APPLICATION OF VECTOR DIFFERENTIATION

\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]

is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of  (abbreviated as 'Div ') is defined as, 

\[Div\ \overrightarrow{F} = \nabla\ . \overrightarrow{F}\]
\[ = (\overrightarrow{i}\frac{\partial}{\partial\ x} + \overrightarrow{j}\frac{\partial}{\partial\ y} + \overrightarrow{k}\frac{\partial}{\partial\ z})\ . \ ({F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k})\]
\[ = (\frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z})\]

Basic properties of Divergence:

If A, B are vector functions and ‘f’ is a scalar function, then

\[1) \nabla\ . ( A + B) = \nabla\ .A + \nabla\ .B\]
\[2) \nabla\ . ( fA) = (\nabla\ f) . A + (f.\nabla\ A)\]
\[3)\overrightarrow{F}\ is\ solenoidal\ if\ \nabla\ . \overrightarrow{F}=0\]

Example:

\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\]

Soln:

\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (xy^2 - zy^3)\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(xyz)}+ \frac{\partial}{\partial\ y} {(3x^2y)}+ \frac{\partial}{\partial\ z} {(xy^2-xy^3)}\]

=   yz  + 3x2 ( 1 )  + ( 0 - y3 )

=  yz  + 3x2 - y3

Example:

\[If\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}, then\ find\ div\ \overrightarrow{F}\ at\ the\ point\ (1,-1,2)\]

Soln:

\[Let\ \overrightarrow{F}=x^2y\overrightarrow{i} + xy^2z\overrightarrow{j} + xyyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(x^2y)}+ \frac{\partial}{\partial\ y} {(xy^2z)}+ \frac{\partial}{\partial\ z} {(xyyz)}\]
\[\nabla\ . \overrightarrow{F} = 2xy + 2xyz + xy\]

At ( 1, -1, 2)

\[\nabla\ . \overrightarrow{F} = 2(1)(-1) + 2(1)(-1)(2) + (1)(-1)\]

=    -2  - 4  - 1

=   -7

Example:

\[Show\ that\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k} is\ solenoidal\]

Soln:

\[Let\ \overrightarrow{F}=3y^4z^2\overrightarrow{i} + 4x^3z^2\overrightarrow{j} + 6x^2y^3\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(3y^4z^2)}+ \frac{\partial}{\partial\ y} {(4x^3z^2)}+ \frac{\partial}{\partial\ z} {(6x^2y^3)}\]
\[\nabla\ . \overrightarrow{F} = 0+ 0 + 0 = 0\]
\[\overrightarrow{F}\ is\ solenoidal\]

Example:

\[If\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} - 3pyz\overrightarrow{k} is\ solenoidal\ at\ (1,1,1)\, find\ 'p'\]

Soln:

\[Let\ \overrightarrow{F}=2xy\overrightarrow{i} + 3x^2y\overrightarrow{j} - 3pyz\overrightarrow{k}\]
\[\nabla\ . \overrightarrow{F} = \frac{\partial {F_1}}{\partial\ x} + \frac{\partial {F_2}}{\partial\ y} + \frac{\partial {F_3}}{\partial\ z}\]
\[ = \frac{\partial }{\partial\ x} {(2xy)}+ \frac{\partial}{\partial\ y} {(3x^2y)}- \frac{\partial}{\partial\ z} {(3pyz)}\]
\[\nabla\ . \overrightarrow{F} = 2y + 3x^2- 3py\]

At ( 1, 1, 1)

\[\nabla\ . \overrightarrow{F} = 2(1) + 3(1)^2- 3p(1)\]
\[\nabla\ . \overrightarrow{F} = 2+ 3 - 3p\]
\[Given\ \overrightarrow{F}\ is\ solenoidal\]
\[i.e\ \nabla\ . \overrightarrow{F} =0\]

2  + 3 – 3p =   0

5 – 3p = 0

3p  = 5

p  =  5/3

Curl of a vector function:

\[If\ \overrightarrow{F}={F_1}\overrightarrow{i} + {F_2}\overrightarrow{j} + {F_3}\overrightarrow{k}\]

is a vector function,

\[then\ curl\ \overrightarrow{F}=\nabla\ × \overrightarrow{F}\]
\[curl\ \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {F_1} & {F_2} & {F_3}\\ \end{vmatrix}\]

Irrotational vector  : 

\[A\ vector\ \overrightarrow{F}\ is\ said\ to\ be\ irrotational\ if\]
\[curl\ \overrightarrow{F} =0\]
\[i.e\ \nabla\ × \overrightarrow{F} =0\]

Example:

\[If\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} +(x y ^2- zy^3)\overrightarrow{k}\ then\ find\ curl\ \overrightarrow{F}\]

