2.2 PRODUCT OF VECTORS

Multiplication of vectors -Study Material for IIT JEE | askIITians
Definition: 
\[Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.\] \[Then\ the\ scalar\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by\]\[\overrightarrow{a}.\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\cos\theta \]
Properties of Scalar Product:
\[1. \overrightarrow{a}\ and\overrightarrow{b}are\ perpendicular\ vectors\ if\ and\ only\ if \overrightarrow{a}.\overrightarrow{b}=0. \]
\[2. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively \]
\[Then\ i) \overrightarrow{i}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1 \]
\[ ii) \overrightarrow{i}.\overrightarrow{j}= \overrightarrow{j}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{i}= \overrightarrow{k}.\overrightarrow{j}=\overrightarrow{i}.\overrightarrow{k}=0 \]
\[3. \ Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[4. \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[If\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[ \overrightarrow{a}.\overrightarrow{b}= a_1b_1 + a_2b_2 + a_3b_3\ (using\ property\ 2)\]

Example  :

\[Find\ the\ scalar\ product\ of\ \overrightarrow{i}+\overrightarrow{j}\ ,\ \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= \overrightarrow{i} + \overrightarrow{j} \]
\[\overrightarrow{b}= \overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+\overrightarrow{j}) .(\overrightarrow{i}+\overrightarrow{j}+ 3\overrightarrow{k})\]

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

= 1 + 1

= 2

\[ \overrightarrow{a}.\overrightarrow{b}= 2\]

Example  :

\[Show\ that\ the\ vectors\ 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k} and\ 3\overrightarrow{i}+ 2\overrightarrow{j}+6\overrightarrow{k} are\ perpendicular\ to\ each\ other \]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i}+ 3\overrightarrow{j}- 2\overrightarrow{k}) .(3\overrightarrow{i}+2\overrightarrow{j}+ 6\overrightarrow{k})\]

= 2(3) + 3(2) - 2 (6)

= 6 + 6 -12

= 0

\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[2.\ Find\ the\ value\ of\ m\ if\ the\ vectors\ 2\overrightarrow{i}+ m\overrightarrow{j}- 3\overrightarrow{k} and\ 3\overrightarrow{i}+ 1\overrightarrow{j}+4\overrightarrow{k} are\ perpendicular\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ m\overrightarrow{j}- 3\overrightarrow{k} \]
\[ \overrightarrow{b}= 3\overrightarrow{i}+ 1\overrightarrow{j}+4\overrightarrow{k} \]
\[Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other \]
\[i.e\ \overrightarrow{a}.\overrightarrow{b}= 0\]
\[(2\overrightarrow{i} + m\overrightarrow{j}-3k).(3\overrightarrow{i}+ \overrightarrow{j}+4\overrightarrow{k}) = 0 .\]

2(3) + m(1) - 3 (4) = 0

6 + m -12 = 0

m - 6 = 0

m = 6.

\[3.\ Prove\ that\ the\ vectors\ \overrightarrow{i}-\overrightarrow{j}+ 2\overrightarrow{k},\ \overrightarrow4{j}+ 2\overrightarrow{k}and\ -10\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k} are\ mutually\ perpendicular.\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k} \]
\[ \overrightarrow{b}= + 4\overrightarrow{j}+2\overrightarrow{k} \]
\[\overrightarrow{c}= -10\overrightarrow{i}- 2\overrightarrow{j}+ 4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}- \overrightarrow{j}+ 2\overrightarrow{k}) .(4\overrightarrow{j}+ 2\overrightarrow{k})\]

= 1(0) - 1(4) + 2 (2)

= 0

\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{b}.\overrightarrow{c}= (0\overrightarrow{i}+ 4\overrightarrow{j}+ 2\overrightarrow{k}) .(-10\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})\]

= 0(-10) + 4(-2) + 2

= 0 - 8 + 8

= 0

\[\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{c}.\overrightarrow{a}= (-10\overrightarrow{i}- 2\overrightarrow{j}+ 4\overrightarrow{k}) .(\overrightarrow{i}-\overrightarrow{j}+ 2\overrightarrow{k})\]

= - 10(1) -2 (-1) + 4 (2)

= - 10 + 2 + 8

= 0

\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]

\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b}\]

