# If $$\overrightarrow a = \hat i + \hat j + \hat k \: and \: \overrightarrow b = \hat j - \hat k$$, then find a vector $$\overrightarrow c$$ such that $$\overrightarrow a$$ x $$\overrightarrow c = \overrightarrow b\: and \: \overrightarrow a.\overrightarrow c = 3$$

Toolbox:
• $\hat i.\hat i=1$
• $\hat j.\hat j=1$
• $\hat k.\hat k=1$
Step 1:
$\overrightarrow a=\hat i+\hat j+\hat k$
$\overrightarrow b=\hat j-\hat k$
Given : $\overrightarrow a\times \overrightarrow c=\overrightarrow b$
$\overrightarrow a.\overrightarrow c=3$
Let $\overrightarrow c=x\hat i+y\hat j+z\hat k$
$\overrightarrow a\times \overrightarrow c=\overrightarrow b$
$\begin{vmatrix}\hat i&\hat j&\hat k\\1 & 1 & 1\\x& y&z\end{vmatrix}=\hat j-\hat k$
(i.e)$(z-y)\hat i-(z-x)\hat j+(y-x)\hat k=\hat j-\hat k$
Step 2:
Now equating the coefficients we get,
$z-y=0$
$-(z-x)=1$
$y-x=-1$
$y=z$
$x-z=1$
$x-y=1$
$\therefore z=x-1$
$y=x-1$
Step 3:
$\overrightarrow a.\overrightarrow c=3$
$\hat i+\hat j+\hat k.(x\hat i+y\hat j+z\hat k)=3$
$x+y+z=3$
$3x-2=3$
$\therefore z=x-1,y=x-1$
$x=\large\frac{5}{3}$
$y=\large\frac{5}{3}$$-1=\large\frac{2}{3} y=z=\large\frac{2}{3} \overrightarrow c=\large\frac{5}{3}$$\hat i+\large\frac{2}{3}$$\hat j+\large\frac{5}{3}$$\hat k$
Hence the required vector $\overrightarrow c=\large\frac{5}{3}$$\hat i+\large\frac{2}{3}$$\hat j+\large\frac{5}{3}$$\hat k$