The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

Put \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
∴ 2a + 3b + 10c = 4  ....(1)
4a- 6b - 5c = 1 .....(2)
6a + 9b -20c = 2 ...(3)
Δ = \(\left| \begin{matrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (120 - 45) -3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720 = 1200
Δ₁ = \(\left| \begin{matrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{matrix} \right| \)
= \(4\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| +10\left| \begin{matrix} 1 & -6 \\ 2 & 9 \end{matrix} \right| \)
= 4 (120 - 45) -3 (-20 - 10) + 10 (9 + 12)
= 300 + 90 + 120 = 600
Δ₂ = \(\left| \begin{matrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{matrix} \right| =2\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| -4\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| \)
= 2 (-2 - 10) - 4 (-80 - 30) + 10 (8 - 6)
= -60 + 440 + 20 = 400
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{matrix} \right| 2\left| \begin{matrix} -6 & 1 \\ 9 & 2 \end{matrix} \right| -3\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (-12 - 9) -3 (8 - 6) + 4 (36 + 36)
= - 42 - 6 + 288 = 240
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 1200 } =\frac { 1 }{ 2 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 2 } \) ⇒ x = 2
∴ b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 400 }{ 1200 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \) ⇒ y = 1 
∴ c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 240 }{ 1200 } =\frac { 1 }{ 5 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 5 } \) ⇒ z = 5
∴ Solution set is {2, 3, 5}

The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

Given lines are \(\frac { x-3 }{ 3 } =\frac { y-3 }{ -1 } \)....(1)
and z-1 = 0 
\(\Rightarrow\) z = 1
and \(\frac { x-6 }{ 2 } =\frac { z-1 }{ 3 } \) ...(2)  
and y-2 = 0\(\Rightarrow\) y = 2
Substituting y = 2 and z = 1 in (1) we get
\(\frac { x-3 }{ 3 } =\frac { 2-3 }{ -1 } =\frac { -1 }{ -1 } =1\Rightarrow x-3=3\Rightarrow x=6\)
The point of intersection is (6, 2, 1)
Let us check whether (6, 2, 1) satisfies (1) and (2)
\((1)\rightarrow \frac { 6-6 }{ 2 } =\frac { 1-1 }{ 3 } \Rightarrow 0=0\)
\((2)\rightarrow \frac { 6-3 }{ 3 } =\frac { 1-3 }{ -1 } \Rightarrow -1=-1\)
Hence, the given two lines intersect and the point of intersection is (6, 2, 1).

Given lines are
\(\vec { r } =(6\hat { i } +\hat { j } +2\hat { k } )+s(\hat { i } +2\hat { j } -3\hat { k } )\)
\(\vec { a } =6\hat { i } +\hat { j } +2\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \)
and \(\vec { r } =(3\hat { i } +2\hat { j } -2\hat { k } )+t(2\hat { i } +4\hat { j } -5\hat { k } )\)
\(\vec { c } =3\hat { i } +2\hat { j } -2\hat { k } \quad and\quad \vec d = 2\hat { i } +4\hat { j } -5\hat { k } \)
Since \(\vec { b } \neq \vec { d } \), they are not parallel and they do not intersect.
Hence the given lines are skew lines.
Shortest distance between the two skew lines
\(\delta =\frac { |(\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )| }{ |(\vec { b } \times \vec { d } )| } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{matrix} \right| \)
\(=\hat { i } (-10+12)-\hat { j } (-5+6)+\hat { k } (4-4)\)
\(=2\hat { i } -\hat { j } \Rightarrow |\vec { b } \times \vec { d } |\quad \sqrt { { 2 }^{ 2 }++(-1)^{ 2 } } =\sqrt { 5 } \)
\(\vec { c } =\vec { a } =(3\hat { i } +2\hat { j } -2\hat { k } )-(6\hat { i } +\hat { j } +2\hat { k } )\)
\(=-3\hat { i } +\hat { j } -4\hat { k } \)
\(\therefore (\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(-3\hat { i } +\hat { j } -4\hat { k } ).(2\hat { i } -\hat { j } )\)
= -6-1 = -7
\(\therefore \delta =\frac { |-7| }{ \sqrt { 5 } } =\frac { 7 }{ \sqrt { 5 } } units\)

