3.1 PRODUCT OF THREE AND FOUR VECTORS

Scalar triple Product

$Let\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} be\ any\ three\ vectors,$$their\ scalar\ triple\ product\ is\ denoted\ by\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]$

Properties of Scalar triple Product:

$1.\ Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ ,\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k} and\ \overrightarrow{c}= c_1\overrightarrow{i}\ + c_2\overrightarrow{j}+ c_3\overrightarrow{k}$
$Then\ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}$
$2.\ The\ three\ vectors\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ if\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0$
$3.\ The\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}, \overrightarrow{OC}\ and\ \overrightarrow{OD}\ are\ coplanar\ if\ [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = 0$

Example  :

$Evaluate:\ [\overrightarrow{i} + \overrightarrow{j}\ \overrightarrow{j} + \overrightarrow{k}\ \overrightarrow{k} + \overrightarrow{i}]$

Soln:

$\overrightarrow{a}= \overrightarrow{i}\ + \overrightarrow{j}$
$\overrightarrow{b}= \overrightarrow{j}\ + \overrightarrow{k}$
$\overrightarrow{c}= \overrightarrow{k} + \overrightarrow{i}$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ \end{vmatrix}$

=    1  ( 1 – 0 )  – 1 ( 0 – 1 )  + 0 (  0 – 1 )

=   1 ( 1 )  -1 (- 1 ) +  0

=   1 + 1

=   2

Example  :

$Prove\ that\ the\ vectors\ 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k},\ 3\overrightarrow{i} + 4\overrightarrow{j}+ \overrightarrow{k}\ and\ \overrightarrow{i} – 2\overrightarrow{j}+ \overrightarrow{k}\ are\ coplanar.$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= 3\overrightarrow{i}+ 4\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{c}= \overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 2 & 1 & 1\\ 3 & 4 & 1\\ 1 & -2 & 1\\ \end{vmatrix}$

=    2  ( 4 + 2 )  – 1 (3 – 1 )  + 1 ( -6 – 4 )

=   2 ( 6 )  – 1 (2 ) +  1 ( – 10 )

=   12 – 2  –  10   =    10  –  10

$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] = 0$

The given points are coplanar.

Example  :

$Find\ the\ value\ of\ m\ if\ the\ vectors\ 2\overrightarrow{i}-\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + 2\overrightarrow{j}- 3\overrightarrow{k}\ and\ 3\overrightarrow{i} + m\overrightarrow{j}+ 5\overrightarrow{k}\ are\ coplanar.$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}$
$\overrightarrow{c}= 3\overrightarrow{i}+ m\overrightarrow{j}+ 5\overrightarrow{k}$
$Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ \implies [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0$
$\begin{vmatrix} 2 &- 1 & 1\\ 1 & 2 & -3\\ 3 & m & 5\\ \end{vmatrix}=0$

2  ( 10 + 3m )  + 1 ( 5 + 9 )  + 1 ( m – 6 ) = 0

20 + 6m + 14 + m – 6 =  0

7m + 28 = 0

7m   =   -28

m =  – 4

Example  :

$Show\ that\ the\ points\ whose\ position\ vectors\ 4\overrightarrow{i}+ 5\overrightarrow{j}+ \overrightarrow{k},\ – \overrightarrow{j}- \overrightarrow{k},\ 3\overrightarrow{i} + 9\overrightarrow{j}+ 4\overrightarrow{k}$$and\ -4\overrightarrow{i} + 4\overrightarrow{j}+ 4\overrightarrow{k}\ lie\ on\ the\ same\ plane.\ (or)\ Coplanar.$

Soln:

