Given \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\)] ⇒ \(\overset { \rightarrow }{ a } \) 丄 \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ a } \)丄 \(\overset { \rightarrow }{ c } \) 
⇒ \(\overset { \rightarrow }{ a } \) 丄^r to the plane containing \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \) 
Also, \(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ c } \right| \) Since \(\overset { \wedge }{ n } \) [where θ is the angle between \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)]
= 1 \(\times\) 1. sin \(\frac { \pi }{ 6 } \).\(\overset { \rightarrow }{ a } \)
[Since \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) are unit vectors \(\overset { \rightarrow }{ a } \) 丄 both \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)  \(\overset { \rightarrow }{ { n } } \) = \(\overset { \rightarrow }{ a } \)]
\(=\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\(\Rightarrow \overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)

Given lines are \(\frac { x+4 }{ 3 } =\frac { y-7 }{ 4 } =\frac { z+5 }{ 5 } \)
⇒ \(\vec { b } =3\hat { i } +4\hat { j } +5\hat { k } \)
and \(\vec { r } =4\hat { k } +t(2\hat { i } +\hat { j } +\hat { k } )\)
⇒ \(\vec { d } =2\hat { i } +\hat { j } +\hat { k } \)
\( |\vec{b}| =\sqrt{9+16+25}=\sqrt{50} \\ =5 \sqrt{2} \)
∴ cos θ =\(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)=\(\frac { (3\hat { i } +4\hat { j } +5\hat { k } ).(2\hat { i } +\hat { j } +\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } .\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } \)
 \( = \frac{15}{5 \sqrt{2} \times \sqrt{6}}=\frac{3}{\sqrt{2} \cdot \sqrt{2} \sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\ \theta=\frac{\pi}{6} \)

Equation of the circle is
2x² + 2y² - 6x + 4y + 2 = 0
Dividing by 2, we get
x² + y² - 3x + 2y + 1 = 0
Here 2g = -3 ⇒ g = \(\frac { -3 }{ 2 } \)
2f = 2 ⇒ f = 1
and c = 1
∴ Centre is (-g, -f) = \(\left( \frac { 3 }{ 2 } ,-1 \right) \)
and r = \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \) = \(\sqrt { { \left( \frac { 3 }{ 2 } \right) }^{ 2 }+{ 1 }^{ 2 }-1 } \)
= \(\sqrt { \frac { 9 }{ 4 } } =\frac { 3 }{ 2 } \) units.

We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus
\({ sin }^{ -1 }\left( sin\left( \frac { 5\pi }{ 6 } \right) \right) \)= \({ sin }^{ -1 }\left( sin\left( \frac { \pi }{ 6 } \right) \right) \) = \({ sin }^{ -1 }\left( sin\frac { \pi }{ 6 } \right) \), since \(\frac{\pi}{6}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)

We observe that the straight line \(\frac { x-1 }{ 4 } =\frac { 2-y }{ 6 } =\frac { z-4 }{ 12 } \) is parallel to the vector \(4\hat { i } -6\hat { j } +12\hat { k } \) and the straight line \(\frac { x-3 }{ -2 } =\frac { y-3 }{ 3 } =\frac { 5-z }{ 6 } \) is parallel to the vector \(2\hat { i } +3\hat { j } -6\hat { k } \)
Since \(4\hat { i } -6\hat { j } +12\hat { k } =-2(-2\hat { i } +3\hat { j } -6\hat { k } )\), two vectors are parallel, and hence the two straight lines are parallel.

Reducing 9x²-16y² = 144 to the standard form,
we have, \(\frac { { x }^{ 2 } }{ 16 }- \frac { { y }^{ 2 } }{ 9 } =1\)
With the transverse axis is along x-axis vertices are (−4, 0) and (4, 0); and c² = a²+b² = 16 + 9 = 25, c = 5  
Hence the foci are (−5, 0) and (5, 0) 

\({ tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }(-2)\)
Let \({ tan }^{ 1 }\left( \sqrt { 3 } \right) =x\Rightarrow \sqrt { 3 } =tanx\)
\(\Rightarrow tanx=tan\frac { \pi }{ 3 } \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
Let sec^-1(-2) = y
\(\Rightarrow -2=sec\ y\Rightarrow cos\ y=\frac { -1 }{ 2 } \Rightarrow cos\ y=-cos\left( \frac { \pi }{ 3 } \right) \)
\(\Rightarrow cos\ y=cos\left( \pi -\frac { \pi }{ 3 } \right) \)
\(\Rightarrow cos\ y=cos\left( \frac { 2\pi }{ 3 } \right) \Rightarrow y=3\frac { 2\pi }{ 3 } \)
\(\therefore { tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }\left( -2 \right) \)
= \(\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } \)
= \(\frac { \pi -2\pi }{ 3 } =-\frac { \pi }{ 3 } \)
\(\therefore\) \({ tan }^{ -1 }\left( \sqrt { 3 } \right) -{ sec }^{ -1 }(-2)\)= \(-\frac { \pi }{ 3 } \)

