Let \(\overset { \rightarrow }{ B } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ A } \times \overset { \rightarrow }{ B } =\overset { \rightarrow }{ C } \Rightarrow \left| \begin{matrix} \overset { \wedge }{ i } \\ 1 \\ x \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 1 \\ y \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ z \end{matrix} \right| =\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\Rightarrow \overset { \wedge }{ i } (z-y)-\overset { \wedge }{ j } (z-x)+\overset { \wedge }{ k } (y-x)\quad \overset { \wedge }{ j } -\overset { \wedge }{ k } \)
Equating the like components on both sides, we get
z - y = 0   .....(1)
x - y = 1   .....(2)
y - x = -1  .....(3)
Also, \(\overset { \rightarrow }{ A } .\overset { \rightarrow }{ B } =3\Rightarrow \left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) .\left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) =3\)
⇒ x + y + z = 3   ....(4)
Solving (1), (2), (3) and (4), we get \(x=\frac { 5 }{ 3 } ,y=\frac { 2 }{ 3 } \)and \(z=\frac { 2 }{ 3 } \)
\(\therefore \overset { \rightarrow }{ B } =\frac { 5 }{ 3 } \overset { \wedge }{ i } +\frac { 2 }{ 3 } \overset { \wedge }{ j } +\frac { 2 }{ 3 } \overset { \wedge }{ k } \)

LHS = \(\left( \frac { i+\sqrt { 3 } }{ -i+\sqrt { 3 } } \right) ^{ 2\omega }+\left( \frac { i-\sqrt { 3 } }{ i+\sqrt { 3 } } \right) ^{ 2\omega }\)
= \(\left( \frac { \sqrt { 3 } +i }{ \sqrt { 3 } -i } \times \frac { \sqrt { 3 } +i }{ \sqrt { 3 } +i } \right) ^{ 2\omega }+\left( \frac { -\sqrt { 3 } +i }{ \sqrt { 3 } +i } \times \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 3-1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }+\left( \frac { -3+1+2\sqrt { 3 } i }{ 3+1 } \right) ^{ 2\omega }\)
= \(\left( \frac { 2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 4 } \right) ^{ 2\omega }\)
= \(\left( \frac { 1+\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }+\left( \frac { -2+2\sqrt { 3 } i }{ 2 } \right) ^{ 2\omega }\)
=\(\left[ -\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) \right] ^{ 2\omega }+\left[ \frac { -1+\sqrt { 3 } i }{ 2 } \right] ^{ 2\omega }\)
= (-ω²)^2ω+(ω)^2ω
[∴ ω = \(\frac { -1+i\sqrt { 3 } }{ 2 } \), ω² = \(\frac { -1-i\sqrt { 3 } }{ 2 } \)]
= ω^4ω+ω^2ω
= (ω³)¹³³. ω¹ + (ω³)⁶⁶.ω²
= 1.ω+1.ω²  [∴ 1+ω+ω² = 0 & ω³ = 1]
= ω + ω²
= -1 = RHS

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

\({ tan }^{ -1 }\left( { \frac { x-1 }{ x-2 } } \right) +{ tan }^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }\left( \cfrac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\left( \frac { x-1 }{ x-2 } \right) \left( \frac { x+1 }{ x+2 } \right) } \right) =\frac { \pi }{ 4 } \)

\(\Rightarrow \frac { 2{ x }^{ 2 }-4 }{ { x }^{ 2 }-4-{ x }^{ 2 }+1 } =1\)
\(\Rightarrow\) 2x²- 4 = -3
\(\Rightarrow \) 2x²- 4 = -3
\(\Rightarrow\) 2x² = -3 + 4 = 1
\(\Rightarrow\) \({ x }^{ 2 }=\frac { 1 }{ 2 } \)
\(\Rightarrow\) \(x=\frac { 1 }{ \sqrt { 2 } } \)

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
\(\therefore \delta =\frac { |-7| }{ \sqrt { 5 } } =\frac { 7 }{ \sqrt { 5 } } units\)

