Examples On Scalar Triple Product Of Vectors Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 27

For four points with position vectors  \(\vec a,\;\;\vec b,\;\;\vec c,\;\;\vec d\) prove that if  \(\left[ {\vec d\,\vec b\,\vec c} \right] + \left[ {\vec d\,\vec c\,\vec a} \right] + \left[ {\vec d\,\vec a\,\vec b} \right] = \left[ {\vec a\,\vec b\,\vec c} \right]\) then the four points must be coplanar.

Solution: Upon expansion, this relation gives

\[\begin{align}&\quad\quad\quad\vec d \cdot \left( {\vec b \times \vec c} \right) + \vec d \cdot \left( {\vec c \times \vec a} \right) + \vec d \cdot \left( {\vec a \times \vec b} \right) = \vec a \cdot \left( {\vec b \times \vec c} \right) \hfill \\\\&\Rightarrow \quad  \vec d \cdot \left\{ {\left( {\vec b \times \vec c} \right) + \left( {\vec c \times \vec a} \right) + \left( {\vec a \times \vec b} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right) \hfill \\\\& \Rightarrow \quad   \vec d \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\} = \vec a \cdot \left( {\vec b \times \vec c} \right)\quad\qquad\qquad...\left( 1 \right) \hfill \\ \end{align} \]

Note that the RHS of (1) can also be written as

\[\vec a \cdot \left( {\vec b \times \vec c} \right) = \vec a \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\}\]

You must verify why this can be done. Using this in (1), we have

\[\left( {\vec d - \vec a} \right) \cdot \left\{ {\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right)} \right\} = 0\]

This implies that the vector \(\left( {\vec d - \vec a} \right)\) is perpendicular to the cross product of the vectors \(\left( {\vec b - \vec a} \right)\) and \(\left( {\vec c - \vec a} \right)\) which in turn means that \(\left( {\vec d - \vec a} \right)\) must lie in the plane of \(\left( {\vec b - \vec a} \right)\) and \(\left( {\vec c - \vec a} \right)\)

\[\begin{align}&  \Rightarrow  \quad \left( {\vec b - \vec a} \right),\left( {\vec c - \vec a} \right)\,\,and\left( {\vec d - \vec a} \right)\,are{\text{ }}coplanar{\text{ }}vectors \hfill \\\\& \Rightarrow \quad \vec a,\,\vec b,\,\vec c,\,\vec d\,\,are{\text{ }}coplanar{\text{ }}points. \hfill \\ \end{align} \]

Example – 28

Prove that, for any three vectors \(\vec a,\,\vec b,\,\vec c,\)

(a) \(\left[ {\left( {\vec a + \vec b} \right)\,\,\left( {\vec b + \vec c} \right)\,\,\,\left( {\vec c + \vec a} \right)} \right] = 2\left[ {\vec a\,\vec b\,\vec c} \right]\)

(b) \(\left[ {\left( {\vec a - \vec b} \right)\,\,\,\left( {\vec b - \vec c} \right)\,\,\,\left( {\vec c - \vec a} \right)} \right] = 0\)

Solution:(a) The left hand side is

\[\begin{align}& \left( {\vec a + \vec b} \right) \cdot \left\{ {\left( {\vec b + \vec c} \right) \times \left( {\vec c + \vec a} \right)} \right\} \hfill \\\\&= \left( {\vec a + \vec b} \right) \cdot \left\{ {\vec b \times \vec c + \vec b \times \vec a + \vec c \times \vec a} \right\} \hfill \\ \end{align} \]

     \(= 2\left[ {\vec a\,\vec b\,c} \right]\)

This relation incidentally proves that \(\vec a + \vec b,\,\vec b + \vec c\)and \(\vec c + \vec a\) are coplanar if  only if \(\vec a,\,\vec b\) and \(\vec c\)are coplanar.

(b) Since \(\left( {\vec a - \vec b} \right) + \left( {\vec b - \vec c} \right) + \left( {\vec c - \vec a} \right) = \vec 0,\) these three vectors are the sides of a triangle, implying that they are coplanar vectors. Thus, their STP must be zero.

Example – 29

For any three non-coplanar vectors \(\vec a,\,\vec b,\,\vec c\) and any other vector \(\vec r\) prove that the relation

\[\vec r = \frac{{\left[ {\vec r\,\vec b\,\vec c} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec a + \frac{{\left[ {\vec r\,\vec c\,\vec a} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec b + \frac{{\left[ {\vec r\,\vec a\,\vec b} \right]}}{{\left[ {\vec a\,\vec b\,\vec c} \right]}}\vec c\]

is satisfied.

Solution: Assume \(\vec r = x\vec a + y\vec b + z\vec c\) where  \(x,\,y,\,z \in \mathbb{R}\quad\qquad\qquad...(1)\)             

This can be done since \(\vec a,\,\vec b,\,\vec c\) are non-coplanar vectors and hence any vector in space is expressible as their linear combination. To find x, y, z we do the following:

Take the dot product on both sides of (1) with \(\left( {\vec b \times \vec c} \right):\)

\[ \begin{align}&\Rightarrow \quad [\vec r\;\vec b\;\vec c] = x[\vec a\;\vec b\;\vec c]\\&\Rightarrow \quad x = \frac{{[\vec r\;\;\vec b\;\,\vec c]}}{{[\vec a\;\;\vec b\;\,\vec c]}}\end{align}\]

Similarly, y and z can be determined. Now substituting the values of x, y and z in (1) proves the stated assertion.

Download practice questions along with solutions for FREE:
Vectors
grade 11 | Questions Set 1
Vectors
grade 11 | Answers Set 1
Vectors
grade 11 | Questions Set 2
Vectors
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 2
Three Dimensional Geometry
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 1
Three Dimensional Geometry
grade 11 | Answers Set 1
Download practice questions along with solutions for FREE:
Vectors
grade 11 | Questions Set 1
Vectors
grade 11 | Answers Set 1
Vectors
grade 11 | Questions Set 2
Vectors
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 2
Three Dimensional Geometry
grade 11 | Answers Set 2
Three Dimensional Geometry
grade 11 | Questions Set 1
Three Dimensional Geometry
grade 11 | Answers Set 1
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school