Given equation of the circle is
x² + y² - 5x + 2y - 5 = 0
(i) At (-2, 1), (1) becomes
(-2)² + 1²- 5(-2) + 2(1) - 5
= 4 + 1 + 10 + 2 - 5
= 17 - 5 = 12 > 0
∴ (-2, 1) lies outside the circle.
(ii) At (0, 0), (1) becomes -5 < 0
∴ (0, 0) lies inside the circle.
(iii) At (-4, -3), (1) becomes
(-4)² + (-3)² - 5(-4) + 2(-3) - 5
= 16 + 9 + 20 - 6 - 5
= 45 - 11 = 34 > 0
∴ (-4, -3) lies outside the circle.

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

The given line is parallel to the vector \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and the given plane is normal to the vector \(\overset { \rightarrow }{ n } =3\overset { \wedge }{ i } +4\overset { \wedge }{ j } +\overset { \wedge }{ k } \) 
Let θ be the angle between the line and the plane, then
\(\sin { \theta } =\frac { \overset { \rightarrow }{ b } .\overset { \rightarrow }{ n } }{ \left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ n } \right| } =\frac { 9-4+2 }{ \sqrt { 9+1+4 } .\sqrt { 9+16+1 } } \)
\(=\frac { 7 }{ \sqrt { 14 } .\sqrt { 26 } } =\frac { 7 }{ \sqrt { 2 } \times \sqrt { 7 } \times \sqrt { 26 } } =\frac { \sqrt { 7 } }{ \sqrt { 52 } } \)
\(\therefore \theta ={ sin }^{ -1 }\left( \frac { \sqrt { 7 } }{ \sqrt { 52 } } \right) \)

\(tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) \)
Let \(sin^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\)

\(\Rightarrow \frac { 3 }{ 5 } =sinx\)
\(\therefore tanx=\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\({ cot }^{ -1 }\left( \frac { 3 }{ 2 } \right) =y\)
\(\Rightarrow \frac { 3 }{ 2 } =coty\Rightarrow tany=\frac { 2 }{ 3 } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =tan(x+y)\)
\(\frac { tanx+tany }{ 1-tanxtany } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =\frac { 17 }{ 6 } \)

Every point on the line \(\frac { x-1 }{ 2 } =\frac { y-2 }{ 3 } =\frac { z-3 }{ 4 } \) = s (say) is of the form (2s + 1, 3s + 2, 4s + 3) and every point on the line \(\frac { x-4 }{ 5 } =\frac { y-1 }{ 2 } =z=t\) (say) is of the form (5t + 4, 2t + 1, t).
So, at the point of intersection, for some values of s and t, we have
(2s + 1, 3s + 2, 4s + 3) = (5t + 4, 2t + 1, t)
Therefore, 2s - 5t = 3, 3s - 2t = -1 and 4s - t = -3. Solving the first two equations we get t = -1, s = -1. 
These values of s and t satisfy the third equation. Therefore, the given lines intersect. 
Substituting, these values of t or s in the respective points, the point of intersection is (-1,- 1, -1)

From the graph the vertex is at (0, 0) and the parabola is open down
Equation of the parabola is x² = -4ay
(-20, -15) and (20, -15) lie on the parabola 
20² = -4a(-15)
\(4a=\frac { 400 }{ 15 } \)
x² =\(\frac { -80 }{ 3 } \) x y
Therefore equation is 3x² = -80y

Using the definition of the scalar triple product, we get 
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).[(\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )]\) ....(1)
By treating \((\vec { b } \times \vec { c } )\) as the first vector in the vector triple product, we find
\((\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )\) = \(((\vec { b } \times \vec { c } ).\vec { a } )\vec { c } \) -  \(((\vec { b } \times \vec { c } ).\vec { c } )\vec { a } )\) = \([{ \vec { a } ,\vec { b } ,\vec { c } }]\vec { c } \)
Using this value in (1), we get
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).([\vec { a } ,\vec { b } ,\vec { c } ]\vec { c } )=[\vec { a } ,\vec { b } ,\vec { c } ](\vec { a } \times \vec { b } ).\vec { c } ={ [\vec { a } ,\vec { b } ,\vec { c } ] }^{ 2 }\)

\({ tan }^{ -1 }\left( \frac { x }{ y } \right) -{ { tan }^{ -1 }\left( \frac { x-y }{ x+y } \right) }\)
= \(tan^{ -1 }\left( \frac { \frac { x }{ y } -\frac { x-y }{ x+y } }{ 1+\frac { x }{ y } \left( \frac { x-y }{ x+y } \right) } \right) \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { \frac { x(x+y)-y(x-y) }{ y(x+y) } }{ \frac { y(x+y)+x(x-y) }{ y(x+y) } } \right) \)

= tan^-1(1)
= \(\frac { \pi }{ 4 } \)

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.

