Now the correction of your English: You probably meant to say “stuck” which means “confused” rather than “stocked” which gives the impression that you have acquired a warehouse and filled it with stuff!

In question 5, the vector aspect(i.e a*v =b*v, to show a = b) of the solution is not clear to me. Please sir, can you throw more light into this as I would not like to assume wrongly. Thank you very much for your time.

“throw more light” is too vague a request. What you need to do it tell the particular step you do not understand. I will expand on that step.There is an error in the answer though: I should have said that a and b are collinear rather than say a x b is collinear as I have said. Hope that helps.

Good afternoon sir,
sir i don’t understand this statement:

Note that is symmetric in and , is symmetric in and and
is antisymmetric in , and Because each term is the product of a
symmetric and an antisymmetric object which must vanish.

We will explain “symmetry” and “antisymmetry” in full in the next few lectures. For now, know a term is symmetrical in two indices if swapping them does not alter the value of the term. It is anti-symmetrical if a swap of the indices negates the term. Fir example, AiAj is symmetrical in i and j because a swap does not alter the product as Ai times Aj is the same as Aj times Ai. However, eijk is antisymmetrical by the definition of the alternating symbol. It is possible for a term to be neither symmetrical nor antisymmetrical.

Good afternoon sir,thank you for all your responses so far,they have been quite helpful.
Here is my question: In slide 58, I don’t understand how you moved from

u x(v x w)=(u x z)
where z =(v x w). note that this expression would result to
Eijk.Vi.Wj.Ek( Ek is bold- the base vector).

k can be changed to become EijβViWjEβ.( NOTE: u,v,w,z, Eβ are bold letters)

so we have: ux(vxw)= u x z = Eαβγ.Uα.Zβ.Eγ( note that Eγ is bold)

therfore:
Z= ZβEβ=Eijβ.Vi.Wj.Eβ = ( NOTE: Z is bold but Zβ is not bold),
so we can write that:
Zβ=Eijβ.Vi.Wj,
which can be sustituted into the expression:
ux(vxw)= u x z = Eαβγ.Uα.Zβ.Eγ
and then solved further as
ux(vxw)= u x z = Eαβγ.Uα.(Eijβ.Vi.Wj).Eγ ( the expression in bracket is new value for Zβ.

Going through the 30 problems all over again and i realized that i am not too clear with the question 28. More precisely,the second step after the addition sign, i did not know how the l came about. I would appreciate a little bit assistance on this.

Good Morning Sir,

Thank you very much for the Q & A. But I don’t fully understand Question. 9 and the solution.

I would appreciate a little bit of assistance on this sir.

Thanks

Look at the response on the page.

Thank you very much sir. This was very helpful. Would be back, if i encounter any difficulty while going through the problems.

Good afternoon sir.

Thanks for d solved problems, but am stocked in number 9. will appreciate , if u can make it clearer.

The answer to your question is as shown in the newer post above.

Now the correction of your English: You probably meant to say “stuck” which means “confused” rather than “stocked” which gives the impression that you have acquired a warehouse and filled it with stuff!

Good morning sir,

In question 5, the vector aspect(i.e a*v =b*v, to show a = b) of the solution is not clear to me. Please sir, can you throw more light into this as I would not like to assume wrongly. Thank you very much for your time.

“throw more light” is too vague a request. What you need to do it tell the particular step you do not understand. I will expand on that step.There is an error in the answer though: I should have said that a and b are collinear rather than say a x b is collinear as I have said. Hope that helps.

Thank you very much sir. This helps.

Good afternoon sir,

sir i don’t understand this statement:

Note that is symmetric in and , is symmetric in and and

is antisymmetric in , and Because each term is the product of a

symmetric and an antisymmetric object which must vanish.

We will explain “symmetry” and “antisymmetry” in full in the next few lectures. For now, know a term is symmetrical in two indices if swapping them does not alter the value of the term. It is anti-symmetrical if a swap of the indices negates the term. Fir example, AiAj is symmetrical in i and j because a swap does not alter the product as Ai times Aj is the same as Aj times Ai. However, eijk is antisymmetrical by the definition of the alternating symbol. It is possible for a term to be neither symmetrical nor antisymmetrical.

This was actually my next question. Thanks

Good afternoon sir,thank you for all your responses so far,they have been quite helpful.

Here is my question: In slide 58, I don’t understand how you moved from

eαβγ uαzβeγ to eijβeγαβuαviwjeγ

Remember that you are free to change any dummy index so as to ensure an index is not repeated more than once.

George Oladayo,

Note this

ux(vxw)=(uxz)

where z=(vxw). note that this expression would result to

EijkViWjEk( Ek is bold- the base vector).

k can be changed to become EijβViWjEβ.

so you can continue from there.

Note this

u x(v x w)=(u x z)

where z =(v x w). note that this expression would result to

Eijk.Vi.Wj.Ek( Ek is bold- the base vector).

k can be changed to become EijβViWjEβ.( NOTE: u,v,w,z, Eβ are bold letters)

so we have: ux(vxw)= u x z = Eαβγ.Uα.Zβ.Eγ( note that Eγ is bold)

therfore:

Z= ZβEβ=Eijβ.Vi.Wj.Eβ = ( NOTE: Z is bold but Zβ is not bold),

so we can write that:

Zβ=Eijβ.Vi.Wj,

which can be sustituted into the expression:

ux(vxw)= u x z = Eαβγ.Uα.Zβ.Eγ

and then solved further as

ux(vxw)= u x z = Eαβγ.Uα.(Eijβ.Vi.Wj).Eγ ( the expression in bracket is new value for Zβ.

thank you. I hope this would help

Thank you very much Mr Adeyeye. That was helpful.

Pls can you help with Q9; from the step

= √

, =√/

=√/() ×()/

=1/2√ * [ + ] = / (pls how this answer gotten?)

Also question 17b and question 17c

Good evening sir,

I don’t understand problem 8,starting from the third step.

Thank you again for your time.

Hi everyone, please I don’t seem to understand number 18?!?

Gud evening sir:

will appreciate if u throw more light on number 18.

Tried getting it. But didnt.

from question 6, the next steps were quite confusing after finding the determinant.

Oh!..yea yea!!!….

i totally ignored what was on the slide.

Thanks by the way John.

i’m expanding one by one tho….

Good day sir,

Going through the 30 problems all over again and i realized that i am not too clear with the question 28. More precisely,the second step after the addition sign, i did not know how the l came about. I would appreciate a little bit assistance on this.

Thank you very much fro your time sir.