We have to find \((\sqrt{3}+1)^{\frac{1}{3}}\). Let \(z=(\sqrt{3}+i)^{\frac{1}{3}}\). Then \({ z }^{ 3 }=\sqrt { 3 } +i=r\left( cos\theta +isin\theta \right) \)
Then, \(r=\sqrt { 3+1 } =2\) and \(\alpha =\theta =\frac { \pi }{ 6 } \) (\(\because \sqrt{3}+i\) lies in the first quadrant)
Therefore, \({ z }^{ 3 }=\sqrt { 3 } +i=2\left( cos\frac { \pi }{ 6 } +isin\frac { \pi }{ 6 } \right) \)
\(\Rightarrow z=\sqrt [ 3 ]{ 2 } \left( cos\left( \frac { \pi +12k\pi }{ 18 } \right) +isin\left( \frac { \pi +12k\pi }{ 18 } \right) \right) \), k = 0, 1, 2.
Taking k = 0, 1, 2, we get
k = 0, z \(={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 1, \(z={ z }^{ \frac { 1 }{ 3 } }\left( cos\frac { \pi }{ 18 } +sin\frac { \pi }{ 18 } \right) \)
k = 2, \(z={ 2 }^{ \frac { 1 }{ 3 } }\left( cos\frac { 25\pi }{ 18 } +sin\frac { 25\pi }{ 18 } \right) ={ 2 }^{ \frac { 1 }{ 3 } }\left( -cos\frac { 7\pi }{ 18 } -sin\frac { 7\pi }{ 18 } \right) \)

Let the of the parbola be x² = 4ay (1)

Since (30, 16) is a point on (1),
we get 30² = 4 \(\times\) a \(\times\) 16
⇒ a = \(\frac { 30\times 30 }{ 4\times 16 } =\frac { 225 }{ 16 } \)
∴ becomes, x² = \({ x }^{ 2 }=\frac { 4\times 225 }{ 16 } y=\frac { 225 }{ 4 } y\)
Let AC = h m and BD = lm
∴ A(6, h) is a point on the parabola [∵ OD = 6]
∴ \({ 6 }^{ 2 }=\frac { 225 }{ 4 } \times h\)
⇒ \(h=\frac { 36\times 4 }{ 225 } \Rightarrow h=0.52\)
∴ AD = 3 + h = 3 + 0.52 = 3.52 m
Also (12, 1) is a point on the parabola
[∵ ON = 6 + 6 = 12]
∴ \({ 12 }^{ 2 }=\frac { 225 }{ 4 } \times l\)
⇒ l = \(\frac { 12\times 12\times 4 }{ 225 } =\frac { 576 }{ 225 } =2.08\) = 5.08 m
Hence the length of first two vertical cables are 3.52 m and 5.08 m.

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

LHS = \(\left( \cfrac { 1+sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } }{ 1+sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } } \right) ^{ 10 }\)
Let z = \(sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \)
∴ \(\frac { 1 }{ z } =sin\frac { \pi }{ 10 } -icos\frac { \pi }{ 10 } \)
∴ LHS =\(\left[ \frac { 1+z }{ 1+\frac { 1 }{ z } } \right] ^{ 10 }=\left[ \frac { 1+z }{ \frac { z+1 }{ z } } \right] ^{ 10 }\)= z¹⁰
= \(\left[ sin\frac { \pi }{ 10 } +icos\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos\frac { \pi }{ 10 } -isin\frac { \pi }{ 10 } \right] ^{ 10 }\)
= \({ i }^{ 10 }\left[ cos10\frac { \pi }{ 10 } -isin10\frac { \pi }{ 10 } \right] \) [By De Moivers theorum]
= i¹⁰[cos π- i sin π] = -1(-1-i(0)) = 1
Aliter: 
Let \(z=\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\)
\(\frac{1}{z}=\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}\)
\(\left[\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right]^{10}=\left[\frac{1+z}{1+1 / z}\right]\)
\( =\left[\frac{(1+z)}{(z+1)} \cdot z\right]^{10} =z^{10} \)
\( =\left(\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\right)^{10} \)
\( =\left[\cos \left(\frac{\pi}{2}-\frac{\pi}{10}\right)+i \sin \left(\frac{\pi}{2}-\frac{\pi}{10}\right)\right]^{10} \)
\( =\left[\cos \frac{4 \pi}{10}+i \sin \frac{4 \pi}{10}\right]^{10} \)
\( =\cos \frac{4 \pi}{10}(10)+i \sin \frac{4 \pi}{10}(10) \)
\( =\cos 4 \pi+i \sin 4 \pi \)   [\( \because\) By de Moiwe's theorem]  
= 1+ i(0) = 1