$\overrightarrow{OA}= 4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{OB}= -\overrightarrow{j} – \overrightarrow{k}$
$\overrightarrow{OC}= 3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}$
$\overrightarrow{OD}= -4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}$
$\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$
$=-\overrightarrow{j} – \overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})$
$=-\overrightarrow{j} – \overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j} – \overrightarrow{k}$
$\overrightarrow{AB}= -4\overrightarrow{i} -6\overrightarrow{j} -2\overrightarrow{k}$
$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}$
$=3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})$
$=3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{AC}= -\overrightarrow{i} +4\overrightarrow{j} +3\overrightarrow{k}$
$\overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA}$
$=-4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})$
$=-4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j}-\overrightarrow{k})$
$\overrightarrow{AD}= -8\overrightarrow{i} -\overrightarrow{j} +3\overrightarrow{k}$
$[\overrightarrow{AB}\ \overrightarrow{AC} \overrightarrow{AD}] =\begin{vmatrix} -4 & -6 & -2\\ -1 & 4 & 3\\ -8 & -1 & 3\\ \end{vmatrix}$

=    -4 ( 12 + 3)  +  6 ( -3 + 24 )  – 2 ( 1 + 32 )

=   -4 (15)  + 6 (21) – 2 ( 33 )

=   -60 + 126 – 66

=   -126 + 126

$[\overrightarrow{AB}\ \overrightarrow{AC} \overrightarrow{AD}] =0$

The given points lie in the same plane.

Vector Triple Product

$Let\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} be\ any\ three\ vectors,$ $then\ the\ product \overrightarrow{a}×( \overrightarrow{b}×\overrightarrow{c})\ and\ ( \overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}$$are\ called\ vector\ triple\ product\ of\ \overrightarrow{a},\overrightarrow{b},\ \overrightarrow{c}$

Note:

$\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c}) = (\overrightarrow{a}. \overrightarrow{c}) \overrightarrow{b} – (\overrightarrow{a}.\overrightarrow{b}) \overrightarrow{c}$

Product of Four vectors

$Let\ \overrightarrow{a},\overrightarrow{b} ,\overrightarrow{c}, \overrightarrow{d} be\ any\ four\ vectors,$ $then\ the$$scalar\ product\ of\ these\ four\ vectors\ is\ defined\ as\ ( \overrightarrow{a} × \overrightarrow{ b} ) . ( \overrightarrow{c} × \overrightarrow{ d} )$
$( \overrightarrow{a} × \overrightarrow{ b} ) × ( \overrightarrow{c} × \overrightarrow{ d} ) = [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{d}]\ \overrightarrow{c} – [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ \overrightarrow{d}$

Example  :

$If\ \overrightarrow{a} = 2\overrightarrow{i}- 3\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} +2\overrightarrow{j}+ 4\overrightarrow{k}\ and\ \overrightarrow{c} = \overrightarrow{i} +3\overrightarrow{j} +\overrightarrow{k}$ $find\ \overrightarrow{a} × (\overrightarrow{ b} × \overrightarrow{c})$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}- 3\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i} +2\overrightarrow{j}+ 4\overrightarrow{k}$
$\overrightarrow{c} = \overrightarrow{j} +3\overrightarrow{j} +\overrightarrow{k}$
$\overrightarrow{b}×\overrightarrow{c} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 2 & 4\\ 1 & 3 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 2- 12) -\overrightarrow{j}(1-4)+\overrightarrow{k}(3-2)$
$= \overrightarrow{i}(-10) -\overrightarrow{j}(-3)+\overrightarrow{k}(1)$
$\overrightarrow{b}× \overrightarrow{c}= -10\overrightarrow{i}+3\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{a} × (\overrightarrow{ b} × \overrightarrow{c}) = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 2 & -3 & 1\\ -10 & 3 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( -3 – 3) -\overrightarrow{j}(2 +10)+\overrightarrow{k}(6 -30)$
$= \overrightarrow{i}(-6) -\overrightarrow{j}(12)+\overrightarrow{k}(-24)$
$\overrightarrow{a} × (\overrightarrow{ b} × \overrightarrow{c})= -6\overrightarrow{i}-12\overrightarrow{j}-24\overrightarrow{k}$

