Vector Algebra — Class 12 Maths Solution

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

$\therefore {\overrightarrow {\rm{a}} ^2} = {x^2} + {y^2} + {z^2}$

$\therefore$ $\overrightarrow {\rm{a}} \times \widehat {\rm{i}} = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\x&y&z\\1&0&0\end{array}} \right|$

$= \widehat {\rm{i}}[0] - \widehat {\rm{j}}[ - z] + \widehat {\rm{k}}[ - y]$
$= z\widehat {\rm{j}} - y\widehat {\rm{k}}$

$\therefore$ ${(\overrightarrow {\rm{a}} \times \widehat {\rm{i}})^2} = (z\widehat {\rm{j}} - y\widehat {\rm{k}})(z\widehat {\rm{j}} - y\widehat {\rm{k}})$

$= {y^2} + {z^2}$

Similarly, ${(\overrightarrow {\rm{a}} \times \widehat {\rm{j}})^2} = {x^2} + {z^2}$

and ${(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2} = {x^2} + {y^2}$

$\therefore {(\overrightarrow {\rm{a}} \times \widehat {\rm{i}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{j}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2} = {y^2} + {z^2} + {x^2} + {z^2} + {x^2} + {y^2}$

$= 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2{\overrightarrow {\rm{a}} ^2}$