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, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

The given line \(\frac { x-1 }{ 6 } =\frac { y+3 }{ 2 } =\frac { z-2 }{ 3 } \) is parallel to the vector \(\overset { \rightarrow }{ b } =6\check { i } +2\check { j } +3\check { k } \)
The normal to the given plane in
\(\overset { \rightarrow }{ n } =2\check { i } +\check { j } +2\check { k } \)
\(sin\theta =\frac { \overset { \rightarrow }{ b } .\overset { \rightarrow }{ n } }{ \left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ n } \right| } \)
\(=\frac { (6\check { i } +2\check { j } +3\check { k } )(2\check { i } -\check { j } +2k) }{ \sqrt { 39+4+9 } \sqrt { 4+1+4 } } \)
\(=\frac { 12-2+6 }{ \sqrt { 49 } \sqrt { 9 } } \)
\(\Rightarrow sin\theta =\frac { 16 }{ 7\times 3 } =\frac { 16 }{ 21 } \)
\(\therefore \theta =sin^{ -1 }\left( \frac { 16 }{ 21 } \right) \)

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

Since ABCD is a parallelogram and diagonal are bisect each other.
Let A(3, - 1, - 1), B(5, - 4, 0), C(2, 3, - 2), \(D(x_{ 2 },y_{ 2 },z_{ 2 })\)
Midpoint of AC \(\left( \frac { 2+3 }{ 2 } ,\frac { 3-1 }{ 2 } .\frac { -2-1 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,\frac { 2 }{ 2 } ,\frac { -3 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,1,\frac { -3 }{ 2 } \right) \)
Mid-point of BD is same let P be the mid-point of ACand BD
\(\frac { 5 }{ 2 } =\frac { 5+x_{ 2 } }{ 2 } ,1=\frac { -4+y_{ 2 } }{ 2 } \)
\(\frac { -3 }{ 2 } =\frac { 0+z_{ 2 } }{ 2 } \)
\(\Rightarrow x_{ 2 }=0,y_{ 2 }=6,z_{ 2 }=-3\)
D(0,6,-3)

Sample space ={B₁B₂, B₁G₂, G₁B₂, G₁G₂}, B₁ and G₁ are the older boy and girl respectively.
Let   E₁ = both the children are boys;
E₂ = one of the children are boys;
E₃ = the older child is a boy
Then, (a) P(E₁/E₂) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 2 } }{ { E }_{ 2 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \)
(b) P(E₁/E₃) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 3 } }{ { E }_{ 3 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 2 }{ 4 } } =\frac { 1 }{ 2 } \)

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
Let \(\alpha \) makes with x-axis, y makes with z-axis and \(\beta \) makes with y-axis
\(\alpha =cos\quad 60^{ \circ },m=cos\quad \beta ,y=cos45^{ \circ }\)
\(\alpha ^{ 2 }+\beta ^{ 2 }+\gamma ^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }60^{ \circ }+cos^{ 2 }\beta +cos^{ 2 }45^{ \circ }\)
\(\Rightarrow \frac { 1 }{ 4 } +cos^{ 2 }\beta +\frac { 1 }{ 2 } =1\)
\(\Rightarrow cos^{ 2 }\beta =1-\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \)
\(=\frac { 4-1-2 }{ 4 } \)
\(=\frac { 1 }{ 4 } \)
\(\Rightarrow cos \beta =\pm \frac { 1 }{ 2 } \)
\(\beta =\frac { \pi }{ 3 } or\ \frac { 2\pi }{ 3 } \)
But angle between two lines in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
Hence required angles is \(\frac { \pi }{ 3 } \)

Total number of cards = 2 + 6 + 8 = 16
(a) Number of red balls = 2
\(\therefore\)  Required probability = \(\frac { 2 }{ 16 } =\frac { 1 }{ 8 } \)
(b) Number of black balls = 6
\(\therefore\)  Required probability = \(\frac { 6 }{ 16 } =\frac { 3 }{ 8 } \)
(c) number of green balls = 8
\(\therefore\)  Required probability = \(\frac { 8 }{ 16 } =\frac { 1 }{ 2 } \)
(d) Number of non-red balls = 14
 \(\therefore\) Required probability = \(\frac { 14 }{ 16 } \)

