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Mathematics: Post your doubts here!
in Question 9i) just equate both the equations equation of l : r1 = (3-λ)i + (-2+2λ)j + (1+λ)k equation of m : r2 = (4 +µa)i + (4 +µb)j + (2-µ)k since they intersect r1 = r2 (3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k compare the coeficients and after some substitutions u shall...- MustafaMotani
- Post #9,007
- Forum: A-Levels
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Mathematics: Post your doubts here!
bro i did Q6(i) but its too long and too complicated for me to type in the post ... i shall give u little starting point tan(2x+x) = (tan2x + tanx)/ (1 - tan(2x)tan(x)) = k tanx i then substituted tan2x = (2tanx)/ (1-tan^2(x)) and after doing some algebraic i got the answer- MustafaMotani
- Post #9,006
- Forum: A-Levels
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Mathematics: Post your doubts here!
from part i) u would have got y = 70 e^(e^(-3t) -1) p = mass at time t / initial mass x 100 p = 70 e^(e^(-3t) -1) / 70 x 100 p = e^(e^(-3t)-1)) x 100 now if yu see if t becomes very large the expression e^(-3t) would approach zero . look at it this way e^(-3t) = 1/e^(3t) see if...- MustafaMotani
- Post #9,000
- Forum: A-Levels
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Mathematics: Post your doubts here!
thanx for the prayers.. :)- MustafaMotani
- Post #8,988
- Forum: A-Levels
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Mathematics: Post your doubts here!
sis marking scHeme tells us everything ... :) Calc w^2 ... find mid point of w and w^2 ...that will be centre of the circle .... Find |w-w^2| (length of the diameter) divide it by 2 to get radius .. now |z - (centre of the circle)| = radius- MustafaMotani
- Post #8,927
- Forum: A-Levels
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Mathematics: Post your doubts here!
u just have to squre root all three ansers in ii)a ... i did similar question just few posts above...- MustafaMotani
- Post #8,905
- Forum: A-Levels
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Mathematics: Post your doubts here!
Q.10 Crossing OC and AB shall give perpendicular vector ... AB = (-1,1,3) AB x OC = (12,0,4) Equation : r = (x,y,z) r.n=a.n 12x + 4z = 52 3x + z = 13 ii) its quite difficult to explain this part on forum ... there are several ways you can solve this question First we have to finD CF make...- MustafaMotani
- Post #8,901
- Forum: A-Levels
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Mathematics: Post your doubts here!
in 8ii) yu just have to rearrange the equation so that it resembles your equation in question .. ( graph equation) when you will be don rearranging... yu shall see that in place of "y" u will get 0.5 .. ( u have to try it to get it) in 9ii) you got the roots in first part .. from those roots...- MustafaMotani
- Post #8,900
- Forum: A-Levels
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Mathematics: Post your doubts here!
aah yu got confuse in functions :) .... p(z) = z^3 + 6z^2 +24z +32 p(1) = 1^3 + 6(1)^2 +24(1) +32 =63 similarly p(z^2) = (z^2)^3 + 6(z^2)^2 + 24z^2 + 32 (just replaced z by z^2) now if yu make substitution x=z^2 to solve the equation ... u shall get the same equation as in part ii)a x^3 +...- MustafaMotani
- Post #8,899
- Forum: A-Levels
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Mathematics: Post your doubts here!
like i would say no ..:ROFLMAO::ROFLMAO:- MustafaMotani
- Post #8,886
- Forum: A-Levels
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Mathematics: Post your doubts here!
when you cross two vectors you get a vector that is perpendicular to both of them...Since OA and OB lie on the same plane their cross will give the vector that is perpendicular to the plane .... to get it more clearly just try to imagine such stuff in silence with full concentration ...- MustafaMotani
- Post #8,882
- Forum: A-Levels
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Mathematics: Post your doubts here!
yu got OA and OB ryt..? just find the cross product of OA and OB to get its normal- MustafaMotani
- Post #8,881
- Forum: A-Levels
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Mathematics: Post your doubts here!
I know its little untidy..but hope u get it..- MustafaMotani
- Post #8,880
- Forum: A-Levels
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Mathematics: Post your doubts here!
tip: whenever examiner asks to find the angle between the two planes just dot the normal vectors of both planes... its equally possible for you to get acute or obtuse angle in the process... if you find obtuse and question asks for acute just subtract it from 180- MustafaMotani
- Post #8,877
- Forum: A-Levels
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Mathematics: Post your doubts here!
u sure yu got it 100%..?- MustafaMotani
- Post #8,876
- Forum: A-Levels
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Mathematics: Post your doubts here!
Oh its easy... if you have done part i u must have found direction vector (normal vector) of the plane ABC .. in part (ii) just find normal of OAB and dot it ... if u can visualise u shall see that normal of OAB is (0,0,1) ( just for simplicity) dont fret if yu dont get what I said in my...- MustafaMotani
- Post #8,874
- Forum: A-Levels
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