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'
\(cos\theta =\frac { { \vec { n } }_{ 1 }.{ \vec { n } }_{ 2 } }{ \left| { \vec { n } }_{ 1 } \right| \left| { { \vec { n } }_{ 2 } } \right| } =\frac { \left| 1(2)+1(-2)-2(1) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 2 }^{ 2 }+\left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } } } \)
= \(\frac { \left| -2 \right| }{ \sqrt { 6 } .\sqrt { 9 } } =\frac { 2 }{ \sqrt { 6 } (3) } =\frac { 2 }{ 3\sqrt { 6 } } \)
\(\theta ={ { cos }^{ -1 }\left( \frac { 2 }{ 3\sqrt { 6 } } \right) }\)

Equation of given plane is
\(\vec { r } .\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =8\)
\(\therefore { \vec { n } }_{ 1 }=6\hat { i } +3\hat { j } +2\hat { k } \)
and the line is \(\vec { r } =\left( 2\hat { i } -\hat { j } +\hat { k } \right) +t\left( \hat { i } +2\hat { j } -2\hat { k } \right) \)
\(\therefore \vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
Angle between a line and a plane is
\(sin\theta =\frac { \left| \vec { b } .\vec { n } \right| }{ \left| \vec { b } \right| \left| \vec { n } \right| } =\cfrac { \left| 6(1)+3(2)+2(-2) \right| }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 }+{ 2 }^{ 2 } } } } \)
= \(\frac { 6+6-4 }{ \sqrt { 9 } .\sqrt { 36+9+4 } } =\frac { 8 }{ \sqrt { 9 } .\sqrt { 49 } } =\frac { 8 }{ 3\left( 7 \right) } =\frac { 8 }{ 21 } \)
\(\therefore \theta ={ sin }^{ -1 }\left( \cfrac { 8 }{ 21 } \right) \)

The equation of the plane having intercepts a, b, c on the x, y, z axes respectively is \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\). 
Since the sum of the reciprocals of the intercepts on the coordinate axes is a constant, we have \(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =k\),  where k is a constant, and which can be written as \(\frac { 1 }{ a } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ b } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ c } \left( \frac { 1 }{ k } \right) =1\)
This shows that the plane \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) passes through the fixed point \(\left( \frac { 1 }{ k } ,\frac { 1 }{ k } ,\frac { 1 }{ k } \right) \)

If the position vector of the given point is \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \), then the equation of the plane passing through a point and normal to a vector is given by \((\vec { r } -\vec { a } ).\vec { n } =0\) or \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
Substituting \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \) in the above equation, we get
\(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )=(4\hat { i } +2\hat { j } -3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
Thus, the required vector equation of the plane is \(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )\)= 3. If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) then
we get the Cartesian equation of the plane 2x − y + z = 3.

Let \(\hat { d } =6\hat { i } +2\hat { j } -3\hat { k } \) and p = 12
If \(\hat { d } \) is the unit normal vector in the direction of the vector \(6\hat { i } +2\hat { j } -3\hat { k } \)
then \(\hat { d } =\frac { \hat { d } }{ \left| \hat { d } \right| } =\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )\)
If \(\hat { r } \) is the position vector of an arbitrary point (x, y, z) on the plane, then using \(\vec { r } .\hat { d } =p\), the vector equation of the plane in normal form is \(\vec { r } .\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) in the above equation, we get \((x\hat { i } +y\hat { j } +z\hat { k } ).\frac { 1 }{ 7 } (6\hat { i } +2\hat { j } -3\hat { k } )=12\)
Applying dot product in the above equation and simplifying, we get 6x + 2y - 3z = 84,  which is the Cartesian equation of the required plane.

Given  \(\hat { a } \times (\hat { b } \times \hat { c } )=\frac { 1 }{ 2 } \hat { b } \)
⇒ \((\hat { a } .\hat { c } )\hat { b } -(\hat { a } .\hat { b } )\hat { c } =\frac { 1 }{ 2 } \hat { b } \)
⇒ \(\lambda \hat { b } -\mu \hat { c } =\frac { 1 }{ 2 } \hat { b } \) [∵ put \(\lambda =\hat { a } .\hat { c } \) & \(\mu =\hat { a } .\hat { b } \)]
⇒ \(\left( \lambda -\frac { 1 }{ 2 } \right) \hat { b } -\mu \hat { c } \) = 0
Since \(\hat { b } \) and \(\hat { c } \) are non-collinear vectors
\(\lambda -\frac { 1 }{ 2 } \) = 0 and μ = 0
∴ \(\lambda =\frac { 1 }{ 2 } \)
⇒ \(\hat { a } .\hat { c } =\frac { 1 }{ 2 } \)
⇒ \(|\hat { a } ||\hat { c } |cos\theta =\frac { 1 }{ 2 } \)  [∵ By the definition of scalar product]
⇒ (1) (1) \(cos\theta =\frac { 1 }{ 2 } \)    \(\left[ \because |\vec { a } |=|\vec { c } |=1 \right] \)
⇒ cos θ = \(\frac{1}{2}\)
⇒ θ = 60⁰ = \(\frac { \pi }{ 3 } \)
Hence angle between \(\vec { a } \) and \(\vec { c } \) is \(\frac { \pi }{ 3 } \).