Example  :

\[Find\ the\ projection\ of\ the\ vector\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}+2\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}+2 \overrightarrow{j}+ 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}+ 2\overrightarrow{k})}{\sqrt{(1)^2 + (2)^2 + (2)^2 }}\]
\[= \frac{3(1)+ 4(2)- 5(2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{3+ 8- 10}{\sqrt{9}}\]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{1}{3}\]
\[Angle\ between\ two\ vectors\ \overrightarrow{a} and\ \overrightarrow{b}: \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]

Example  :

\[Find\ the\ angle\ between\ vectors\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} and\ 2\overrightarrow{i} -3\overrightarrow{j}- 5\overrightarrow{k} \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k}) .(2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k})\]

=   3 ( 2 ) + 4 ( - 3 ) - 2 ( - 5 )

=    6  - 12 + 10

= 4

\[ \overrightarrow{a}.\overrightarrow{b}= 4\]
\[\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2 + (-2)^2 }=\sqrt{(9 + 16 +4 }=\sqrt{29}\]
\[\overrightarrow{|b|} = \sqrt{(2)^2 + (-5)^2 + (-3)^2 }=\sqrt{(4 + 25 +9 }=\sqrt{38}\]
\[\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[= \frac{4}{\sqrt{29}\sqrt{38}}\]
\[\theta = \cos ^-1 ( \frac{4}{\sqrt{29}\sqrt{38}})\]

Example  :

\[Find\ the\ projection\ of\ the\ vector\ 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} on\ 7\overrightarrow{i}+ \overrightarrow{j}+2\overrightarrow{k}. Also\ find\ the\ angle\ between\ them \]

Soln:

\[\overrightarrow{a}= 3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ \overrightarrow{j}- 2\overrightarrow{k}) .(7\overrightarrow{i}+\overrightarrow{j}+ 2\overrightarrow{k})\]

=3 ( 7 ) + 1 ( 1) - 2 ( 2 ) 

   =    21 + 1 - 4

=   18.

\[ \overrightarrow{a}.\overrightarrow{b}= 18\]
\[\overrightarrow{|a|} = \sqrt{(3)^2 + (1)^2 + (-2)^2 }=\sqrt{(9 + 1 +4 }=\sqrt{14}\]
\[\overrightarrow{|b|} = \sqrt{(7)^2 + (1)^2 + (2)^2 }=\sqrt{(49 + 1 + 4 }=\sqrt{54}\]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} = \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}} = \frac{18}{ \sqrt{54}}\]
\[\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[= \frac{18}{\sqrt{14}\sqrt{54}}\]
\[\theta = \cos ^-1 ( \frac{18}{\sqrt{14}\sqrt{54}})\]

Vector Product of Two Vectors

Definition

Cross product - Wikipedia
\[Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.\] \[Then\ the\ vector\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by\]\[\overrightarrow{a}×\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}×\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\sin\theta\ n^\wedge \]

Properties of Vector Product:

\[1. \overrightarrow{a}\ and\overrightarrow{b}are\ parellel\ vectors\ if\ and\ only\ if \overrightarrow{a}× \overrightarrow{b}= 0. \]
\[2.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ any\ two\ vectors\ then \overrightarrow{a}×\overrightarrow{b} = -\overrightarrow{b}×\overrightarrow{a}\]
\[3. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively \]
\[Then\ i) \overrightarrow{i}×\overrightarrow{i}= \overrightarrow{j}×\overrightarrow{j}=\overrightarrow{k}×\overrightarrow{k}=0 \]
\[ ii) \overrightarrow{i}×\overrightarrow{j}= \overrightarrow{k};\overrightarrow{j}×\overrightarrow{k}= \overrightarrow{i};\overrightarrow{k}×\overrightarrow{i}=\overrightarrow{j}\]
\[ iii) \overrightarrow{j}×\overrightarrow{i}= -\overrightarrow{k};\overrightarrow{k}×\overrightarrow{j}= -\overrightarrow{i};\overrightarrow{i}×\overrightarrow{k}= -\overrightarrow{j}\]

4. Vector product in determinant from

\[Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \overrightarrow{a_1} & \overrightarrow{a_2} & \overrightarrow{a_3}\\ \overrightarrow{b_1} & \overrightarrow{b_2} & \overrightarrow{b_3}\\ \end{vmatrix}\]
\[5.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[6.\ If \overrightarrow{d_1}\ and \overrightarrow{d_2}\ are\ two\ diagonals\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = \frac{1}{2}|\overrightarrow{d_1} × \overrightarrow{d_2}|\]
\[7.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ triangle. Then\ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{a} × \overrightarrow{b}|\]
\[8.\ Area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC}\ is\ \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[9.\ \ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[10.\ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}\]