Example  :

$If\ \overrightarrow{a} = 2\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} -2\overrightarrow{j}+ 3\overrightarrow{k}\ and\ \overrightarrow{c} = 3\overrightarrow{i}-\overrightarrow{j} +5\overrightarrow{k}$ $find\ (\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}$

Soln:

$\overrightarrow{a}= 2\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b}= \overrightarrow{i} -2\overrightarrow{j}+ 3\overrightarrow{k}$
$\overrightarrow{c} = 3\overrightarrow{j}-\overrightarrow{j} +5\overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\ 2 & 3 & 1\\ 1 & -2 & 3\\ \end{vmatrix}$
$= \overrightarrow{i}( 9+ 2) -\overrightarrow{j}(6-1)+\overrightarrow{k}(-4-3)$
$= \overrightarrow{i}(11) -\overrightarrow{j}(5)+\overrightarrow{k}(-7)$
$\overrightarrow{a}× \overrightarrow{b}= 11\overrightarrow{i}-5\overrightarrow{j}-7\overrightarrow{k}$
$(\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c} = \begin{vmatrix} \overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\ 11 & -5 & -7\\ 3 & -1 & 5\\ \end{vmatrix}$
$= \overrightarrow{i}( -25 – 7) -\overrightarrow{j}(55 +21)+\overrightarrow{k}(-11+15)$
$= \overrightarrow{i}(-32) -\overrightarrow{j}(76)+\overrightarrow{k}(4)$
$(\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}= 11\overrightarrow{i}-5\overrightarrow{j}-7\overrightarrow{k}$

Example  :

$If\ \overrightarrow{a} = \overrightarrow{i} + \overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = 2\overrightarrow{i} + \overrightarrow{k},\ \overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}\}\ and\ \overrightarrow{d} = 3\overrightarrow{i} +2\overrightarrow{j} + \overrightarrow{k}$$find\ (\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} )$

Soln:

$\overrightarrow{a}= \overrightarrow{i} + \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b} = 2\overrightarrow{i} +\overrightarrow{k}$
$\overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}$
$\overrightarrow{d} = 3\overrightarrow{i} +2\overrightarrow{j} + \overrightarrow{k}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 1\\ 2 & 0 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 1 – 0) -\overrightarrow{j}(1-2)+\overrightarrow{k}(0 – 2)$
$= \overrightarrow{i}(1) -\overrightarrow{j}(-1)+\overrightarrow{k}(-2)$
$\overrightarrow{a}× \overrightarrow{b}= \overrightarrow{i}+\overrightarrow{j} -2\overrightarrow{k}$
$\overrightarrow{c}×\overrightarrow{d} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 2 & 3\\ 3 & 2 & 1\\ \end{vmatrix}$
$= \overrightarrow{i}( 2 – 6) -\overrightarrow{j}(1-9)+\overrightarrow{k}(2 -6)$
$= \overrightarrow{i}(-4) -\overrightarrow{j}(-8)+\overrightarrow{k}(-4)$
$\overrightarrow{c}× \overrightarrow{d}= -4\overrightarrow{i}+8\overrightarrow{j}-4\overrightarrow{k}$
$(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = (\overrightarrow{i}+\overrightarrow{j} -2\overrightarrow{k}) .(-4\overrightarrow{i}+8\overrightarrow{j}-4\overrightarrow{k})$
$=1(-4) + 1(8) -2 (-4)$
$=-4 + 8 +8$
$(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = 12$

Example  :

$If\ \overrightarrow{a} = \overrightarrow{i} + \overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} -\overrightarrow{j}- \overrightarrow{k},\ \overrightarrow{c} = -\overrightarrow{i}+\overrightarrow{j} +2\overrightarrow{k}\}\ and\ \overrightarrow{d} = 2\overrightarrow{i}+\overrightarrow{j}$$verify\ that\ ( \overrightarrow{a} × \overrightarrow{ b} ) × ( \overrightarrow{c} × \overrightarrow{ d} ) = [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{d}]\ \overrightarrow{c} – [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ \overrightarrow{d}$