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors and which make angles α,β respectively with positive x-axis
Draw AL and BM 丄 to x-axis
Then \(|\vec { OL } |=|\vec { OA } |cos\alpha \Rightarrow \vec { OL } =\vec { |OL| } \hat { i } =cos\alpha \hat { i } \)
\(|\vec { LA } |=|\vec { OB } |\) sin α
⇒ \(\vec { LA } =|\vec { OB } |\hat { j } =sin\alpha (-\hat { j } )=-sin\alpha \hat { j } \)
[\(\vec { LA } \) is in the opp direction of y axis]
\(\hat { a } =\vec { OA } =\vec { OL } +\vec { LA } =cos\alpha \hat { i } -sin\alpha \check { j } \) ..(1)
Similarly \(\hat { b } =\vec { OB } =\vec { OM } +\vec { MB } =cos\beta \hat { i } +sin\beta \hat { j } \) ...(2)
Now \(\hat { a } \times \hat { b } =|\hat { a } ||\hat { b } |sin(\alpha +\beta )\hat { k } =sin(\alpha +\beta )\hat { k } \) ....(3)
[\(|\hat { a } |=|\hat { b } |\) = 1]
Also \(\hat { a } \times \hat { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\alpha & -sin\alpha & 0 \\ cos\beta & cos\beta & 0 \end{matrix} \right| \)
= \(\hat { i } (0)-\hat { j } (0)+\hat { k } \)(cosα sinβ + sinα cosβ)
= (sin α cos β + cos α sin β)\(\hat { k } \) .(4)
using (3) and (4), sin(α+β) = sin α cos β + cos α sin β

Rearranging the terms, the equation of ellipse is 4x² + 24x + y²− 2y + 21 = 0

That is, 4(x² + 6x + 9 − 9) + (y² − 2y + 1 − 1) + 21 = 0,
 4(x + 3)² − 36 + (y−1)² −1 + 21 = 0,
 4(x + 3)² + (y − 1)² = 16,
\(\frac { { \left( x+3 \right) }^{ 2 } }{ 4 } +\frac { { \left( y+1 \right) }^{ 2 } }{ 16 } =1\)
Centre is (-3, 1) a = 4, b = 2, and the major axis is parallel to y-axis c² = 16−4 = 12
c = ±2\(\sqrt { 3 } \)
Therefore, the foci are (−3, 2\(\sqrt { 3 } \) +1) and (−3, −2\(\sqrt { 3 } \) +1).
Vertices are (3, ±4 +1).
That is the vertices are (3, 5) and (3, -3) and the length of Latus rectum = \(\frac { { 2b }^{ 2 } }{ a } \) = 2 units.

Let A be the origin, \(\vec { b } \) be the position vector of B and \(\vec {c } \) be the position vector of C .
Now D is the midpoint of BC , and so the position vector of D \(\frac{\vec{b}+\vec{c}}{2}\). There, we get

\({ \left| \vec { AD } \right| }^{ 2 }=\vec { AD } .\vec { AD } \)= \(\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) .\left( \frac { \vec { b } +\vec { c } }{ 2 } \right) \)= \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )\)....(1)
Now, \(\vec { BD } =\vec { AD } -\vec { AB } \) = \(\frac { \vec { b } +\vec { c } }{ 2 } -\vec { b }=\frac { \vec {c } -\vec { b} }{ 2 }\)
Then, we get, =\({ \left| \vec { BD } \right| }^{ 2 }=\vec { BD } .\vec {BD } \) = \(\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) .\left( \frac { \vec { c } -\vec { b} }{ 2 } \right) \) = \(\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )\)....(2)
Now, adding (1) and (2), we get
Therefore, \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }+2\overrightarrow { b } .\vec { c } )+\frac { 1 }{ 4 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 }-2\overrightarrow { b } .\vec { c } )=\frac { 1 }{ 2 } ({ \left| \vec { b } \right| }^{ 2 }+{ \left| \vec { c } \right| }^{ 2 })\)
⇒ \({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 }=\frac { 1 }{ 2 } ({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 })\)
Hence, \({ \left| \vec { AB } \right| }^{ 2 }+{ \left| \vec { AC } \right| }^{ 2 }=2({ \left| \vec { AD } \right| }^{ 2 }+{ \left| \vec { BD } \right| }^{ 2 })\)