Example  :

\[Find\ \overrightarrow{a}× \overrightarrow{b}if\ \overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}+ 3\overrightarrow{k}\]

Soln:

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

Example  :

\[Find\ the\ area\ of\ the\ parellelogram\ whose\ adjacent\ sides\ are\ 3\overrightarrow{i}- \overrightarrow{k} and\ \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k}.\]

Soln:

\[\overrightarrow{a}=3\overrightarrow{i} - \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ 3 & 0 &-1\\ 1 & 1 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 0 + 1) -\overrightarrow{j}(3+1)+\overrightarrow{k}(3-0)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(4)+\overrightarrow{k}(3)\]
\[ \overrightarrow{a}× \overrightarrow{b}= \overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(1)^2 + (-4)^2 + (3)^2 }=\sqrt{(1 + 16 +9 }=\sqrt{26}\]
\[ Area\ of \ parellelogram = \sqrt{26} sq.units\]


Example  :

\[Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ 3\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}\, 2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k} and\ 5\overrightarrow{i}+ \overrightarrow{j}+3\overrightarrow{k}.\]

Soln:

\[\overrightarrow{OA}= 3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}\]
\[\overrightarrow{OC}= 5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- (3\overrightarrow{i} + 2\overrightarrow{j}- \overrightarrow{k})\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- 3\overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{AB}= -\overrightarrow{i} - 5\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- (2\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- 2\overrightarrow{i} + 3\overrightarrow{j} -\overrightarrow{k})\]
\[\overrightarrow{BC}= 3\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ -1 & -5 & 2\\ 3 & 4 & 2\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -10 - 8) -\overrightarrow{j}(-2 - 6)+\overrightarrow{k}(-4 + 15)\]
\[ = \overrightarrow{i}(-18) -\overrightarrow{j}(-8)+\overrightarrow{k}(11)\]
\[\overrightarrow{AB}× \overrightarrow{BC}= -18\overrightarrow{i}-8\overrightarrow{j}+11\overrightarrow{k}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-18 )^2 + (-8)^2 + (11)^2 }=\sqrt{(324 + 64 + 121 }=\sqrt{509}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[=\frac{\sqrt{509}}{2}\ sq. units\]

Example  :

\[ If\ |\overrightarrow{a}| = 2,\ |\overrightarrow{b}|= 7\ and\ |\overrightarrow{a} × \overrightarrow{b}|=7,\ find\ the\ angle\ between\ \overrightarrow{a}\ and\ \overrightarrow{b}.\]

Soln:

\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[ =\frac{7}{(2)(7)} = \frac{1}{2}\]
\[\theta = 30^{\circ} \]

Example  :

\[Find\ the\ unit\ vector\ perpendicular\ to\ each\ of\ the\ vectors\ 2\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k} and\ \overrightarrow{i}+ 2\overrightarrow{j}+\overrightarrow{k}.\]\[Also\ find\ the\ sine\ of\ the\ angle\ between\ the\ vectors .\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\ 2 & 1 & 1\\ 1 & 2 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 1 - 2) -\overrightarrow{j}(2 - 1)+\overrightarrow{k}(4 - 1)\]
\[ = \overrightarrow{i}(-1) -\overrightarrow{j}(1)+\overrightarrow{k}(3)\]
\[\overrightarrow{a}× \overrightarrow{b}= -\overrightarrow{i}-\overrightarrow{j}+3\overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-1 )^2 + (-1)^2 + (3)^2 }=\sqrt{(1 + 1 + 9}=\sqrt{11}\]
\[ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{\overrightarrow{i}\ - \overrightarrow{j}+ 3\overrightarrow{k}}{\sqrt{11}} \]
\[ n^\wedge = \frac{\overrightarrow{i}\ - \overrightarrow{j}+ 3\overrightarrow{k}}{\sqrt{11}} \]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (1)^2 + (1)^2 }=\sqrt{(4 + 1 + 1 }=\sqrt{6}\]
\[\overrightarrow{|b|} = \sqrt{(1)^2 + (2)^2 + (1)^2 }=\sqrt{(1 + 4 + 1 }=\sqrt{6}\]
\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{11}}{\sqrt{6}\sqrt{6}} = \frac{\sqrt{11}}{6}\]
\[\ sin\ \theta =\frac{\sqrt{11}}{6}\]