Question :
Find a vector b such that a.b = 1 , a $\displaystyle \times$ b = j - k and a = i + j + k
I am so sorry but i am still stuck there i tried to do what u adviced me to do but it s not leading me anywhere
This is what i have done till now!!!
$\displaystyle b = ui + vj + wk$ ; $\displaystyle a = i + j + k $
$\displaystyle a.b = 1$ and $\displaystyle a \times b = j - k$
$\displaystyle a.b = (i + j + k)(ui + vj + wk) = ui^2 + vj^2 + wk^2$
Now what should i do with $\displaystyle a.b$
And
$\displaystyle a \times b = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ u & v & w \end{vmatrix}$ = $\displaystyle i(v - w) - j(w - u) + k(v - u)$
therefore
$\displaystyle i(v - w) - j(w - u) + k(v - u) = j - k$
what should i do with this???
ok
$\displaystyle u + v + w = 1 $
Mr fantastic is this the right step or no
$\displaystyle u = 1 - v - w$
Substituting $\displaystyle u$ in $\displaystyle (v - u) = -1$ we get
$\displaystyle v - 1 + v + w = -1$
$\displaystyle 2v + w = 0$
$\displaystyle v = \frac{-w}{2}$
Substituting $\displaystyle v$ in $\displaystyle (v - w) = 0$ weget
$\displaystyle \frac{-w}{2} - w = 0$
$\displaystyle \frac{-w - 2w}{2} = 0 $
$\displaystyle \frac{-3w}{2} = 0$
$\displaystyle w = - \frac{2}{3}$
substituting $\displaystyle w$ in $\displaystyle (v - w) = 0$
$\displaystyle v - \frac{2}{3} = 0 $
$\displaystyle v = \frac{2}{3}$
Now Subtitute $\displaystyle v$ in $\displaystyle (v - u) = 0$
$\displaystyle \frac{2}{3} - u = 0$
$\displaystyle u = \frac{2}{3}$
Am i doing it correctly or no ?????