Given p, q are the roots of lx² + nx + n = 0 
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S 
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S

 \(\sum _{ n=1 }^{ 102 }{ { i }^{ n } } \) = (i¹+i²+i³+i⁴)+(i⁵+i⁶+i⁷+i⁸)+....+(i⁹⁷+i⁹⁸+i⁹⁹+i¹⁰⁰)+i¹⁰¹+i¹⁰²
= (i¹+i²⁺i³+i⁴)+(i¹+i²+i³+i⁴)+...+(i¹+i²+i³+i⁴)+i^-1+i^-2
= {i+(-1)+(-i)+1}+{i+(-1)+(-i)}+......+{i+(-1)+(-i)+1}+i+(-1)
= 0+0+...0+i-1
= -1+i

Let \(\vec { a } \) = \(2\hat { i } +3\hat { j } +\hat { k } \), \(\vec { b } \)= \(\hat { i } -2\hat { j } +2\hat { k } \) and \(\vec { c } \) = \(\hat { 3i } +\hat { j } +3\hat { k } \)
\(\vec { a } ,\vec { b } \) and \(\vec { c } \) are coplanar if  \(\vec { a } .(\vec { b } \times \vec { c } )\)
Consider \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} 2 & 3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & 3 \end{matrix} \right| =2\left| \begin{matrix} -2 & 2 \\ 1 & 3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ 3 & 3 \end{matrix} \right| +1\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| \)
= 2 (-6- 2) -3 (3 - 6) + 1(1 + 6)
= 2(-8) - 3(-3) + 1(7)
= -16 + 9 + 7
= -16+16
= 0.
Hence, the given vectors are co-planar.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

Equation of the circle is x² + (y + 2)² = 0
Compare with(x-h)²+(y-k)² = r²
h = 0, k = -2, r² = 0
Centre (h, k) = (0, -2)
radius is 0.

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

Here Δ = b²−4ac = 0 for equal roots. This implies 4(k + 2)² = 4(9)k. This implies k = 4 or 1.

L. H. S = \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } \right\} \)
\(=\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) .\left\{ \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right\} \ \left[ \because \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ 0 } \right] \)
\(=\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ c } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] +0+0\)
\(\left[ \because \left[ \overset { \rightarrow }{ b } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =\left[ \overset { \rightarrow }{ c } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] =0 \right] \)
\(=\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)= R. H. S
Hence proved

\({ sin }^{ -1 }(\frac { 3 }{ 5 } )-{ cos }^{ -1 (}\frac { 12 }{ 13 } )={ sin }^{ -1 }(\frac { 16 }{ 65 }) \)
Let \({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\Rightarrow sinx=\frac { 3 }{ 5 } \)

and \(cosx=\frac { adj }{ hyp } =\frac { 4 }{ 5 } \)
and \({ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) =y\Rightarrow cosy=\frac { 12 }{ 13 } \)
\(siny=\frac { opp }{ hyp } =\frac { 5 }{ 12 } \)

We know that
sin (x - y) = sin x cos y -cos x siny
= \(\frac { 3 }{ 5 } \times \frac { 12 }{ 13 } -\frac { 4 }{ 5 } \times \frac { 5 }{ 12 } =\frac { 36 }{ 65 } -\frac { 20 }{ 6g } =\frac { 16 }{ 65 } \)
\(\therefore x-y={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
\(\Rightarrow { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) -{ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) ={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
Hence proved

\({ cos }^{ -1 }\left( cos\frac { \pi }{ 7 } cos\frac { \pi }{ 17 } -sin\frac { \pi }{ 17 } sin\frac { \pi }{ 17 } \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 7 } +\frac { \pi }{ 17 } \right) \right) \)
\(\left[ \therefore cosA\ cosB-sinA\ sinB=cos(A+B) \right] \)
= \({ cos }^{ -1 }\left( cos\left( \frac { 24\pi }{ 119 } \right) \right) \)    \(\left[ \therefore \frac { 24\pi }{ 119 } \varepsilon \left[ 0,\pi \right] \right] \)
= \(\frac { 24\pi }{ 119 } \) 

