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
\[Let\ \overrightarrow{a},\overrightarrow{b} ,\overrightarrow{c}, \overrightarrow{d} be\ any\ four\ vectors,\] \[then\ the\]\[ scalar\ product\ of\ these\ four\ vectors\ is\ defined\ as\ ( \overrightarrow{a} × \overrightarrow{ b} ) . ( \overrightarrow{c} × \overrightarrow{ d} )\]
Example :
\[If\ \overrightarrow{a} = 2\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} -2\overrightarrow{j}+ 3\overrightarrow{k}\ and\ \overrightarrow{c} = 3\overrightarrow{j}-\overrightarrow{j} +5\overrightarrow{k}\] \[find\ (\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}\]
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 :
\[If\ \overrightarrow{a} = 2\overrightarrow{i} - \overrightarrow{j}+ 2\overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} +\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}\}\ and\ \overrightarrow{d} = \overrightarrow{i}-\overrightarrow{j} - \overrightarrow{k}\]\[find\ (\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} )\]
Soln:
\[\overrightarrow{a}= 2\overrightarrow{i} - \overrightarrow{j}+ 2\overrightarrow{k}\]
\[\overrightarrow{b} = \overrightarrow{i} +\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}\]
\[\overrightarrow{d} = \overrightarrow{i}-\overrightarrow{j} - \overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
2 & -1 & 2\\
1 & 1 & 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -1 - 2) -\overrightarrow{j}(2-2)+\overrightarrow{k}(2 + 1)\]
\[ = \overrightarrow{i}(-3) -\overrightarrow{j}(0)+\overrightarrow{k}(3)\]
\[ \overrightarrow{a}× \overrightarrow{b}= -3\overrightarrow{i}+3\overrightarrow{k}\]
\[\overrightarrow{c}×\overrightarrow{d} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
1 & 2 & 3\\
1 & -1 & -1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -2 + 3) -\overrightarrow{j}(-1-3)+\overrightarrow{k}(-1 -2)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(-4)+\overrightarrow{k}(-3)\]
\[ \overrightarrow{c}× \overrightarrow{d}= \overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}\]
\[(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = (-3\overrightarrow{i}+3\overrightarrow{k}) .(\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k})\]
= -3 ( 1 ) + 3 ( - 3 )
= -3 -9
= - 12
\[(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = -12\]