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Scalar Triple Product — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSScalar Triple Product1 Mark
Question
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:
  1. $\pm 1$
  2. $0$
  3. $-2$
  4. $2$
✓

Answer

  1. $\pm1$
Solution:
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 $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$  $\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1\text{ and }\cos180^\circ=-1\big)$  $$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$  $\big(\therefore\vec{\text{a}} \text{ is perpendicular to }\vec{\text{b}})$ 
$=1 \text{ or }-1$ $\big(\therefore\big|\vec{\text{a}}\big|=1 \text{ and }\big|\vec{\text{b}}\big|=1\big)$ $$
$=\pm1$

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