Given planes are \(\vec { r } .(\hat { i } +\hat { j } -2\hat { k } )\) and 
\(2x-2y+z=2\Rightarrow \vec { r } .\left( 2\hat { i } -2\hat { j } +\hat { k } \right) =3\)
\(\therefore { \vec { n } }_{ 1 }=\hat { i } +\hat { j } -2\hat { k } \) and \({ \vec { n } }_{ 2 }=2\hat { i } -2\hat { j } +\hat { k } \)
Angle between the plane is'
\(cos\theta =\frac { { \vec { n } }_{ 1 }.{ \vec { n } }_{ 2 } }{ \left| { \vec { n } }_{ 1 } \right| \left| { { \vec { n } }_{ 2 } } \right| } =\frac { \left| 1(2)+1(-2)-2(1) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 2 }^{ 2 }+\left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } } } \)
= \(\frac { \left| -2 \right| }{ \sqrt { 6 } .\sqrt { 9 } } =\frac { 2 }{ \sqrt { 6 } (3) } =\frac { 2 }{ 3\sqrt { 6 } } \)
\(\theta ={ { cos }^{ -1 }\left( \frac { 2 }{ 3\sqrt { 6 } } \right) }\)

The direction ratios of AB is
(6-7, 0- 2, 3 - 1) = (-1, - 2, 2)
[∵ (x₂-x₁), (y₂-y₁), (z₂-z₁)]
Also direction ratios of BC is
(4 - 6, 2 - 0, 4 - 3) = (- 2, 2, 1)
Product of direction ratios is
(- 1) (- 2) + (- 2) (2) + 2 (1)
= 2 - 4 + 2 [∵ if two lines are 丄^r there d.r  d₁b₁ + d₂b₂ + d₃b₃ = 0]
= 4 - 4 = 0
Hence AB 丄 BC
ㄥABC = \(\frac { \pi }{ 2 } \).

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

The latus rectum LL' of the hyperbola passes through S(ae, 0)

∴ L is (ae, y₁)
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } \) = 1 we get,
Substituting L in \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } -\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow { e }^{ 2 }-\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow { e }^{ 2 }-1=\frac { { { y }_{ 1 } }^{ 2 } }{ { b }^{ 2 } } \) \([{ b }^{ 2 }={ a }^{ 2 }({ e }^{ 2 }-1)\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }({ e }^{ 2 }-1)\) \(\Rightarrow \frac { { b }^{ 2 } }{ { a }^{ 2 } } ={ e }^{ 2 }-1]\)
\(\Rightarrow { { y }_{ 1 } }^{ 2 }={ b }^{ 2 }\left( \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right) \) \(\Rightarrow { { y }_{ 1 } }^{ 2 }=\frac { { b }^{ 4 } }{ { a }^{ 2 } } \)
\(\Rightarrow { y }_{ 1 }=\pm \frac { { b }^{ 2 } }{ a } \)
∴ End points oflatus rectum Land L' are
\(\left( ae,\frac { { b }^{ 2 } }{ a } \right) \) and \(\left( ae,-\frac { { b }^{ 2 } }{ a } \right) \)
Hence, the length of latus rectum LL' = \(\frac { { b }^{ 2 } }{ a } +\frac { { b }^{ 2 } }{ a } =\frac { 2{ b }^{ 2 } }{ a } \) units.
Hence proved.

By definition, cot^-1x\(\in(0,\pi)\)
Therefore, cot^-1\((\frac{1}{7})\) = \(\theta\) implies \(\theta \in(0,\pi)\)
But cot^-1\((\frac{1}{7})\) = \(\theta\) implies cot \(\theta\) = \(\frac{1}{7}\) and hence tan \(\theta\) = 7 and \(\theta\) is acute.
Using tan \(\theta\) = \(\frac{7}{1}\), We construct a right triangle as shown .
Then, we have, cos \(\theta\) = \(\frac{1}{5\sqrt2}\).

Given A = \(\left[ \begin{matrix} 1 & \tan { x } \\ -\tan { x } & 1 \end{matrix} \right] \)
|A| = 1 + tan² x
∴ A^-1 = \(\frac { 1 }{ |A| } \)adjA
= \(\frac { 1 }{ 1+tan^{ 2 } } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of elements in the off diagonal]
A^T= \(\\ \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
∴ A^TA^-1
=\(\left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1-tan^{ 2 }x & -tanx-tanx \\ tanx+tanx & -tan^{ 2 }x+1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } & \frac { -2tanx }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tanx }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+tan^{ 2 }x } \end{matrix} \right] \)
A^TA^-1 =\(\left[ \begin{matrix} cos2x & -sin2x \\ sin2x & cos2x \end{matrix} \right] \)
Hence proved.