A) \[\overrightarrow{a}\cdot \overrightarrow{c}\ne 0\]
B) \[\overrightarrow{a}\cdot \overrightarrow{c}\ne \overrightarrow{b}\cdot \overrightarrow{c}\]
C) \[|\overrightarrow{a}|=|\overrightarrow{c}|\]
D) None
Correct Answer: C
Solution :
\[\overrightarrow{a}\cdot (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{a}\cdot \overrightarrow{c}\Rightarrow \overrightarrow{a}\cdot \overrightarrow{c}=0\] \[\overrightarrow{b}\cdot (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{b}\cdot \overrightarrow{c}\Rightarrow \overrightarrow{b}\cdot \overrightarrow{c}=0\] \[\overrightarrow{b}\cdot (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\cdot \overrightarrow{b}=0\] Hence,\[\overrightarrow{a},\overrightarrow{b}\]and\[\overrightarrow{c}\]are mutually perpendicular vectors. \[|\overrightarrow{a}\times \overrightarrow{b}|=\overrightarrow{c}\]and\[|\overrightarrow{b}\times \overrightarrow{c}|=|\overrightarrow{a}|\] \[|\overrightarrow{a}||\overrightarrow{b}|\sin {{90}^{o}}=|\overrightarrow{a}|\]??..(1) and\[|\overrightarrow{b}||\overrightarrow{c}|=|\overrightarrow{a}|\]?(2) Divide (1) and (2)\[\frac{|\overrightarrow{a}|\,|\overrightarrow{b}|}{|\overrightarrow{b}||\overrightarrow{c}|}=\frac{|\overrightarrow{c}|\,}{|\overrightarrow{a}|}\]Hence,\[|\overrightarrow{a}|=\,|\overrightarrow{c}|\]
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