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

y = a cos2x + b sin2x
\(\Rightarrow \frac { dy }{ dx } =-a\sin { 2x } \times 2+b\cos { 2x } \times 2\)
\(\Rightarrow \frac { dy }{ dx } =2\left[ -a\sin { 2x } +b\cos { 2x } \right] \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2\left[ -a\cos { 2x } \times 2-b\sin { 2x } \times 2 \right] \)
\(\Rightarrow \frac { { x }^{ 2 }y }{ { dx }^{ 2 } } =-4\left[ a\cos { 2x } +b\sin { 2x } \right] =-4y\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +4y=0\)

\(\int { { e }^{ x }cotx } dx+\int { { e }^{ x }(logsinx) } dx\)
\(=\int { { e }^{ x } } cotxdx+log(sinx)\int { { e }^{ x }dx } -\int { \left[ \frac { d }{ dx } (logsinx)\int { { e }^{ x }dx } \right] } \)
\(=\int { { e }^{ x }cotx } dx-{ e }^{ x }log(sinx)-\int { cotx{ e }^{ x }dx } \)
\(={ e }^{ x }log\left| sinx \right| +C\)

\(\int { \frac { { e }^{ x } }{ x } } dx-\int { \frac { e^{ x } }{ x^{ 2 } } } dx\)
\(=\frac { 1 }{ x } \int { e^{ x }-\int { \left( \frac { d }{ dx } \left( \frac { 1 }{ x } \right) \int { { e }^{ x }dx } \right) } } dx-\int { \frac { { e }^{ x } }{ x^{ 2 } } } \)
\(=\frac { { e }^{ x } }{ x } -\int { -\frac { 1 }{ { x }^{ 2 } } { e }^{ x }dx } -\int { \frac { { e }^{ x } }{ { x }^{ 2 } } } dx\)
\(=\frac { { e }^{ x } }{ x } +\int { \frac { { e }^{ x } }{ x^{ 2 } } dx } -\int { \frac { { e }^{ x } }{ { x }^{ 2 } } } dx\)
\(=\frac { { e }^{ x } }{ x } +C\)

Let x = cos 2\(\theta \)
\(=sin\left[ 2t{ an }^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \right] \)
\(=sin\left[ 2tan^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \right] \)
\(\left[ \because cos2\theta =1-2{ sin }^{ 2 }\theta \ and\ cos\ 2\theta =2{ cos }^{ 2 }\theta -1 \right] \)
\(=sin\left[ 2{ tan }^{ -1 }\left( tan\quad \theta \right) \right] \)
\(=sin(2\theta )=\sqrt { 1-{ cos }^{ 2 }2\theta } \)
\(=sin\quad 2\theta =\sqrt { 1-{ x }^{ 2 } } \)

Let x denote the side and A denote the area of the equilateral triangle at instant t.
\(\frac{dx}{dt}=5 \ and\ x=10\ cm\)
\(A=\frac{\sqrt3}{4}x^2\)
\(\frac{dA}{dt}=\frac{\sqrt3}{4}2x\frac{dx}{dt}\)
\(\frac{dA}{dt}=\frac{\sqrt3}{4}\times 2\times5\times 10\)
\(\frac{dA}{dt}=\sqrt3\times 5\times 5\)
\(=25\sqrt3\)
Hence required rate of change
\(=25\sqrt3 \ cm^2/sec\)

If A is symmetric matrix then
\(A={ A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & b & -2 \\ 3 & 1 & 3 \\ 2a & 3 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 & 2a \\ b & 1 & 3 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(\therefore\)  By equality of matrices, 
b = 3 and a = - 1

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) \(\in R\) then (a, c) \(\in R\) , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c \(\in A\) such that (a, b)(b, c) \(\in R\)  implies (a, c) . \(\in R\)
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 \(\Rightarrow \)(3, 1)  \(\in R\)
\(\therefore\) A is transitive.