectors)

Divide through:

OP/PQ = 7/2. Yay.

[alchemy]

The above has been the most comprehensive answer so far. 

[alchemy]

1) Find the locus of the point P as it moves such that the difference of its distances from two fixed points A(4,0) and B(-4,0) is 6 units. That is PA-PB=6.
------------------------------- My response
So far I've got:
What is the best method to simplify the above?