A) only\[x\]
B) only\[y\]
C) neither\[x\]nor\[y\]
D) both \[x\] and\[y\]
Correct Answer: C
Solution :
It is given that \[\overset{\to }{\mathop{\mathbf{a}}}\,=\widehat{\mathbf{i}}-\widehat{\mathbf{k}}\] \[\overset{\to }{\mathop{\mathbf{b}}}\,=x\widehat{\mathbf{i}}+\widehat{\mathbf{k}}+(1-x)\widehat{\mathbf{k}}\] and \[\overset{\to }{\mathop{\mathbf{c}}}\,=y\widehat{\mathbf{i}}+x\widehat{\mathbf{j}}+(1+x-y)\widehat{\mathbf{k}}\] Now \[[\overrightarrow{\mathbf{a}}\overrightarrow{\mathbf{b}}\overrightarrow{\mathbf{c}}]=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \\ \end{matrix} \right|\] Using the operation\[{{C}_{3}}\to {{C}_{3}}+{{C}_{1}}\], we get \[[\overrightarrow{\mathbf{a}}\overrightarrow{\mathbf{b}}\overrightarrow{\mathbf{c}}]=\left| \begin{matrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \\ \end{matrix} \right|\] \[=1+x-x=1\] Hence, \[[\overrightarrow{\mathbf{a}}\overrightarrow{\mathbf{b}}\overrightarrow{\mathbf{c}}]\]depends on neither\[x\]nor\[y\].
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