Solve Vector Equation 2x-(xdoti)i for x

[square_imp]

Hi all,

I would like some help with a tricky vector equation, I need to solve for x.

2x-(xdoti)i = i + 5j - (xi)

The dot is a dot product and not multiply. i am not sure how to deal with the dot product, I am little confused. Any help would be excellent. Thanks.

 

[NateTG]

if
\vec{x}=x_i \hat{i} + x_j \hat{j}
and
\vec{y}=y_i \hat{i} + y_j \hat{j}
then
x \dotprod y = x_i y_i + x_j y_j

 

[HallsofIvy]

Hi all,
I would like some help with a tricky vector equation, I need to solve for x.
2x-(xdoti)i = i + 5j - (xi)
The dot is a dot product and not multiply. i am not sure how to deal with the dot product, I am little confused. Any help would be excellent. Thanks.

I'm a little confused myself! (xdoti) implies that x is a vector but then "xi" is not clear. If x= ai+ bj then "x dot i" is just a so "2a- (x dot i)i would be
2ai+ 2bj- ai= ai+ 2bj. But then I have no idea what (xi) means.

 

[square_imp]

The (xi) at the end is actually x cross i, I didnt read it carefully emough. Now, I have had another crack at it working out the dot and cross product sections in the equation, this is my working:

2x-(ai + bj).(i) = i +5j -((ai + bj) cross i)

substituting the x for ai(unit vector) and bj(unit vector)
gives:

2x-ai = i + 5j - bk
x = [(a+1)i + 5j -bk]/2

is this working correct? The bit I am most unsure about is the working out of the cross product using matrices. Thanks for the help, sort of clicked my brain into gear.

 

[HallsofIvy]

Write x as ai+ bj+ ck. Then 2x= 2ai+ 2bj+ 2ck and (x dot i)i= ai so 2x- (x dot i)i= (2a-a)i+ 2bj+ 2ck= ai+ 2bj+ 2ck. x cross i= cj- bk.

2x-(xdoti)i = i + 5j - (x cross i) is
ai+2bj+ 2ck= i+ (5-c)j- bk

We must have a=1, 1b=5- c and 2c= -b.

Can you solve those?

 

[square_imp]

Thanks guys, got it all sorted I think. Realised my mistake with the matrix. Got x to be: (i + 2j + k) which when substitutued back into the original formula all works out. Thanks again.