How Do You Solve for Vector x and Scalar λ in Vector Equations?

[DryRun]

Homework Statement

Find the vector x and the scalar λ which satisfy the equations
[tex]x \wedge b = b-λc,\; x.c=-2[/tex]where b = (-2, 1, -1) and c = (1, -2, 2)

Homework Equations

Vector algebra.

The Attempt at a Solution

First, i worked on x.c=-2
Let vector [itex]x = (x_1, x_2, x_3)[/itex]
So, i got the first equation: [itex]x_1-2x_2+2x_3=-2[/itex]

Now, working with: [itex]x \wedge b = b-λc[/itex]
First, i evaluated the L.H.S. and i found the determinant: [itex](-x_2-x_3)\hat i - (-x_1+2x_3)\hat j +(x_1+2x_2)\hat k[/itex]

Next, i found the R.H.S. and i equated both sides, and added them up to get the second equation: [itex]2x_1+x_2-3x_3=-2-λ[/itex]

But how to solve both equations to get x and λ? Maybe I've overlooked something...

I also tried vector algebra on: [itex]x \wedge b = b-λc[/itex] by first doing scalar multiplication by b and then vector multiplication by b.
I got: [tex]x=\frac{-λ(b\wedge c)}{(b^2-λbc)}[/tex]

Well, i think i got it. I had to look closer at my scalar multiplication, which gives:
[tex]λ=\frac{b^2}{bc}[/tex]
Then, i just have to use the value of λ to find vector x.
λ=-1 and then it seems that i messed up as in finding x, i got the denominator = 0.
Any suggestions?

EDIT: OK... Instead of scalar and vector multiplication by b, i tried the same steps with c.
I got λ=-1 and x=(-5/6, 5/6, 1/6)
Is it correct?

 

[mighty2000]

Yes, your solution is correct! Great job working through the equations and using vector algebra to solve for x and λ. Keep up the good work!