LHS = \([\vec { a } -\vec { b } ,\vec { b } -\vec { c } ,\vec { c } -\vec { a } ]\) = 0
[∵ cross product is distributive]
\((\vec { a } -\vec { b } ).[(\vec { b } -\vec { c } )\times (\vec { c } -\vec { a } )]\)
= \((\vec { a } -\vec { b } ).[(\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { c } \times \vec { c } +\vec { c } \times \vec { a } )\)
= \((\vec { a } -\vec { b } ).[\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -0+\vec { c } \times \vec { a } ]\)
\([\because \vec { c } \times \vec { c } =0]\)
= \([\vec { a } \vec { b } \vec { c } ]-[\vec { a } \vec { b } \vec { a } ]+[\vec { a } \vec { c } \vec { a } ]-[\vec { b } \vec { b } \vec { c } ]+[\vec { b } \vec { b } \vec { a } ]-[\vec { b } \vec { c } \vec { a } ]\)
= \([\vec { a } \vec { b } \vec { c } ]-0+0-0+0-[\vec { b } \vec { c } \vec { a } ]\)
= \([\because [\vec { a } \vec { b } \vec { a } ]=[\vec { b } \vec { b } \vec { c } ]=0]\)
= \([\vec { a } \vec { b } \vec { c } ]-[\vec { a } \vec { b } \vec { c } ]\)
= 0 = RHS.

  \(|\vec { c } |\) = 1 and \(\vec { c } \bot \vec { a } \) & \(\vec { b } \)
Also, angle between \(\vec { a } \) and \(\vec { b } \) is \(\frac { \pi }{ 6 } \)
Consider \([\vec { a } \vec { b } \vec { c } ]=\vec { a } .(\vec { b } \times \vec { c } )\)
= \((\vec { a } \times \vec { b } ).\vec { c } \)
[∵ angle between \(\vec { a } \) and \(\vec { b } \) is \(\frac { \pi }{ 6 } \). \(\vec { c } \) 丄 both a & b]
= \(|\vec { a } ||\vec { b } |sin\frac { \pi }{ 6 } .\vec { c } .\vec { c } \)
= \(|\vec { a } ||\vec { b } |.\frac { 1 }{ 2 } \)(1)
= \(|\vec { a } ||\vec { b } |.\frac { 1 }{ 2 } \) [∵ \(\vec { c } .\vec { c } \) = 1]
∴ \([\vec { a } \vec { b } \vec { c } ]^{ 2 }=|\vec { a } |^{ 2 }|\vec { b } |^{ 2 }.\frac { 1 }{ 4 } =\frac { 1 }{ 4 } |\vec { a } |^{ 2 }|\vec { b } |^{ 2 }\).

Let ABCD be the given parallelogram and ABC¹ D¹ be the new parallelogram with same base AB and between the same parallel lines AB and DC.
∴ Vector area of parallelogram
ABCD =\(\vec { AB } \times \vec { AD } \)
=\(\vec { AB } \times (A\vec { D^{ 1 } } +{ D }^{ 1 }\vec { D } )\)
[By Δ law of addition is ΔADD¹]
= \((\vec { AB } \times \vec { AD^{ 1 } } )+(\vec { AB } \times D^{ 1 }\vec { D } )\) [∵ vector product is distributive]
= \((\vec { AB } \times \vec { AD } )\) +0 [∵ \(\vec { AB } \) and \(\vec { DD^{ 1 } } \) are parallel]
= Vector area of parallelogram ABC¹D¹
∴ Area of parallelogram ABCD = Area of parallelogram ABC¹D¹.
Hence, the parallelogram on the same base andabetween the same parallels are equal in area.