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

Given A =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \), B=\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+3+6 & -5+2+3 & -10+1+9 \\ 7+3-10 & 7+2-3 & 14+1-15 \\ 1-3+2 & 1-2+1 & 2-1+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃
BA =\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+7+2 & 1+1-2 & 3-5+2 \\ -15+14+1 & 3+2-1 & 9-10+1 \\ -10+7+3 & 2+1-3 & 6-5+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃.
So, we get AB = BA = 4. I₃
⇒ \(\left( \frac { 1 }{ 4 } A \right) B=B\left( \frac { 1 }{ 4 } A \right) =1\)
⇒ B^-1 = \(\frac { 1 }{ 4 } \) = 1
Writing the given set of equations in matrix form we get,
\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(B=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] =\left[ \frac { 1 }{ 4 } A \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 8 \\ 4 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ -1 \end{matrix} \right] \)
∴ x = 2, y = 1, z = -1
Hence, the solution set is {2, 1, - 1}.

Since the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

We find that |A| = \(\left| \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & 4 \\ 2 & -4 & 3 \end{matrix} \right| \) = 8(21 - 16) + 6(-18 + 8) + 2(24 - 14) = 40 - 60 + 20 = 0
By the definition of adjoint, we get
adj A = \({ \left[ \begin{matrix} \left( 21-16 \right) & -\left( -18+8 \right) & \left( 24-14 \right) \\ -\left( -18+8 \right) & \left( 24-4 \right) & -\left( 32+12 \right) \\ \left( 24-14 \right) & -\left( -32+12 \right) & \left( 56-36 \right) \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
So, we get
A(adj A) = \(\left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & 80-120+40 & 80-120+40 \\ -30+70-40 & -60+140-80 & -60+140-80 \\ 10-40+30 & 20-80+60 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃,
Similarly, we get
(adj A)A = \(\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & -30+70-40 & 10-40+30 \\ 80-120+40 & -60+140-80 & 20-80+60 \\ 80-120+40 & -60+140-80 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃.
Hence, A(adj A) = (adj A)A = |A|I₃.

i²⁵ = (i⁴)⁶ \(\times\) i¹ = i⁶ \(\times\) i = i
∴ |i²⁵| = |i| = 1

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

The parametric form of vector equation of a line passing through a point \(\left( \overset { \rightarrow }{ a } \right) \) and parallel to \(\overset { \rightarrow }{ b } \) is
\({ \overset { \rightarrow }{ r } }=\overset { \rightarrow }{ a } +t\overset { \rightarrow }{ b } \), t ∈ R
⇒ Here \(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\therefore { \overset { \rightarrow }{ r } }=\left( 2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) +t\left( 2\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \right) \), t ∈ R which is the required equation of a line.

Here A = 3, B = 0, C = 3, D = -4, E = 3 and F = 10
A = C and B = 0 (No xy term)
Hence, the given equations represents a circle.

Suppose A is symmetric. Then, A^T = A and so, by theorem (vi), we get
adj(A^T) = (adj A)^T ⇒ adj A = (adj A)^T ⇒ adj A is symmetric.

Given line in Ix + my + n = 0 ....(1)
tangent at (x₁, y₁) to the circle x² + y² = 9² is
xx₁ + yy₁= a₂ ...(2)
Comparing the co-efficients of like terms in (1)
and (2), we get, \(\frac { { x }_{ 1 } }{ l } =\frac { { y }_{ 1 } }{ m } =\frac { -{ { a }^{ 2 } } }{ n } \)
\({ x }_{ 1 }=\frac { -{ a }^{ 2 }l }{ n } \), and \({ y }_{ 1 }=\frac { -{ a }^{ 2 }m }{ n } \)
Since (x₁ , y₁) is a point on the circle, x²₁ + y²₁ = a²
\(\left( \frac { -{ a }^{ 2 }l }{ n } \right) +\left( \frac { -{ a }^{ 2 }m }{ n } \right) ={ a }^{ 2 }\)
\(\frac { -{ a }^{ 4 }{ l }^{ 2 } }{ { n }^{ 2 } } +\frac { { a }^{ 4 }{ m }^{ 2 } }{ { n }^{ 2 } } ={ a }^{ 2 }\)
⇒ \(-{ a }^{ 4 }{ l }^{ 2 }+{ a }^{ 4 }{ m }^{ 2 }={ a }^{ 2 }\)
\({ a }^{ 2 }({ l }^{ 2 }+{ m }^{ 2 })=1\)

