(A) By considering the scalar product a.b, where a=a1i+a2j+a3k and b=b1i+b2j+b3k , prove that
(a1b1+a2b2+a3b3)^2 <=
(a1^2+a2^2+a3^2)( b1^2+b2^2+b3^2)
(B) Three vectors p, q and r are such that p×3q=2p×r. Find a linear relationship between p, q and r.
Please help me for this question.Just started vectors as the first topic. Question adopted from mjc h2 math tutorial.
(A) (a.b)^2 = (a1b1+a2b2+a3b3)^2
also (a.b)^2 = (|a||b| cos theta)^2
and |a|^2 = (a1^2+a2^2+a3^2)
|b|^2 = b1^2+b2^2+b3^2
So (a1b1+a2b2+a3b3)^2 = (a1^2+a2^2+a3^2)( b1^2+b2^2+b3^2) (cos theta)^2
since cos theta <= 1, hence, (a1b1+a2b2+a3b3)^2 <= (a1^2+a2^2+a3^2)( b1^2+b2^2+b3^2)
(B) px3q = 2pxr
px3q = px2r
px3q - px2r = 0
px(3q-2r) = 0
p is parallel to 3q - 2r
Thank you
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Can anyone help me with this qns? Qns 11 (iii)
www.instagram.com/p/BD2_tjlSTcC/
Q11 or Q12?
Sry is qns 11 part iii
I found this on Google.
https://education.ti.com/sites/ASIA/downloads/pdf/2013_GCE_A_Level_Solution_H2_Math_Paper_1.pdf
Not sure if you understand the solution.
Sorry for hijacking this thread but I'm a new user and can't create new topics yet and would appreciate some help with the following qns in the link http://imgur.com/a/aRESy
Hi tobypuff
Q1
Given a=6, S7 = 234, S13 = 822
a = 6 means S1 = 6
Sn is a quadratic polynomial means Sn = an^2 + bn + c
Sub n = 1, 7 and 13, and the values in. You will get 3 equations and 3 unknowns. Solve using GC.
Q6 is pretty simple. I am wondering which part you are stuck at?
No point for me to give the full answer as you won't really learn well. I want to see where you are stuck at. :)
Q7, your T3 and T4 is already wrong. You have misunderstood the question.
Basically, T3 = (1 + r/100) * T2 = 1500 (1 + r/100)^2
T4 = (1 + r/100) * T3 = 1500 (1 + r/100)^3
So Tn = 1500 (1 + r/100)^n-1
And Sn = 1500 ( 1 + (1+r/100) + (1+r/100)^2 + (1+r/100)^3 + ... + (1+r/100)^n-1 )
------- Note that there are n terms. 1 is also a term. It's (1+r/100)^0
Sn = 1500 [ ((1+r/100)^n - 1) / ((1+r/100) - 1) ]
= 1500 [ ((1+r/100)^n - 1) / (r/100) ]
= 150000/r [ (1+r/100)^n - 1]
I just finished preparing for my lesson tomorrow, will reply more when there's time.
Hi eagle,
Thanks for the explanations! Managed to solve q6 easily, turns out I made the silly mistake of equating the trig ratio of sin 45 with the lengths, lol. Noted on q7, I could solve part (i) via GC but still stuck with part (ii).
That said I have a couple of doubts to clear in this link http://imgur.com/a/R3bw9
For Q9 part (i), I'm getting two solutions of D for plane 2. Why is D = -20 rejected? D is equal to 40 in this case.
For Q10 part (iii), I'm getting two solutions of t but there can be only 1 coordinate of Q. How do I know which to reject? Both t lies in the interval stated in the qn. Also need help with part (iv).
Q7
for 7 ii, u need to set up an inequality of the sum of savings using B and A. Should be a pretty simple and straightforward solution.
Q9
For Q9, the problem with your workings is because of usage of formulas without understanding the potential consequences. Took me a while to understand because your workings did not define D, and for vectors, by convention, it is actually small d. Your school teachers may not have problems with your answers, but I think this will give problems to Cambridge examiners who will be marking your papers in A level. So take note on that.
The consequence in this scenario is that modulus introduce additional answers, and it's because the second plane could be either on the other side of A or the same as plane 1, but still 3 times the distance.
One solution is to find the foot of perpendicular from A to plane 1. Then use ratio theorem to find a point on plane 2, so that you can get the correct direction. Since they are parallel, you already have the normal vector, and you can find the value of D without the implications of a modulus.
Q10
For iii. Typically, for such questions, there will be two solutions. One refers to the original point where you calculated the normal from. This answer don't need reject, but it isn't the answer we are looking for.
The other value will be the answer we are looking for.
For iv.
θ = tan inverse (y/x)
so
dθ/dt = difference [tan inverse (y/x)] implicitly w.r.t. t
Think you can try liao.
Noted on Q9 and Q10, understood and could solve both qns easily now :)
I'm still stuck at Q7 (ii) :( Some help on how to proceed? http://imgur.com/Tg91oiL
for 7ii, on your second line, you can stop liao
Then bring all to one side, so you get an inequality >0
Now, just put into GC and solve.
To solve using GC, one method is to use the Y= sign. Then instead of plot, use [2nd][Table] and see the table of values. Make sure to set delta T = 1 for the table so that x is in integers. Slowly go down the table till you find the first positive value. The value of X is the answer.
for question A, you should consider the factor that square of cosine is less than 1 and furthermore, the dot product.
px3q = 2pxr
px3q = px2r
px3q - px2r = 0
px(3q-2r) = 0
when two vectors are parallel to each other, their cross product is a zero vector.
Anyone need help on H2 maths please feel free to what's app me with question at 94316237
Would appreciate some help with part (iii) in the link http://imgur.com/a/Imf3U
How do I go about finding the volume generated by R?
Hi tobypuff, please look at
http://sgforums.com/forums/2297/topics/491916 for suggested solution, and further the discussion on this if needed