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Question
If \[\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}\] and \[\vec{a} \times \vec{b} = \vec{a} \times \vec{c,} \vec{a} \neq 0,\] then
Options
\[\vec{b} = \vec{c}\]
\[\vec{b} = \vec{0}\]
\[\vec{b} + \vec{c} = \vec{0}\]
none of these
Solution
\[\vec{b} = \vec{c}\]
\[\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \]
\[ \Rightarrow \vec{a}^{} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 0\]
\[ \Rightarrow \vec{a .} \left( \vec{b} - \vec{c} \right) = 0 \]
\[\text { Let } \theta \text { be the angle between} \ \vec{ a } \text { and }\left( \vec{b} - \vec{c} \right) \]
\[\left| \vec{a} \right|\left| \left( \vec{b} - \vec{c} \right) \right|\cos \theta . . . (1)\]
\[\text { and } \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \]
\[ \Rightarrow \vec{a} \times \vec{b} - \vec{a} \times \vec{c} = 0\]
\[ \Rightarrow \vec{a} \times \left( \vec{b} - \vec{c} \right) = 0\]
\[\text { Then } , \left| \vec{a} \right| \left| \left( \vec{b} - \vec{c} \right) \right| \sin \theta = 0 . . . (2)\]
\[\text { Here, it is given that} \ \vec{a} \neq 0\]
\[\text { Therefore, for eq (1) and eq (2) to be 0 }\]
We have ,
\[\left| \left( \vec{b} - \vec{c} \right) \right| \cos \theta = 0 \]
\[\text { For } \left| \left( \vec{b} - \vec{c} \right) \right| \cos \theta = 0 , \text { one of } \left| \left( \vec{b} - \vec{c} \right) \right| \text { or }\cos \theta \text { must be } 0\]
Case 1:
\[\text { Let } \cos \theta = 0\]
\[ \Rightarrow \theta = 90^\circ \]
\[ \Rightarrow \sin \theta = 1\]
\[\text { & if } \left| \left( \vec{b} - \vec{c} \right) \right| \sin \theta = 0 \text { and } \sin \theta = 1 \]
\[\text { Then } \left| \left( \vec{b} - \vec{c} \right) \right| = 0\]
\[ \Rightarrow \vec{b} = \vec{c} \]
Case 2:
\[\text { Let } \left| \left( \vec{b} - \vec{c} \right) \right| = 0\]
\[ \Rightarrow \vec{b} = \vec{c} \]
\[\text { Hence }, \vec{b} = \vec{c} \]
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