Let ABCD be a parallelogram such that its diagonals AC and BD are equal.
Taking A as the origin, let the p.v. of B and D be \(\vec { b } \)and \(\vec { d } \) respectively.
Then \(\vec { AB } =\vec { b } \) and \(\vec { AD } =\vec { d } \)
Using triangle law of addition of vectors is ΔABC, we get
\(\vec { AB } +\vec { BC } =\vec { AC } \)
⇒ \(\vec { AB } +\vec { AD } =\vec { AC } \)
⇒ \(\vec { b } +\vec { d } =\vec { AC } \)
Using triangle law of addition of vectors in ΔABD, we get \(\vec { AB } +\vec { BD } =\vec { AD } \)
⇒ \(\vec { b } +\vec { BD } =\vec { d } \)
⇒ \(\vec { BD } =\vec { d } -\vec { b } \)
In parallelogram ABCD we have AC = BD
⇒ \(|\vec { AC } |=|\vec { BD } |\)
⇒ \(|\vec { AC } |^{ 2 }=|\vec { BD } |^{ 2 }\)
⇒ \(|\vec { b } +\vec { d } |^{ 2 }=|\vec { d } -\vec { b } |^{ 2 }\)

⇒ \(4(\vec { b } .\vec { d } )=0\Rightarrow \vec { b } .\vec { d } =0\)=0
⇒ \(\vec { b } \bot \vec { d } \)
⇒ \(\vec { AB } \bot \vec { AD } \)
Hence, ABCD is a rectangle.

Let OACB be a rhombus. Taking O as the origin, let the position vectors of A and B be \(\vec { a } \) and \(\vec { b } \) respectively.
Then \(\vec { OA } =\vec { a } \) and \(\vec { OB } =\vec { b } \)  [∵ \(\vec { AC } =\vec { OB } \)]
So, the p.v. of C is \(\vec { a } +\vec { b } \)
∴ Position vector O f the miid-point of OC is \(\frac { \vec { a } +\vec { b } }{ 2 } \)
Similarly, the position vector of mid-point of AB is \(\frac { \vec { a } +\vec { b } }{ 2 } \).
Hence, the mid-point of OC coincides with the mid-point of AB.
Now, \(\vec { OC } .\vec { AB } =(\vec { a } +\vec { b } ).(\vec { b } -\vec { a } )=|\vec { b } |^{ 2 }-|\vec { a } |^{ 2 }\)
= OB²- OA² = 0   [∵ OB = OA]
⇒ \(\vec { OC } \bot \vec { AB } \).
Hence, the diagonals of a rhombus bisect each other at right angles.

Let O be the centre of the semi-circle and AA¹ be the diameter.
Let P be any point on the circumference of the semi circle.
Taking O as the origin, let the position vectors of A and P be a and \(\vec { r } \) respectively.
Let us prove that \(\angle A P B=90^{\circ}\)
W.K.T OA = OB = OP ( because of radius)
\( \overrightarrow{P A} =\overrightarrow{P O}+\overrightarrow{O A} \)
\(\overrightarrow{P B} =\overrightarrow{P O}+\overrightarrow{O B} \)
\( =\overrightarrow{P O}-\overrightarrow{O A} \)
\(\overrightarrow{P A} \cdot \overrightarrow{P B} =(\overrightarrow{P O}+\overrightarrow{O A})(\overrightarrow{P O}-\overrightarrow{O A}) \)
\( =\overrightarrow{P O}^{2}-\overrightarrow{O A}^{2}=0 \)
\( \overrightarrow{P A} \perp \overrightarrow{P B} \)
\( \Rightarrow \ \angle A P B=90^{\circ} \). Hence proved.

Let ABC be an isosceles triangle with AB = AC and let AD is the median 
D is mid-point of BC.
\( \overrightarrow{A D}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}) \)
\( \overrightarrow{B C}=\overrightarrow{B A}+\overrightarrow{A C} \)
\( \overline{D A} \cdot \overrightarrow{D B}=-\overrightarrow{A D} \cdot\left(\frac{-1}{2} \overrightarrow{C B}\right) \)
\(=-\overrightarrow{A D} \cdot\left(\frac{1}{2} \overrightarrow{B C}\right) \)
\(=\frac{1}{2} \overrightarrow{A D} \cdot \overrightarrow{B C} \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{B A}+\overrightarrow{A C}) \)
\(=\frac{1}{4}(\overrightarrow{A B}+\overrightarrow{A C}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \)
\(=\frac{1}{4}[(\overrightarrow{A C} \cdot \overrightarrow{A C})-(\overrightarrow{A B} \cdot \overrightarrow{A B})] \)
\(=\frac{1}{4}\left(A C^{2}-A B^{2}\right) \)
\(=\frac{1}{4}(0)=0 \)
\(\overrightarrow{D A} \cdot \overrightarrow{D B}=0\)
\(\overrightarrow{D A} \perp \overrightarrow{D B}\)