Soln:

$\overrightarrow{a}= \overrightarrow{i} + \overrightarrow{j}+ \overrightarrow{k}$
$\overrightarrow{b} = \overrightarrow{i} -\overrightarrow{j}- \overrightarrow{k}$
$\overrightarrow{c} = -\overrightarrow{i}+\overrightarrow{j} +2\overrightarrow{k}$
$\overrightarrow{d} = 2\overrightarrow{i} +\overrightarrow{j}$
$\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 1 & 1\\ 1 & -1 & -1\\ \end{vmatrix}$
$= \overrightarrow{i}( -1 + 1) -\overrightarrow{j}(-1-1)+\overrightarrow{k}(-1 – 1)$
$= \overrightarrow{i}(0) -\overrightarrow{j}(-2)+\overrightarrow{k}(-2)$
$\overrightarrow{a}× \overrightarrow{b}= +2\overrightarrow{j}-2\overrightarrow{k}$
$\overrightarrow{c}×\overrightarrow{d} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ -1 & 1 & 2\\ 2 & 1 & 0\\ \end{vmatrix}$
$= \overrightarrow{i}( 0 – 2) -\overrightarrow{j}(0-4)+\overrightarrow{k}(-1 – 2)$
$= \overrightarrow{i}(-2) – \overrightarrow{j}(-4) +\overrightarrow{k}(-3)$
$\overrightarrow{c}× \overrightarrow{d}= -2\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}$
$( \overrightarrow{a} × \overrightarrow{ b} ) × ( \overrightarrow{c} × \overrightarrow{ d} ) =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 0 & 2 & -2\\ -2 & 4 & -3\\ \end{vmatrix}$
$= \overrightarrow{i}( -6 + 8) -\overrightarrow{j}(0-4)+\overrightarrow{k}(0 + 4)$
$= \overrightarrow{i}(2) – \overrightarrow{j}(-4) +\overrightarrow{k}(4)$
$( \overrightarrow{a} × \overrightarrow{ b} ) × ( \overrightarrow{c} × \overrightarrow{ d} )= 2\overrightarrow{i}+4\overrightarrow{j} +4\overrightarrow{k}———-(1)$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{d}] =\begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & -1\\ 2 & 1 & 0\\ \end{vmatrix}$
$= 1( 0 + 1) -1(0 +2)+1(1 + 2)$
$= 1( 1) -1(2)+1(3)$
$= 1 – 2 + 3$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{d}] = 2$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & -1\\ -1 & 1 & 2\\ \end{vmatrix}$
$= 1( -2 + 1) -1(2 -1)+1(1 – 1)$
$= 1( -1) -1(1)+1(0)$
$= -1 – 1 + 0$
$[\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] = -2$
$[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{d}]\ \overrightarrow{c} – [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ \overrightarrow{d} =$
$2(-\overrightarrow{i}+\overrightarrow{j} +2\overrightarrow{k}) – (-2)( 2\overrightarrow{i} +\overrightarrow{j})$
$=-2\overrightarrow{i}+2\overrightarrow{j} +4\overrightarrow{k} + 4\overrightarrow{i} +2\overrightarrow{j}$
$= 2\overrightarrow{i}+4\overrightarrow{j} +4\overrightarrow{k}$
$[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{d}]\ \overrightarrow{c} – [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ \overrightarrow{d} =2\overrightarrow{i}+4\overrightarrow{j} +4\overrightarrow{k}————- (2)$
$From\ (1)\ and\ (2)\ ( \overrightarrow{a} × \overrightarrow{ b} ) × ( \overrightarrow{c} × \overrightarrow{ d} ) = [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{d}]\ \overrightarrow{c} – [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\ \overrightarrow{d}$