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplifying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

Equation of tangent to the circle at P(x₁, y₁ ) is xx₁ yy₁ = a²
That is, x(−3) + y(4) = 25
−3x + 4y = 25
Equation of normal is xy₁ - yx₁ = 0
That is, 4x + 3y = 0 

Given \(f(x)=sin^{ -1 }\left( \frac { x^{ 2 }+1 }{ 2x } \right) \le 1\)
We know that the domain of sin^-1(x) is [-1, 1]
\(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \le 1\)
Consider \(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \)
\(\Rightarrow 0\le \frac { { x }^{ 2 }+1 }{ 2x } +1\)
\(\Rightarrow \frac { { x }^{ 2 }+1+2x }{ 2x } \ge 0\)
\(\Rightarrow \frac { \left( x+1 \right) ^{ 2 } }{ 2x } \ge 0\)
\(\Rightarrow \) x = -1 and x < 0
Consider \(\cfrac { { x }^{ 2 }+1 }{ 2x } \le 1\)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ 2x } -1\le 0\)
\(\Rightarrow \frac { { x }^{ 2 }-2x+1 }{ 2x } \le 0\)
\(\Rightarrow \frac { \left( x-1 \right) ^{ 2 } }{ 2x } \le 0\)
From (1) and (2) Domain {-1, 1}

9x²- y²- 36x - 6y + 18 = 0
Given equation is 9x²- y²- 36x - 6y + 18 = 0
⇒ 9x² - 36x - (y² + 6y) = -18
⇒ 9(x²-4x)-(y²+6y) =-18
⇒ 9(x² - 4x + 4 - 4) - (y² + 6y + 9 - 9) = -18
⇒ 9(x-2)²-36-(y+3)²+9 =-18
⇒ 9(x-2)² - (y+3)² = -18+36-9
⇒ 9(x - 2)² - (y + 3)² = 9
Dividing by 9 we get, \(\frac { { (x-2) }^{ 2 } }{ 1 } -\frac { ({ y+3) }^{ 2 } }{ 9 } =1\)
This is an equation of the hyperbola whose transverse axis is parallel to x-axis.
a² = 1, b² = 9
∴ c² = a² + b² = 1 + 9 = 10 ⇒ c = \(\sqrt { 10 } \)
\(e=\sqrt {1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }} =\sqrt { 1-\frac { 9 }{ 1 } } =\sqrt { 10 } \)
a) Center is (2, -3)
⇒ h = 2, k = -3
(b) Foci are (h + c, k), (17 - c, k)
⇒ (2 +\(\sqrt { 10 } \), -3), (2 - \(\sqrt { 10 } \), -3)
(c) Vertic ar (h + a, k) (h - a, k)
⇒ (2+ 1,-3), (2-1,-3)
⇒ (3, -3) (1, -3)
(d) Equation of directrices are x - 2 = \(\pm \frac { a }{ e } \)
⇒ \(x-2=\pm \frac { 1 }{ \sqrt { 10 } } \)
\(x=2\pm \frac { 1 }{ \sqrt { 10 } } \)
⇒ \(x=2+\frac { 1 }{ \sqrt { 10 } } \) and \(x=2-\frac { 1 }{ \sqrt { 10 } } \)

From the line  \(\frac { x-3 }{ 3 } =\frac { y-8 }{ -1 } =\frac { z-3 }{ 1 } \), we get
\(\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +8\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
From the line \(\frac { x+3 }{ -3 } =\frac { y+7 }{ 2 } =\frac { z-6 }{ 4 } \) we get
\(\overset { \rightarrow }{ c } =-3\overset { \wedge }{ i } -7\overset { \wedge }{ j } +6\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ d } =-3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
Since the given lines are not parallel, the shortest distance between the line is 
\(d=\left| \frac { \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) }{ \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| } \right| \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } =\left| \begin{matrix} \overset { \wedge }{ i } \\ 3 \\ -3 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ -1 \\ 2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ 4 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } (-4-2)-\overset { \wedge }{ j } (12+3)+\overset { \wedge }{ k } (6-3)\\ \)
\(=-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| =\sqrt { 36+225+9 } \)
\(=\sqrt { 270 } \)
\(\therefore \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) =\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) \)
= -6 (-6) + 15 _(15) + 3(3)
= 270 ≠ 0
Since the given lines are neither intersecting, nor parallel they are skew lines
\(\therefore d=\frac { 270 }{ \sqrt { 270 } } =\sqrt { 270 } units\)

