MCQ
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}} $ are three non $-$ coplanar mutually perpendicular unit vectors, then $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big],$ is :
  • $\pm 1$
  • B
    $0$
  • C
    $-2$
  • D
    $2$

Answer

Correct option: A.
$\pm 1$
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|\big|\vec{\text{c}}\big|\cos0^\circ$ or $\big|\vec{\text{a}}\times{\vec{\text{b}}}\big|\big|\vec{\text{c}}\big|\cos180^\circ$
$\big(\therefore\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other$)$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ or $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1$ and $\cos180^\circ=-1\big)$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$
$\big(\therefore\vec{\text{a}} $ is perpendicular to $\vec{\text{b}})$
$=1 $ or $-1 \big(\therefore\big|\vec{\text{a}}\big|=1$ and $\big|\vec{\text{b}}\big|=1\big)$
$=\pm1$

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