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Scalar Triple Product — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsScalar Triple Product1 Mark
Question
The value of $\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big],$ where $\big|\vec{\text{a}}\big|=1,\big|\vec{\text{b}}\big|=5,\big|\vec{\text{c}}\big|=3,$ is:
  1. 0
  2. 1
  3. 6
  4. None of these.
✓

Answer

  1. 0

Solution:

We have

$\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big]$

$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big).\big(\vec{\text{c}}.\vec{\text{a}}\big)$ (By definition of scalar triple product)

$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{b}}-\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{b}}\times{\vec{\text{b}}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}-0-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}-\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{a}}-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{c}}+\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{a}}+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{c}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)\\.\vec{\text{a}}\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]-\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{c}}\big]-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0+0+0-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$

$=0$

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