Let \(\theta \) be angle between a and b
LHS,\(\left| \overrightarrow {a } \times \overrightarrow {b } \right| ^{2 }=(\overrightarrow a\times\overrightarrow b).(\overrightarrow a \times \overrightarrow b)\)
\(=(ab\sin\theta)\overset\wedge n.(ab\sin\theta)\overset\wedge n\)
\(=(a^{ 2 }b^{2 }\sin^{2 }\theta)(\overset\wedge n.\overset\wedge n)\)
\(=a^{2 }b^{2 }\sin^{ 2 }\theta\)
\(=a^{2 }b^{ 2}(1-\cos^{ 2 }\theta)\)
\(=a^{ 2 }b^{ 2 }-(ab\cos\theta)^{ 2}\)
\(=(\overrightarrow a.\overrightarrow a)(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b)^{2}\)....(i)
Also, RHS=\(\begin{vmatrix} \overrightarrow { a } .\overrightarrow { a } & \overrightarrow { a } .\overrightarrow { b } \\ \overrightarrow { a } .\overrightarrow { b } & \overrightarrow { b } .\overrightarrow { b } \end{vmatrix}\)
\(=(\overrightarrow a.\overrightarrow a).(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b).(\overrightarrow a.\overrightarrow b)\)
\(=(\overrightarrow a.\overrightarrow a).(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b)^{ 2}\)...(ii)
From eqn.(i) and (ii)
\(\Rightarrow\)\(RHS=LHS \) 
Hence proved.

Given, \(\left| a\right| =\left| b \right| \)
and a.b \(=\frac{1}{2}\)
Let \(\theta \) be thae angle between \(\overrightarrow a\) and \(\overrightarrow b\)
then, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

\(\cos \theta=\frac{\overrightarrow a.\overrightarrow b}{\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| }=\frac{12}{10\times 2}=\frac{3}{5}\)
\(\cos \theta =\frac{3}{5}\)
\(\sin \theta=\sqrt{1-\cos ^{2}\theta} =\sqrt{1-\frac{9}{25}}\)
\(\sin \theta=\frac{\left| \overrightarrow { a } \times \overrightarrow { b } \right| }{\left| \overrightarrow {a } \right| \left| \overrightarrow {b } \right| }\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =\left| \overrightarrow { a } \right| \left| \overrightarrow { b} \right| \sin\theta\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =10\times2\times\frac{4}{5}=16\)

\(\overset\rightarrow a \times \overset\rightarrow b\)= \(\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & -1 & 2 \\ 2 & -1 & -1 \end{matrix} \right| \)
\(=i(1+2)-j(-1-4)+k(1-2)\)
= 3i + 5j + k
Area of \(\left| \overrightarrow { a } \times \overrightarrow { b } \right| =\sqrt{9+25+1}\)
\(=\sqrt{35} sq.units\)

\(\overrightarrow a \times \overrightarrow b=\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & 2 & -1 \\ 3 & 1 & -1 \end{matrix} \right| \)
\(=\overset\wedge i(-2+1)-\overset\wedge j(-1+3)+\overset\wedge k(1-6)\)
\(=-i-2j-5k\)
\(\left| \overrightarrow { a} \times \overrightarrow { b } \right| =\sqrt{1^{ 2}+2^{ 2}+5^{ 2}}\)
\(=\sqrt{1+4+25}=\sqrt{30}\)

\((2\overset\rightarrow a-3\overset\rightarrow b).(3 \overset\rightarrow a+\overset\rightarrow b).=6\overset\rightarrow a.\overset\rightarrow a+2\overset\rightarrow a.\overset\rightarrow b-9\overset\rightarrow b.\overset\rightarrow a-3\overset\rightarrow b.\overset\rightarrow b\)
\(=6\left| \overset { \rightarrow }{ a } \right| ^{ 2}-(\overset\rightarrow a.\overset\rightarrow b)-3\left| \overset { \rightarrow }{ b } \right| ^{ 2}\)
\(=6(3)^{ 2}-7(2)-3(1)^{2 }\)
= 6(9) - 14 - 3
= 54 - 17 = 37

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=\overset\wedge i+2\overset\wedge j+\overset\wedge k+3\overset\wedge i+\overset\wedge j-\overset\wedge k\)
\(\Rightarrow \overset\rightarrow c=4\overset\wedge i+3\overset\wedge j\)
\(\Rightarrow \overset\rightarrow d=\overset\rightarrow a-\overset\rightarrow b\)
=(i+2j+k)-(3i+j-k)
\(\Rightarrow\overset\rightarrow d =-2\overset\wedge i+\overset\wedge j+2\overset\wedge k\)
Projection \(\overset\rightarrow c \) on \(\overset\rightarrow d\)\(=\frac{\overset\rightarrow c.\overset\rightarrow d}{\left| \overset\rightarrow d \right| }\)
\(=\frac{(4\overset\wedge i+3\overset\wedge j).(-2\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{\left|-2\overset\wedge i+\overset\wedge j+2\overset\wedge k \right| }\)
\(=\frac{-8+3}{\sqrt{4+1+4}}=-\frac{5}{3}\)

