If $\overrightarrow {\rm{a}} = \widehat {\rm{i}} - \widehat {\rm{j}} + \widehat {\rm{k}}$ and $\overrightarrow {\rm{b}} = \widehat {\rm{j}} - \widehat {\rm{k}},$ then find a vector $\overrightarrow {\rm{c}}$ such that $\overrightarrow {\rm{a}} \times \overrightarrow {\rm{c}} = \overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{c}} = 3$.

As we know, for any two vectors
$\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$

and $\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{b}} = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}$,

where $\overrightarrow {\rm{a}} = {a_1}\widehat {\rm{i}} + {a_2}\widehat {\rm{j}} + {a_3}\widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = {b_1}\widehat {\rm{i}} + {b_2}\widehat {\rm{j}} + {b_3}\widehat {\rm{k}}$.

Let $\overrightarrow {\rm{c}} = x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}}$

Also, $\overrightarrow {\rm{a}} = \widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}}$

and $\overrightarrow {\rm{b}} = \widehat {\rm{j}} - \widehat {\rm{k}}$

For $\overrightarrow {\rm{a}} \times \overrightarrow {\rm{c}} = \overrightarrow {\rm{b}}$

$\left| {\begin{array}{llllllllllllllllllll}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\1&1&1\\x&y&z\end{array}} \right| = \widehat {\rm{j}} - \widehat {\rm{k}}$

$\Rightarrow$ $\widehat {\rm{i}}(z - y) - \widehat {\rm{j}}(z - x) + \widehat {\rm{k}}(y - x) = \widehat {\rm{j}} - \widehat {\rm{k}}$

…..(i)
$\therefore$ $z - y = 0$

…….(ii)
$x - z = 1$

……(iii)
$x - y = 1$
Also, $\overrightarrow {\rm{a}} \cdot \overrightarrow {\rm{c}} = 3$

$(\widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}}) \cdot (x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}}) = 3$

$\Rightarrow$ $x + y + z = 3$ …..(iv)
On adding Eqs. (ii) and (iii),

we get
$2x - y - z = 2$

……(v)
On solving Eqs. (iv) and (v),

we get
$x = \frac{5}{3}$
$\therefore$ $y = \frac{5}{3} - 1 = \frac{2}{3}$ and $z = \frac{2}{3}$

Now, $\overrightarrow {\rm{c}} = \frac{5}{3}\widehat {\rm{i}} + \frac{2}{3}\widehat {\rm{j}} + \frac{2}{3}\widehat {\rm{k}}$

$= \frac{1}{3}(5\widehat {\rm{i}} + 2\widehat {\rm{j}} + 2\widehat {\rm{k}})$

OBJECTIVE T YPE