MHB Help Needed: I'm Stuck on Steps and Not Sure If They're Correct

[Joe20]

I have done up some of the steps. I got stuck and not sure how to continue. I am not sure if those steps are correct. Need help on that.

View attachment 7976

View attachment 7975

 

[GJA]

Hi, Alexis87.

I have done up some of the steps. I got stuck and not sure how to continue. I am not sure if those steps are correct. Need help on that.

I did not check the details of the work you posted, so I am not suggesting that anything you did there is incorrect. The intent of this post is to suggest an alternate method that avoids the need for computing tedious cross products using vector components.

Using the equality $p\times q = 3p\times r,$ take the cross product on both left hand sides with $p$; i.e.,

$p\times q = 3p\times r\qquad\Longrightarrow\qquad p\times(p\times q)=3p\times(p\times r)$

and now use the "BAC-CAB" BAC-CAB Identity -- from Wolfram MathWorld rule and some algebra to get your desired result (noting that the various dot products you obtain from the BAC-CAB rule are constants).

 

[Joe20]

Hi, Alexis87.
I did not check the details of the work you posted, so I am not suggesting that anything you did there is incorrect. The intent of this post is to suggest an alternate method that avoids the need for computing tedious cross products using vector components.

Using the equality $p\times q = 3p\times r,$ take the cross product on both left hand sides with $p$; i.e.,

$p\times q = 3p\times r\qquad\Longrightarrow\qquad p\times(p\times q)=3p\times(p\times r)$

and now use the "BAC-CAB" BAC-CAB Identity -- from Wolfram MathWorld rule and some algebra to get your desired result (noting that the various dot products you obtain from the BAC-CAB rule are constants).

Continuing from your advice:

p x (p x q) = 3p x (p x r)

p(p.q) - q(p.p) = p(3p.r) - r(3p.p)

p(p.q) - p(3p.r) = q(p.p) - 3r(p.p)

p[(p.q)-(3p.r)] = (q - 3r) (p.p)

p [(p.q)-(3p.r)] /(p.p) = q-3r => Is it correct ? then [(p.q)-(3p.r)] /(p.p) will be the scalar or lamda?

 

[GJA]

That's correct.