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

Equation of the plane passing through the point
\(=\vec { a } =\hat { i } -2\hat { j } +4 \) .......(1)
Equation of the given plane is x + 2y - 3z = 11
\(\Rightarrow \vec { r } .(\hat { i } +2\hat { j } -3\hat { k } )=11\)
The given plane is perpendicular to the vector \(\hat { i } +2\hat { j } -3\hat { k } \)
∴ The required plane is parallel to the vector 
\(\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \).......(2)
The given plane is parallel to the line 
\(\frac { x+7 }{ 3 } =\frac { y+3 }{ -1 } =\frac { z }{ 1 } \) Whose direction ratios are  3,-1,1
The required plane is parallel to the vector
 \(\vec { c } =3\hat { i } -\hat { j } +\hat { k } \)     (3)
∴  The non-parametric vector equation of the plane passing through a point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{matrix} \right| \)
\(=\hat { i } (2-3)-\hat { j } (1+9)+\hat { k } (-1-6)\)
\(=-\hat { i } -10\hat { j } -7\hat { k } \)
\(\therefore (\vec { r } -(\hat { i } -2\hat { j } +4\hat { k } )).(-\hat { i } -10\hat { j } -7\hat { k } )=0\)
\([\vec { r } -(-\hat { i } -10\hat { j } -7\hat { k } )]-[(\hat { i } -2\hat { j } +4\hat { k } ).(-\hat { i } -10\hat { j } -7\hat { k } )]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )-[-1+20-28]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )\times 9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )-9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )=9\)
Let \(\Rightarrow \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\therefore (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +10\hat { j } +7\hat { k } )=9\)
⇒ x+10y+7z = 9 which is the required Cartesian equation of the plane.

Given plane is passing through the points
\(\vec { a } =2\hat { i } +2\hat { j } +2\hat { k }, \vec { b } =9\hat { i } +3\hat { j } +6\hat { k } \)
Equation of the given plane is 2x + 6y + 6z = 9. It can be written as \(\vec { r } .(2\hat { i } +6\hat { j } +6\hat { k } )=9\)
Since the given plane is perpendicular to \(2\hat { i } +6\hat { j } +6\hat { k } \), the required plane is parallel to \(\vec { c } =2\hat { i } +6\hat { j } +6\hat { k } \). Hence, parametric form of vector equation of plane passing through two points and parallel to a vector is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R\)
\(\vec { r } =2\hat { i } +2\hat { j } +\hat { k } +s(7\hat { i } +\hat { j } +5\hat { k } )+t(2\hat { i } +6\hat { j } +6\hat { k } ),s,t\in R\)
Cartesian equation of the plane is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{matrix} \right| =0\)
⇒ (x-2)(6-30) - (y-2)(42-10) + (z-1)(42-2) = 0
⇒ (x - 2)(-24) - (y - 2)(32) + (z - 1)(40) = 0
⇒ 24x + 48 - 32y + 64 + 40z - 40 = 0
⇒ -24x - 32y + 40z + 72 = 0
\(\div\) - 8 we get
3x+ 4y - 5z - 9 = 0 is the Cartesian form.
∴ The parametric form of vector equation is
\(\vec { r } =\vec { r } (3\vec { i } +4\vec { j } -5\vec { k } )=9\)