Let AOB be right angled triangle, right angled at O.
Take O as origin.
Then \(\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ OB } =\overset { \rightarrow }{ b } \)
\(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \)
\(\therefore \overset { \rightarrow }{ OA } \bot \overset { \rightarrow }{ OA } =\overset { \rightarrow }{ a } \bot \overset { \rightarrow }{ b } \Rightarrow \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\)
\(\therefore { AB }^{ 2 }={ \left| \overset { \rightarrow }{ AB } \right| }^{ 2 }=\overset { \rightarrow }{ AB } .\overset { \rightarrow }{ AB } \)
\(=0=\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } \right) \)
\(=\overset { \rightarrow }{ b } .\overset { \rightarrow }{ b } -\overset { \rightarrow }{ b } .\overset { \rightarrow }{ a } -\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } \)
\(={ \left| \overset { \rightarrow }{ b } \right| }^{ 2 }-2\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +{ \left| \overset { \rightarrow }{ a } \right| }^{ 2 }\)
= OB² - 0 + OA²    \(\left[ \because \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0 \right] \)
⇒ AB² = OA² + OB²
Hence the Pythagoras theorem

The matrix from of the system of equations is
\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \) ⇒ AX = B where
A =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \)
Transforming augmented matrix to row-echelon form we get
[A|B] =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -3 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ 0 \end{matrix} \right] \)
Here \(\rho\) (A) = \(\rho\) [A|B] = 1
∴ \(\rho\) (A) = \(\rho\) [A|B] = 1 < the number of unknowns, the system is consistent and has one parameter family of solutions
∴ Put y = t, where t \(\in \) R
Writing the row-echelon form to equations we get
 2x - 3y = 7
∴ 2x - 3t = 7
⇒  2x = 7 + 3t
⇒ x = \(\frac { 1 }{ 2 } \)(7 + 3t)
∴ Solution set is {\(\frac { 1 }{ 2 } \)(7+3t), t} where t \(\in \) R.

x² = 24y
The given parabola is open upward parabola and 4a = 24 ⇒ a = 6.
(a) Vertex is (0, 0)
⇒ h = 0, k = 0
(b) focus is (0 + h, a + k)
⇒ (0 + 0, 6 + 0) = (0, 6)
(c) Equation of directrix is y = k - a
⇒ y = 0 - 6 ⇒ y = -6
(d) Length of latus rectum is 4a = 24

The distance between origin to z = i, −2 + i, and 3 are
|z| = |i| = 1
|z| = |−2+i| = \(\sqrt { \left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
|z| = |3| = 3
Since \(1<\sqrt { 5 } <3\), the farthest point from the origin is 3 .

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 \(\theta\) = \(\frac{1}{5\sqrt2}\).

We get AB = \(\left[ \begin{matrix} 0 & -3 \\ 1 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0+0 & 0+3 \\ -2+0 & -3-4 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 \\ -2 & -7 \end{matrix} \right] \)
(AB)^-1 = \(\frac { 1 }{ \left( 0+6 \right) } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \)...(1)
A^-1 = \(\frac { 1 }{ \left( 0+3 \right) } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] \)
B^-1 = \(\frac { 1 }{ \left( 2-0 \right) } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \)
B^-1A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \) ...(2).
As the matrices in (1) and (2) are same, (AB)⁻¹ = B⁻¹A⁻¹ is verified.