Finding the Value of m and Expression of Vector x in Vectors Coursework Question

[gtfitzpatrick]

if the vector x and the scalar m satisfy the eqs.
a x x = m + b and a.x = 1

where a=i+2j, b=2i+j-2k

Find the value of m and the expression ofr the vector x in terms of i,j and k

 

[tiny-tim]

Hi gtfitzpatrick!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[gtfitzpatrick]

i got the cross product of the leftside which gave me 2(x3)i - (x3)j + ((x2)-2(x1))k and cleaned up the right side which gave (m+2)i + (2m+1)j - 2k and let the i, j ,k parts equal each other
2(x3) =(m+2)
-(x3) = (2m+1)
(x2)-2(x1) = -2

and tried to solve but I'm getting 0

 

[gtfitzpatrick]

wait a sec i actually got m= -4/5 and vector x = -3/5k

i got both the i and j parts = 0
from a.x = 1 gives (x1)+2(x2)=1
and eq 1 i got 2(x2)+ (x1) =1 and (x2)-2(x1) = -2 which gives (x1)=(x2)=0

 

[gtfitzpatrick]

i think I'm doing this right, But any hints would be greatly appreciated
Thanks

 

[tiny-tim]

if the vector x and the scalar m satisfy the eqs.
a x x = m + b and a.x = 1

where a=i+2j, b=2i+j-2k

Find the value of m and the expression ofr the vector x in terms of i,j and k

Hi gtfitzpatrick!

I was a little confused at first, but I think you meant

a x x = ma + b

Hint: whenever you see a cross-product, try dotting it with something …

in this case, try dot-producting this equation with a, and you should get an easy equation for m.

(and then dot-product it with … ?)

 

[gtfitzpatrick]

Thanks Tiny-Tim,
That is what i meant well spoted. Dot producting both side is a good tip, It was a quicker way of finding M, I'm still not sure of the second part though...

 

[tiny-tim]

Thanks Tiny-Tim,
That is what i meant well spoted. Dot producting both side is a good tip, It was a quicker way of finding M, I'm still not sure of the second part though...

show us what you've got so far

 

[gtfitzpatrick]

Hello,
I crossed the left side giving me 2X_{3}i - X_{3}j + (X_{2} - 2X_{1})i

(1,2,3 are subsets but i can't get them to work)

and the right side cleans up to give (m+2)i + (2m+1)j - 2k

and then let the i's, j's and k's equal each other an i ended up with 3 eqs
2X_{3} = m +2
-X_{3} = 2m +1
(X_{2}-2X_{1}) = -2
and from a.x = 1 i got X_{1} + 2X_{2} 1

from the first 2 eqs i get m = -4/5 and x3 = 3/5

and from eqs 3&4 i get x2 = 0 and x1 = 1

 

[tiny-tim]

Hello gtfitzpatrick!

I'm sorry I've taken so long to reply.

…from the first 2 eqs i get m = -4/5 and x3 = 3/5

and from eqs 3&4 i get x2 = 0 and x1 = 1

Yes, that looks fine! …

x = (1,0,3/5), and so x x a = (6/5,-3/5,-2) = (-4/5,8/5,0) + (2,1,-2) = -4/5 a + b.

 

[gtfitzpatrick]

cheers tiny tim,
all the help is much appreciated