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:

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