For any vector $\overrightarrow {\rm{a}}$, the value of ${(\overrightarrow {\rm{a}} \times \widehat {\rm{i}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{j}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2}$ is
For any vector $\overrightarrow {\rm{a}}$, the value of ${(\overrightarrow {\rm{a}} \times \widehat {\rm{i}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{j}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2}$ is
Official 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 ${(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2} = {x^2} + {y^2}$
$\therefore {(\overrightarrow {\rm{a}} \times \widehat {\rm{i}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{j}})^2} + {(\overrightarrow {\rm{a}} \times \widehat {\rm{k}})^2} = {y^2} + {z^2} + {x^2} + {z^2} + {x^2} + {y^2}$
$= 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2{\overrightarrow {\rm{a}} ^2}$
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