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Question

If $$a = a _ { 1 } i + a _ { 2 } j + a _ { 3 } k , b = b _ { 1 } i + b _ { 2 } j + b _ { 3 } k \mathrm { c } = c _ { 1 } \mathrm { i } + \mathrm { c } _ { 2 } \mathrm { j } + \mathrm { c } _ { 3 } \mathrm { k } , | \mathrm { c } | = 1$$ and $$( a \times b ) \times c = 0$$ then $$\left| \begin{array} { l l l } { a _ { 1 } } & { a _ { 2 } } & { a _ { 3 } } \\ { b _ { 1 } } & { b _ { 2 } } & { b _ { 3 } } \\ { c _ { 1 } } & { c _ { 2 } } & { c _ { 3 } } \end{array} \right| ^ { 2 }$$ is equal to


A
0
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B
1
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C
|a|2|b|2
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D
|a×b|2
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Solution

The correct option is B 0
Given that $$ (\begin{matrix} a\times & b)\times  & c \end{matrix}=0$$

$$ { \left( \begin{matrix} a & b & c \end{matrix} \right)  }^{ 2 }=\left( \begin{matrix} a & b & c \end{matrix} \right) \times \left( \begin{matrix} a & b & c \end{matrix} \right) ----(1)$$

Using cross product of two vectors,
$$ \left( \begin{matrix} { u }_{ 1 } & { u }_{ 2 } & { u }_{ 3 } \end{matrix} \right) \times \left( \begin{matrix} v_{ 1 } & v_{ 2 } & v_{ 3 } \end{matrix} \right) =\times \left( \begin{matrix} { u }_{ 2 }{ v }_{ 3 }-{ u }_{ 3 }v_{ 2 } & { u }_{ 3 }v_{ 1 }-{ u }_{ 1 }{ v }_{ 3 } & { u }_{ 1 }{ v }_{ 2 }-{ u }_{ 2 }{ v }_{ 1 } \end{matrix} \right) $$

 Now applying this in the equation (1), we get

$$ { \left( \begin{matrix} a & b & c \end{matrix} \right)  }^{ 2 }=\left( \begin{matrix} a & b & c \end{matrix} \right) \times \left( \begin{matrix} a & b & c \end{matrix} \right) =\left( \begin{matrix} 0 & 0 & 0 \end{matrix} \right) \quad (given)$$

Hence option (a) is correct.

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