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Mathematics: Post your doubts here!

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holoholo

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Can someone please explain how to get a second vector for the question (ii) for the vector product ?

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D

Devinky

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holoholo said:
Can someone please explain how to get a second vector for the question (ii) for the vector product ?

View attachment 59259
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You need two vectors that lie in the plane. Then cross both of them to find the normal vector.
The first vector is i + 2j + k (the direction vector of l)
The second vector is (4i-2j+2k)-(2i+2j+k), since both points are in the plane the vector between them is in the plane.
Note: cross-product is NOT required for Maths P3 (only required for F. Maths). You only need to find a vector that is perpendicular to both of them and this is one of the techniques.
 
nehaoscar

nehaoscar

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Konstantino Nikolas said:
View attachment 59269
Could anyone please help out with the second part?
ms:
View attachment 59270
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So first part you have 3/3-2x + -x-2/x^2+4
For binomial expansion as mentioned in data booklet we need the bracket like (1+x)^n
So change 3/3-2x ==> 3/3(1-(2x/3)) ==> (1-(2x/3))^-1
  • 1) And expand (1-(2x/3))^-1
Similarly
-x-2/x^2+4 ==> -x-2/4((x^2/4)+1) ==> 4^-1(-x-2)((x^2/4)+1)^-1
  • 2) Expand ((x^2/4)+1)^-1 and then multiply the expansion by 4^-1(-x-2)
Now add 1 and 2 to obtain full expansion of 3/3-2x + -x-2/x^2+4
 
Konstantino Nikolas

Konstantino Nikolas

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nehaoscar said:
So first part you have 3/3-2x + -x-2/x^2+4
For binomial expansion as mentioned in data booklet we need the bracket like (1+x)^n
So change 3/3-2x ==> 3/3(1-(2x/3)) ==> (1-(2x/3))^-1
  • 1) And expand (1-(2x/3))^-1
Similarly
-x-2/x^2+4 ==> -x-2/4((x^2/4)+1) ==> 4^-1(-x-2)((x^2/4)+1)^-1
  • 2) Expand ((x^2/4)+1)^-1 and then multiply the expansion by 4^-1(-x-2)
Now add 1 and 2 to obtain full expansion of 3/3-2x + -x-2/x^2+4
Click to expand...
Ohh such a stupid mistake i made ... I'm soo sorry for making you do it :p Thank you! I realised my mistake ... instead of substituting value of A, i had substituted value of B ... I can only pray to God I don't make such mistakes in my exam.
Thank you. :D
 
Konstantino Nikolas

Konstantino Nikolas

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Copy Cat said:
View attachment 59296
(ii) & (iii) part
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Konstantino Nikolas

Konstantino Nikolas

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Copy Cat said:
Can you explain the "cross product....."again.
Thanks
Click to expand...

Let the new plane -> q
It is given that 'p' is perpendicular to 'q'. That means the normal to 'p' will also be perpendicular to the normal to 'q'. So we now just have to find a vector that is perpendicular to the normal of 'p'.

The cross product of two vectors will give you a third vector that is perpendicular to both the original vectors. So if we want to find the vector perpendicular to 'p', we need another vector. That will be the line 'l'. However, to get the vector and not an equation of line, we can take the vector parallel to 'l' which you can find in its equation itself. (the 'b' in r = a + tb)

Now take the cross product of both to get the normal of 'q'. (you know how to do that right?)

Once you do this, well, then you do the substitutions into r.n=d and get the final answer.

Hope you got it.
 
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