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Question

Let $$\overrightarrow u  = \widehat i + \widehat j,\overrightarrow v  = \widehat i - \widehat j\,\,and\,\,\overrightarrow w  = \widehat i + 2\widehat j + 3\widehat k$$. If $$\widehat n$$ is a unit vector such that $$\overrightarrow u .\widehat n = 0\,\,and\,\,\overrightarrow v .\widehat n = 0$$ then $$\left| {\overrightarrow w .\widehat n} \right|$$ is equal to


A
1
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B
2
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C
3
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D
0
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Solution

The correct option is D 3
$$\because \overset { \wedge  }{ n } $$ is perpendicular to both 
$$\overline { u } \quad and\quad \overline { v } \\ \implies\quad \overset { \wedge  }{ n } =\cfrac { \overline { u } \times \overline { v }  }{ \left| \overline { u } \times \overline { v }  \right|  } \\ \implies\quad \overline { u } \times \overline { v } =\left| \begin{matrix} \overset { \wedge  }{ i }  & \overset { \wedge  }{ j }  & \overset { \wedge  }{ k }  \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{matrix} \right| \\ \implies\quad \overset { \wedge  }{ i } (0)-\overset { \wedge  }{ j } (0)+\overset { \wedge  }{ k } (-1-1)\\ \implies\quad -2\overset { \wedge  }{ k } \\ \therefore \overset { \wedge  }{ n } =\cfrac { -2\overset { \wedge  }{ k }  }{ 2 } =-\overset { \wedge  }{ k } \\ \overline { w } .\overset { \wedge  }{ n } =(\overset { \wedge  }{ i } +2\overset { \wedge  }{ j } +3\overset { \wedge  }{ k } ).(-\overset { \wedge  }{ k } )\\ =-3\\ \implies\quad \left| \overline { w } .\overset { \wedge  }{ n }  \right| =3$$

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