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
B
2
C
3
D
0

Solution

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

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