The magnitude of the projection of the vector
\[\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge \] on the vector perpendicular to the plane
containing the vectors \[\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge \] and \[\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge \] , is :
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Official solution
Let vector \[\overrightarrow p \] is perpendicular to the both vectors \[\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge \] and \[\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge \].
\[ \therefore \] \[\overrightarrow p \] = (\[\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge \]) \[ \times \] (\[\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge \])
= \[\left| {i11j12k13 } \right|\]
= \[\widehat i - 2\widehat j + \widehat k\]
Now a vector \[\overrightarrow a \] = \[\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge \] is given and we have to findout projection of vector \[\overrightarrow a \] on \[\overrightarrow p \].
\[ \therefore \] Projection of vector \[\overrightarrow a \] on \[\overrightarrow p \]
= \[\left| {\overrightarrow a } \right|\cos \theta \]
= \[\left| {\overrightarrow a } \right| \times {{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow p } \right|}}\]
= \[{{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}\]
= \[{{\left( {2\widehat i + 3\widehat j + \widehat k} \right).\left( {\widehat i - 2\widehat j + \widehat k} \right)} \over {\sqrt {1 + 4 + 1} }}\]
= \[{{2 - 6 + 1} \over {\sqrt 6 }}\]
= \[{{ - 3} \over {\sqrt 6 }}\]
Magnitude of projection of vector \[\overrightarrow a \] on \[\overrightarrow p \]
= \[\left| {{{ - 3} \over {\sqrt 6 }}} \right|\] = \[{{3 \over {\sqrt 6 }}}\] = \[{{\sqrt 3 } \over {\sqrt 2 }}\]