Question
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, such that $\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0,$ then show that $\vec{\text{d}}$ is the null vector.

Answer

Given that
$\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors such that
$\vec{\text{d}}.\vec{\text{a}}=\vec{\text{d}}.\vec{\text{b}}=\vec{\text{d}}.\vec{\text{c}}=0$
Given that
$\vec{\text{d}}.\vec{\text{a}}=0$
⇒ $\vec{\text{d}}$ perpendicular to $\vec{\text{a}}$
or $\vec{\text{d}}=0\dots(1)$
$\vec{\text{d}}.\vec{\text{b}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{b}}$ or $\vec{\text{d}}=0\dots(2)$
$\vec{\text{d}}.\vec{\text{c}}=0$
⇒ $\vec{\text{d}}$ is perpendicular to $\vec{\text{c}}$ or $\vec{\text{d}}=0\dots(3)$
From (1), (2), (3), we get
$\vec{\text{d}}$ is perpendicular to $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ or $\vec{\text{d}}=0,$ but $\vec{\text{d}}$ can not be perpendicular to $​​\vec{\text{a}},​​\vec{\text{b}}$ and $​​\vec{\text{c}},$ because $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, so
$\vec{\text{d}}=0,$

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