Let the position vectors of the parts A and B on the circle lie \(\vec { a } \) and \(\vec { b } \) respectively.
Since O is the centre of the circle
\(|\vec { OA } |=|\vec { OB } |\Rightarrow |\vec { a } |=|\vec { b } |\) ....(1)
Also D is the mid-point of AB,
⇒ \(\vec { OD } =\frac { \vec { a } +\vec { b } }{ 2 } \) (mid-point formula)
\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { OB } -\vec { OA } )\)
=\(\left( \frac { \vec { a } +\vec { b } }{ 2 } \right) .(\vec { Ob } -\vec { Oa } )\)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { a}| ^{ 2 } \right] \)
=\(\left[ \because (\vec { a } +\vec { b } ).(\vec { b } -\vec { a } )=|\vec { b } |^{ 2 }-|\vec { a } |^{ 2 } \right] \)
= \(\frac { 1 }{ 2 } \left[ |\vec { b } |^{ 2 }-|\vec { b| } ^{ 2 } \right] \) (using (1))
= \(\frac{1}{2}\)(0) = 0
⇒ \(\vec { OD } .\vec { AB } =0\Rightarrow \vec { OD } \bot \vec { AB } \)
Hence, if a line is drawn from the centre of a to the mid-point of a chord, then that line is perpendicular to the chord.

Resultant of the given forces is \(\vec { F } \) = \((\hat { 3i } -\hat { 2j } +\hat { 2k } )\) + \((\hat { 2i } +\hat { j } -\hat { k } )\) = \(​​\hat { 5i } -\hat { j } +\hat { k } \)
The displacement of the particle is given by
\(\vec { d } \) =  \((\hat { 4i } -\hat { j } +\hat { \lambda k } )-(\hat { i } +3\hat { j } -\hat { k } )\) =  \((3\hat { i } -\hat { 4j } +(\lambda +1)\hat { k } )\)
As the work done by the forces is 16 units, we have
\(\vec { F } \).\(\vec { d } \) = 16
That is \((\hat { 5i } -\hat { j } +\hat { k } ).(3\hat { i } -\hat { 4j } +(\lambda +1))\hat { k } \)  = 16 ⇒ λ + 20 = 16
So, λ = - 4

With usual notations in triangle ABC, let \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \)  and  \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Applying dot product, we get
\(\vec { BC } .\vec { BC } =-\vec { BC } .\vec { CA }-\vec { BC }. \vec { AB } \)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }=-\left| \vec { BC } \right| \left| \vec { CA } \right| \) cos(兀-c)-\(\left| \vec { BC } \right| \left| \vec { AB} \right| \)cos(兀-B)
⇒ a² = ab cos C + ac cos B
Therefore a = b cos C + c cos B
The results (ii) and (iii) are proved in a similar way

With usual notations in triangle ABC, we have \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \) \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Then applying dot product, we get
\(\vec { BC } .\vec { BC } =(-\vec { CA } -\vec { AB } ).(-\vec { CA } -\vec { AB } )\)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }={ \left| \vec { CA } \right| }^{ 2 }+{ \left| \vec { AB } \right| }^{ 2 }+\vec { 2CA } .\vec { AB } \)
⇒ a² = b²+c²+2bc cos (\(\pi\) - A)
⇒ a² = b²+c²−2bc cos A.
The results (ii) and (iii) are proved in a similar way.

Here (x₁, y₁, z₁) = (3, -4, -3) and direction ratios of the given straight line are (a, b, c) = (-4, -7, 12). 
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1, 1).
We observe that, the given point (x₁, y₁, z₁) = (3, 4, -3) satisfies the given plane 5x-y+z = 8
Next, aA+bB+cC = (-4)(5)+(-7)(-1)+(12)(1) = -1 \(\neq \) 0.
So, the normal to the plane is not perpendicular to the line.
Hence, the given line does not lie in the plane.

If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) is the position vector of an arbitrary point (x, y, z) on the plane, then the given equation can be written as \((x\hat { i } +y\hat { j } +z\hat { k } ).(3\hat { i } -4\hat { j } +3\hat { k } )=-8\) or \((x\hat { i } +y\hat { j } +z\hat { k } ).(-3\hat { i } +4\hat { j } -3\hat { k } )=8\). 
That is, \(\hat { r } .(-3\hat { i } +4\hat { j } -3\hat { k } )=8\) which is the vector equation of the given plane in standard form.

