Scalar Triple Product And Vector Triple Product

Vector Algebra / Thursday, October 31st, 2019

Scalar Triple Product

If α, β and γ be three vectors then the product (α X β) . γ is called triple scalar product (or, box product) of . It is denoted by [ α β γ].

Note: [ α β γ] is a scalar quantity.

Properties of Triple Scalar Product

$(1)~~If~~\overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }~~be~~three~~vectors~~then~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=~~Volume~~of$

$the~~parallelopiped~~with~~sides~~\overrightarrow{\alpha },\overrightarrow{\beta }~~and~~\overrightarrow{\gamma }.$

$(2)~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=0~~if~~and~~if~~\overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }~~are~~coplanar.$

$(3)~~\frac{1}{6}\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=~~Volume~~of~~the~~tetrahedron~~with~~sides~~\overrightarrow{\alpha },\overrightarrow{\beta }~~and~~\overrightarrow{\gamma }.$

$(4)~~If~~\overrightarrow{\alpha }={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k},~~\overrightarrow{\beta }={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k},~~\overrightarrow{\gamma }={{x}_{3}}\hat{i}+{{y}_{3}}\hat{j}+{{z}_{3}}\hat{k}$

$(5)~~In~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]~~if~~any~~two~~vectors~~are~~\text{interchanged}~~then~~the~~sign$

$is~~altered~~but~~the~~value~~remains~~same,~~e.g~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=-\left[ \overrightarrow{\alpha }~~\overrightarrow{\gamma }~~\overrightarrow{\beta } \right]$

$(6)~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=0~~if~~any~~two~~vectors~~are~~identical.$

$(7)~~If~~any~~of~~the~~vectors~~\overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }~~is~~sum~~of~~two~~vectors~~then~~is~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]$

$sum~~of~~two~~triple~~scalar~~product~~e.g~~\left[ \overrightarrow{{{\alpha }_{1}}}+\overrightarrow{{{\alpha }_{2}}}~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=\left[ \overrightarrow{{{\alpha }_{1}}}~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]+\left[ \overrightarrow{{{\alpha }_{2}}}~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]$

$(8)~~If~~any~~of~~the~~vector~~\overrightarrow{\alpha },\overrightarrow{\beta ~}~~and~~\overrightarrow{\gamma }~~is~~a~~scalar~~multiple~~of~~a~~vector$

$then~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]~~is~~a~~scalar~~multiple~~of~~\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]~~e.g~~\left[ \overrightarrow{\alpha }~~\lambda \overrightarrow{\beta }~~\overrightarrow{\gamma } \right]=\lambda \left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\gamma } \right]$

$(9)~~\left[ \hat{i}~~\hat{j}~~\hat{k} \right]=\left[ \hat{j}~~\hat{k}~~\hat{i} \right]=\left[ \hat{k}~~\hat{i}~~\hat{j} \right]=1$

Vector Triple Product

If α, β and γ be three vectors then the product

$\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)\times \overrightarrow{\gamma }$

is called vector triple product of α, β and γ.

Properties of Vector Triple Product

$(1)~~\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)\times \overrightarrow{\gamma }=\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\overrightarrow{\beta }-\left( \overrightarrow{\beta }.\overrightarrow{\gamma } \right)\overrightarrow{\alpha }$

$(2)~~\overrightarrow{\alpha }\times \left( \overrightarrow{\beta }\times \overrightarrow{\gamma } \right)=\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\overrightarrow{\beta }-\left( \overrightarrow{\alpha }.\overrightarrow{\beta } \right)\overrightarrow{\gamma }$

$(3)~~\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)\times \overrightarrow{\gamma }\ne \overrightarrow{\alpha }\times \left( \overrightarrow{\beta }\times \overrightarrow{\gamma } \right)$

Associative law for cross product fails.

 Example 01

Prove that

$\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right).\left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right)=\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\left( \overrightarrow{\beta }.\overrightarrow{\delta } \right)-\left( \overrightarrow{\alpha }.\overrightarrow{\delta } \right)\left( \overrightarrow{\beta }.\overrightarrow{\gamma } \right)$

Solution:

$L.H.S=\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right).\left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right)=\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right).\overrightarrow{\rho }~~where~~\overrightarrow{\rho }=\left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right)$

$=\left[ \overrightarrow{\alpha }~~\overrightarrow{\beta }~~\overrightarrow{\rho } \right]=-\left[ \overrightarrow{\alpha }~~\overrightarrow{\rho }~~\overrightarrow{\beta } \right]~=-\left( \overrightarrow{\alpha }\times \overrightarrow{\rho } \right).\overrightarrow{\beta }$

$=-\left\{ \overrightarrow{\alpha }\times \left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right) \right\}.\overrightarrow{\beta }$

$=-\left\{ \left( \overrightarrow{\alpha }.\overrightarrow{\delta } \right)\overrightarrow{\gamma }-\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\overrightarrow{\delta } \right\}.\overrightarrow{\beta }$

$=\left\{ \left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\overrightarrow{\delta } \right\}.\overrightarrow{\beta }-\left\{ \left( \overrightarrow{\alpha }.\overrightarrow{\delta } \right)\overrightarrow{\gamma } \right\}.\overrightarrow{\beta }$

$=\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\left( \overrightarrow{\delta }.\overrightarrow{\beta } \right)-\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\left( \overrightarrow{\gamma }.\overrightarrow{\beta } \right)$

$=\left( \overrightarrow{\alpha }.\overrightarrow{\gamma } \right)\left( \overrightarrow{\beta }.\overrightarrow{\delta } \right)-\left( \overrightarrow{\alpha }.\overrightarrow{\delta } \right)\left( \overrightarrow{\beta }.\overrightarrow{\gamma } \right)=R.H.S$

 Example 02

Prove that

$\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)\times \left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right)=\overrightarrow{\beta }\left[ \overrightarrow{\gamma }~~\overrightarrow{\delta }~~\overrightarrow{\alpha } \right]-\overrightarrow{\alpha }\left[ \overrightarrow{\gamma ~}~~\overrightarrow{\delta }~~\overrightarrow{\beta } \right]$

Hence prove that if the α, β, γ, δ are coplanar vectors

$\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)\times \left( \overrightarrow{\gamma }\times \overrightarrow{\delta } \right)=\overrightarrow{0}$