Given lines are x+1 = 2y = -12z
\(\Rightarrow \frac { x+1 }{ 1 } =\frac { y-0 }{ 1 } =\frac { z-0 }{ \frac { -1 }{ 12 } } \)
and x = y + 2 = 6z - 6
\(\Rightarrow \frac { x-0 }{ 1 } =\frac { y+2 }{ 1 } =\frac { z-1 }{ \frac { 1 }{ 6 } } \)
\(\therefore \vec { a } =-\hat { i } ,\vec { b } =\hat { i } +\frac { 1 }{ 2 } \vec { j } -\frac { 1 }{ 12 } \hat { k } \)
\(\vec { c } =-2\hat { j } +\hat { k } \ and\ \vec { d } =\hat { i } +\hat { j } +\frac { 1 }{ 6 } \hat { k } \)
\(\vec { c } -\vec { a } =2\hat { j } +\hat { k } -(-\hat { i } )=\hat { i } -2\hat { j } +\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \frac { 1 }{ 2 } & -\frac { 1 }{ 12 } \\ 1 & 1 & \frac { 1 }{ 6 } \end{matrix} \right| \)
\(=\hat { i } \left( \frac { 1 }{ 12 } +\frac { 1 }{ 12 } \right) -\hat { j } \left( \frac { 1 }{ 6 } +\frac { 1 }{ 12 } \right) +\hat { k } \left( 1-\frac { 1 }{ 12 } \right) \)
\(=\frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \)
Now \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)
\(=\left( \hat { i } -2\hat { j } +\hat { k } \right) .\left( \frac { 1 }{ 6 } \hat { i } -\frac { 1 }{ 4 } \hat { j } +\frac { 1 }{ 2 } \hat { k } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 2 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1 }{ 6 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 6 } +1=\frac { 7 }{ 6 } \)
Since \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)0, the given lines are skew lines. 
\(\therefore |\vec { b } \times \vec { d } |=\sqrt { \frac { 1 }{ 36 } +\frac { 1 }{ 16 } +\frac { 1 }{ 4 } } \)
\(\sqrt { \frac { 4+9+36 }{ 144 } } =\sqrt { \frac { 49 }{ 144 } } =\frac { 7 }{ 12 } \)
Shortest distances between the skew lines

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

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.

The parametric equation of tangent at 't₁' to the parabola y² = 4ax is yt₁ = x + at₁²  ...(1)
Also, the parametric equation of tangent at 't₂' to the parabola y = 4ax is yt₂ = x+ at₂² ...(2)
(1) ➝ yt₁ = x + at₁²
(2) ➝ yt₂ = x + at₂²
(1) - (2) y(t₁ -t₂) = a(t₁² - t₂²)

⇒ y = a(t₁ + t₂)
Substitutingy = a(t₁ + t₂) in (1) we get,
a(t₁ + t₂)t₁ = x + at₁²

⇒ x = at₁t₂
Hence, the point of intersection of two lengths is
[at₁t₂, a(t₁ + t₂)]

x²+3y² = 12
\(\div 12\) we get, \(\frac { { x }^{ 2 } }{ 12 } +\frac { { y }^{ 2 } }{ 4 } =1\)
∴ a² = 12, b² = 4
The line x-y+ 4 = 0 can be rewritten as y = x+4.
∴ m = 1, c = 4
The condition for y = mx + 4 to be a tangent to the ellipse is c² = a²m² + b²
∴ (4)² = 12(1)² + 4
⇒ 16 = 12+4
⇒ 16 = 16
Since the condition is satisfied, the line x - y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12.
Also, the point of contact is \(\left( -\frac { { a }^{ 2 }m }{ c } ,\frac { { b }^{ 2 } }{ c } \right) \)
\(\Rightarrow \left( -\frac { 12(1) }{ 4 } ,\frac { 4 }{ 4 } \right) \Rightarrow (-3,1)\)
∴The point of contact is (-3, 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 = \(\pm\frac{3}{4}\)

Now, \(tan^{ -1 }\left( \frac { x-1 }{ x-2 } \right) +tan^{ -1 }\left( \frac { x+1 }{ x+2 } \right) =tan^{ -1 }\left[ \frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \right] =\frac { \pi }{ 4 } \)
Thus, \(\frac { \frac { x-1 }{ x-2 } +\frac { x+1 }{ x+2 } }{ 1-\frac { x-1 }{ x-2 } \left( \frac { x+1 }{ x+2 } \right) } \) = 1, which on simplification gives 2x²−4 = −3
Thus, x² = \(\frac{1}{2}\)gives x = \(\pm \frac { 1 }{ \sqrt { 2 } } \)

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ........... (1)
(1) passes through (1, 0)
⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0
⇒ 2g + c = -1 ................(2)
(1) passes through (-1, 0)
⇒ (-1)² + 0 + 2g(-1) + 2f(0)+ c = 0
⇒ -2g + c = -1 ..............(3)
Also (1) passes through (0, 1)
⇒ 0 + 1²+ 2g(0) +2f(1) + c = 0
⇒ 2f + c = -1 .................(4)
(2) + (3) ⇒ 2c = -2
⇒ c = -1
Substituting c = -1 in (2), we get
2g-1 = -1
⇒ 2g = 0
⇒ g = 0
Substituting c = -1 in (4) we get,
2f -1 = -1
⇒ 2f  = 0
⇒ f = 0
∴ The required equation of the circle 
x² + y² - 1 = 0