Let \(\vec { a } ={ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } \)
∴ LHS = \(\hat { i } \times (\vec { a } \times \hat { i } )+\hat { j } \times (\vec { a } \times \hat { j } )+\hat { k } \times \vec { a } \times \hat { k }\)
\((\hat { i }. \hat { i } )\vec { a } -(\hat { i } .\hat { a } )\hat { i } +(\hat { j } .\hat { j } )\vec { a } -(\hat { j } .\vec { a } )\hat { j } +(\hat { k } .\hat { k } )\vec { a } -(\hat { k } .\vec { a } )\hat { k } \)
\([\because \vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
\(1.\vec { a } -{ a }_{ 1 }\hat { i } +1.\vec { a } -{ a }_{ 2 }\hat { j } +1.\vec { a } -{ a }_{ 3 }\hat { k } ]\)
\([\because \hat { i } .\hat { i } =\hat { j } .\hat { j } =\hat { k } .\hat { k } =1\)and
\(\hat { i } \vec { a } =\hat { i } ({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )={ a }_{ 1 }\hat { j } .\vec { a } ={ a }_{ 2 }\quad \hat { k } .\vec { a } ={ a }_{ 3 }\)
\(3\vec { a } -({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } )\)
= \(3\vec { a } -\vec { a } =2\vec { a } \)
= RHS .
∴ LHS = RHS. Hence proved

  \(\vec { a } \)= \(a\hat { i } +a\hat { j } +c\hat { k } , \vec{b}=\hat { i } +\hat { k } \), \(\vec { c } \)= \(c\hat { i } +c\hat { j } +b\hat { k } \)
Given \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 0
⇒ \(\left| \begin{matrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{matrix} \right| \) = 0
⇒ \(a\left| \begin{matrix} 0 & 1 \\ c & b \end{matrix} \right| -a\left| \begin{matrix} 1 & 1 \\ c & b \end{matrix} \right| +c\left| \begin{matrix} 1 & 0 \\ c & c \end{matrix} \right| \) = 0
⇒ a(0-c)-a(b-c)+c(c-0) = 0
\(\Rightarrow-\not a c-a b+\not a c+c^{2}=0\)
⇒ c² = ab ⇒ c =\(\sqrt { ab } \).
Hence c is the geometric mean of a and b.

Given \(\vec { a } =\hat { i } -\hat { k } ,\vec { b } =x\hat { i } +\hat { j } +(1-x)\hat { k } ,\vec { c } =y\hat { i } +x\hat { j } +(1+x+y)\hat { k } \)
\([\vec { a } ,\vec { b } \vec { c } ]=\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & 1-x \\ x & 1+x-y \end{matrix} \right| +0-y\left| \begin{matrix} x & y \\ y & x \end{matrix} \right| \)
= [(1+x-y)-x(1-x)]-[x²-y]
\(=1+\not x-\not y-\not x+\not x^{2}-\not x^{x}+\not y\)
= 1
∴ \([\vec { a } \vec { b } \vec { c } ]\) = 1 for all values of x and y
∴ \([\vec { a } \vec { b } \vec { c } ]\) depends on neither x nor y.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

Taking \(\vec { p } =(\vec { a } \times \vec { b } )\) as a single vector and using the vector triple product expansion, we get 
\((\vec { a } \times \vec { b } )\times(\vec{c}\times\vec{d})=\vec{p}\times(\vec{c}\times\vec{d})\)
= \((\vec { p } .\vec { d } )\vec { c } -(\vec { p } .\vec { c } )\vec { d } \)
= \(((\vec { a } \times \vec { b } ).\vec { d } )\vec { c } -((\vec { a } \times \vec { b } ).\vec { c } )\vec { d } =[\vec { a } ,\vec { b } ,\vec { d } ]\vec { c } -[\vec { a } \vec { b } \vec { c } ]\vec { d } \)
Similarly, taking \(\vec { q } \) = \(\vec { c } \times \vec { d } \)
\((\vec { a } \times \vec { b } )\times (\vec { c } \times \vec { d } )=(\vec { a } \times \vec { b } )\times \vec { q } \)
= \((\vec { a } .\vec { q } )\vec { b } -(\vec { b } .\vec { q } )\vec { a } \)
\(= [\vec { a } ,\vec { c }, \vec d ]\vec { b } -[\vec { b } ,\vec { c },\vec d ]\vec { a } \)