12.0 Scalar Triple Product

It is defined for three vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ as $\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$ and can also be written as

\[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]\]

It is the dot product of one vector with the resultant of cross product of other two vectors. It is called the scalar product because the resultant of the triple product gives the scalar value.

It denotes the volume of parallelopiped which is calculated, with sides $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $, as the area of base (parallelogram) $\left| {\overrightarrow b \times \overrightarrow c } \right|$ multiplied by its height (the component of $\overrightarrow a $ in the direction of ${\overrightarrow b \times \overrightarrow c }$.

Volume $V$ of parallelopiped = magnitude of $\overrightarrow a .(\overrightarrow b \times \overrightarrow c )$ which is

Property 1: If $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are given in the component form as $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$, $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$, $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$ then, the scalar triple product is written as the determinant of $3 \times 3$ matrix i.e.,

\[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = \left( {\begin{array}{c} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right)\]

Property 2: The scalar triple product is invariant under a circular shift of its three operands $(a,b,c)$ i.e.,$$\vec a.(\vec b \times \vec c) = \vec b.(\vec c \times \vec a) = \vec c.(\vec a \times \vec b)$$Hence, it is represented as \[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = \left[ {\begin{array}{c} {\overrightarrow b }&{\overrightarrow c }&{\overrightarrow a } \end{array}} \right] = \left[ {\begin{array}{c} {\overrightarrow c }&{\overrightarrow a }&{\overrightarrow b } \end{array}} \right]\]

Property 3: If we swap any two of the three operands, triple product becomes negative i.e., $$\vec a.(\vec b \times \vec c) = - \vec a.(\vec c \times \vec b) = - \vec b.(\vec a \times \vec c) = - \vec c.(\vec b \times \vec a)$$

Property 4: \[\left[ {\begin{array}{c} {k\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = k\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]\]

Property 5:\[\left[ {\begin{array}{c} {\overrightarrow a + \overrightarrow b }&{\overrightarrow c }&{\overrightarrow d } \end{array}} \right] = \left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow c }&{\overrightarrow d } \end{array}} \right] + \left[ {\begin{array}{c} {\overrightarrow b }&{\overrightarrow c }&{\overrightarrow d } \end{array}} \right]\]

Property 6: $\overrightarrow a $ ,$\overrightarrow b $ and $\overrightarrow c $ are taken in this order, then they form a right handed system if \[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] > 0\]

Property 7: If the scalar triple product is equal to zero, i.e., \[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = 0\] then the three vectors $\overrightarrow a $ ,$\overrightarrow b $ , $\overrightarrow c $ are coplanar.

Property 8: If any two vectors in scalar triple product are equal, then the value is equal to zero i.e.,$$\vec a.(\vec a \times \vec c) = \vec b.(\vec b \times \vec a) = \vec c.(\vec c \times \vec b) = 0$$

Question 20. For any three vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $. Prove that \[\left[ {\begin{array}{c} {\overrightarrow a + \overrightarrow b }&{\overrightarrow b + \overrightarrow c }&{\overrightarrow c + \overrightarrow d } \end{array}} \right] = 2\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]\]

Solution: $$\begin{equation} \begin{aligned} \{ (\overrightarrow a + \overrightarrow b ) \times (\overrightarrow b + \overrightarrow c )\} .(\overrightarrow c + \overrightarrow a ) \\ = \{ \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow b + \overrightarrow b \times \overrightarrow c \} .(\overrightarrow c + \overrightarrow a ) \\ = \{ \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c \} .(\overrightarrow c + \overrightarrow a ) \\\end{aligned} \end{equation} $$ as $$(\overrightarrow b \times \overrightarrow b = 0)$$$$\begin{equation} \begin{aligned} = (\overrightarrow a \times \overrightarrow b ).\overrightarrow c + (\overrightarrow a \times \overrightarrow c ).\overrightarrow c + (\overrightarrow b \times \overrightarrow c ).\overrightarrow c + (\overrightarrow a \times \overrightarrow b ).\overrightarrow a + (\overrightarrow a \times \overrightarrow c ).\overrightarrow a + (\overrightarrow b \times \overrightarrow c ).\overrightarrow a \\ = (\overrightarrow a \times \overrightarrow b ).\overrightarrow c + 0 + 0 + 0 + 0 + (\overrightarrow b \times \overrightarrow c ).\overrightarrow a \\\end{aligned} \end{equation} $$

\[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] + \left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = 2\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]\]

Question 21. If vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are coplanar then show that \[\left[ {\begin{array}{c} {\overrightarrow a + \overrightarrow b }&{\overrightarrow b + \overrightarrow c }&{\overrightarrow c + \overrightarrow a } \end{array}} \right]\] are coplanar.

Solution: As we have seen in the above example that \[\left[ {\begin{array}{c} {\overrightarrow a + \overrightarrow b }&{\overrightarrow b + \overrightarrow c }&{\overrightarrow c + \overrightarrow a } \end{array}} \right] = 2\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]\]

And it is given that vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are coplanar i.e \[\left[ {\begin{array}{c} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right] = 0\]

So \[\left[ {\begin{array}{c} {\overrightarrow a + \overrightarrow b }&{\overrightarrow b + \overrightarrow c }&{\overrightarrow c + \overrightarrow a } \end{array}} \right] = 0\] which means that vectors $\overrightarrow a+b $ ,$\overrightarrow b+c $ and $\overrightarrow c+a $ are coplanar.