\((\overset\rightarrow a-\overset\rightarrow b)^{2}=(\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a+\overset\rightarrow b)\)
\(=\overset\rightarrow a.\overset\rightarrow a+\overset\rightarrow a.\overset\rightarrow b+\overset\rightarrow b.\overset\rightarrow a+\overset\rightarrow b.\overset\rightarrow b\)
(as \(\overset\rightarrow a\bot\overset\rightarrow b\), then \(\overset\rightarrow a. \overset\rightarrow b=\overset\rightarrow b.\overset\rightarrow a=0)\)
\(=\left|\overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right|\) ... (i)
\((\overset\rightarrow a-\overset\rightarrow b)^{2}=(\overset\rightarrow a-\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)\)
\(=\overset\rightarrow a.\overset\rightarrow a-\overset\rightarrow a.\overset\rightarrow b-\overset\rightarrow b.\overset\rightarrow a+\overset\rightarrow b.\overset\rightarrow b\)
\(=\left|\overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right|^{2 }\)...(ii)
By (i) and (ii) both are equal,
\((\overset\rightarrow a+\overset\rightarrow b)^{2} =(\overset\rightarrow a-\overset\rightarrow b)^{2}\)

\(\left|\overset\rightarrow a+\overset\rightarrow b \right|^{ 2 } +\left|\overset\rightarrow a-\overset\rightarrow b \right|^{ 2 }=2{(\left| \overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right| ^{ 2 })}\)
\(\Rightarrow\left| \overset\rightarrow b \right| ^{ 2 }=2,116\)
\(\Rightarrow\left| \overset\rightarrow b \right| =46\)

\(\overset\rightarrow a=3\overset\wedge i+7\overset\wedge j+2\overset\wedge k\)
\(\left|\overset\rightarrow a \right| =\sqrt{9+49+4}=\sqrt{62}\)
and 
\(b=7\overset\wedge i-2\overset\wedge j+3\overset\wedge k\)
\(\left|\overset\rightarrow b \right| =\sqrt{9+49+4}=\sqrt{62}\)
\(\left|\overset\rightarrow a \right| =\left|\overset\rightarrow b \right| \)
but \(\overset\rightarrow a \neq\overset\rightarrow b\)because their corresponding components are different.

\(\overset\rightarrow {BA}\)= Position vector of A- Position vector of B
\(\overset\rightarrow{BA}=(\overset\wedge i+2\overset\wedge j-3\overset\wedge k)-(-\overset\wedge i-2\overset\wedge j+\overset\wedge k)\)
\(\overset\rightarrow {BA}=2\overset\wedge i+4\overset\wedge j-\overset\wedge k \)
\(\left| \overset\rightarrow {BA}\right| =\sqrt{2^{2}+4^{2}+(-4)^{2}}\)
\(=\sqrt{4+16+16}\)
\(=\sqrt{36}=6 \)
Direction cosines of the vector \(\overset\rightarrow{BA}\) are \((\frac{2}{6},\frac{4}{6},-\frac{4}{6})\)
\(=(\frac{1}{3},\frac{2}{3},-\frac{2}{3})\)

Positive vector of \(C=\frac{2(\overset\rightarrow a-3\overset\rightarrow b)+3(2\overset\rightarrow a+\overset\rightarrow b)}{2+3}\)
\(=\frac{2\overset\rightarrow a-6\overset\rightarrow b+6\overset\rightarrow a+3\overset\rightarrow b}{5}\)
Position vector of \(C=\frac{8\overset\rightarrow a-3\overset\rightarrow b}{5}\)

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=(2\overset\wedge i+\overset\wedge j+3\overset\wedge k)+(\overset\wedge i+2\overset\wedge j-\overset\wedge k)\)
\(=3\overset\wedge i+3\overset\wedge j+2\overset\wedge k\)
\(\overset\wedge c=\frac{\overset\rightarrow c}{\left|c \right| }\)
\(=\frac{2i+3j+2k}{\sqrt{9+9+4}}\)
\(=\frac{3}{\sqrt{22}}\overset\wedge i+\frac{3}{\sqrt{22}}\overset\wedge j+\frac{2}{\sqrt{22}}\overset\wedge k\)

\(\overset\rightarrow {AB}=\overset\rightarrow {OB}-\overset\rightarrow {OA}\).....(i)
and \(\overset\rightarrow {BC}=\overset\rightarrow {OC}-\overset\rightarrow {OB}\)....(ii)
\(\therefore \overset\rightarrow {CA}=\overset\rightarrow {OA}-\overset\rightarrow {OC}\)....(iii)
By adding eqn (i),(ii) and (iii)
\(\overset\rightarrow {AB}+\overset\rightarrow {BC}+\overset\rightarrow {CA}=0\)

Scalar quantity: Work done
Vector quantity: Force, velocity, displacement.