Soln:

\[Let\ \overrightarrow{F}=xyz\overrightarrow{i} + 3x^2y\overrightarrow{j} + (x y ^2- zy^3)\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {xyz} & {3x^2y} & {x y ^2- zy^3}\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(x y ^2- zy^3)} - \frac{\partial}{\partial\ z} {(3x^2y)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(x y ^2- zy^3)} - \frac{\partial}{\partial\ z} {(xyz)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(3x^2y)} - \frac{\partial}{\partial\ y} {(xyz)})\]
\[ = \overrightarrow{i}( 2xy - 3zy^2 ) -\overrightarrow{j}((y^2 - 0) - xy)+\overrightarrow{k}(6xy - xz))\]
\[\nabla\ × \overrightarrow{F} = [2xy - 3zy^2] \overrightarrow{i} - [y^2 - xy] \overrightarrow{j} + [ 6xy - xz] \overrightarrow{k}\]

Example:

\[Show\ that\ \overrightarrow{F}= x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\ is\ irrotational\]

Soln:

\[Let\ \overrightarrow{F}=x\overrightarrow{i} +y^2\overrightarrow{j} +z^3\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {x} & y^2 & z^3\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {(z^3)} - \frac{\partial}{\partial\ z} {(y^2)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(z^3)} - \frac{\partial}{\partial\ z} {(x)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(y^2)} - \frac{\partial}{\partial\ y} {(x)})\]
\[ = \overrightarrow{i}[ 0- 0 ] -\overrightarrow{j}[0 - 0]+\overrightarrow{k}[0- 0]\]
\[\nabla\ × \overrightarrow{F} = 0\]
\[ \overrightarrow{F}\ is\ irrotational\]

Example:

\[If\ \overrightarrow{F}=( 2x + 2y + 2z)\overrightarrow{i} - (xy + yz + zx)\overrightarrow{j} + 3xyz\overrightarrow{k}\ then\ find\ \nabla\ × \overrightarrow{F}\ and\ \nabla\ × ( \nabla\ × \overrightarrow{F})\]

Soln:

\[Let\ \overrightarrow{F}=( 2x + 2y + 2z)\overrightarrow{i} - (xy + yz + zx)\overrightarrow{j} + 3xyz\overrightarrow{k}\]
\[\nabla\ × \overrightarrow{F} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {2x + 2y + 2z} & - (xy + yz + zx) & 3xyz \\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {( 3xyz )} + \frac{\partial}{\partial\ z} {(xy + yz + zx)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(3xyz)} - \frac{\partial}{\partial\ z} {(2x + 2y + 2z)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} -{(xy + yz + zx )} - \frac{\partial}{\partial\ y} {(2x + 2y + 2z )})\]
\[ = \overrightarrow{i}[ 3xz + (0 + y + x) ] -\overrightarrow{j}[(3yz – ( 0 + 0 + 2)]+\overrightarrow{k}[- (y + 0 + z ] – ( 0 + 2 + 0)]\]
\[\nabla\ × \overrightarrow{F} = [3xz + y + x] \overrightarrow{i} - [3yz - 2] \overrightarrow{j} + [ - y - z - 2] \overrightarrow{k}\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial\ x} & \frac{\partial }{\partial\ y} & \frac{\partial }{\partial\ z}\\ {3xz + y + x} & - (3yz - 2) & (- y - z - 2) \\ \end{vmatrix}\]
\[ = \overrightarrow{i}( \frac{\partial}{\partial\ y} {( - y- z - 2 )} - \frac{\partial}{\partial\ z} {(-3yz+ 2)})- \overrightarrow{j}( \frac{\partial}{\partial\ x} {(- y- z - 2 )} - \frac{\partial}{\partial\ z} {(3xz – y - x)}) + \overrightarrow{k}( \frac{\partial}{\partial\ x} {(-3yz+ 2)} - \frac{\partial}{\partial\ y} {(3xz+ y+ x)})\]
\[ = \overrightarrow{i}[ (-1 -0 - 0) - (-3y - 0)] -\overrightarrow{j}[(- 0 - 0 - 0) – ( 3x - 0 - 0)]+\overrightarrow{k}[(0 - 0 ) – ( 0 +1 + 0)]\]
\[ = \overrightarrow{i}[ - 1 + 3y] -\overrightarrow{j}[3x]+\overrightarrow{k}[-1]\]
\[\nabla\ × ( \nabla\ × \overrightarrow{F}) = \overrightarrow{i}[ 3y – 1 ] -\overrightarrow{j}[3x]-\overrightarrow{k}\]

VECTOR DIFFERENTIATION

Vector point function and Vector field:

Let P be any point in a region ‘D’ of space.  Let r be the position vector of P.  If there exists a vector function F corresponding to each P,  then such a function F is called a vector function and the region D is called a vector field.

Example:  consider the vector function

\[\overrightarrow{F}= (x-y)\overrightarrow{i}+ xy\overrightarrow{j}+ yz\overrightarrow{k}\ ————(1)\]

Let P be a point whose position vector is

\[\overrightarrow{r}= 2\overrightarrow{i}+ \overrightarrow{j}+ 3\overrightarrow{k}\ in\ the\ region\ D\ of space.\]

At P , the value of F is obtained by putting x = 2, Y = I, z = 3 in F

\[i.e\ at\ P,\ \overrightarrow{F}= \overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k}\]

Thus, to each point P of the region D, there corresponds a vector F given by the  vector function (I). Hence F is a vector point function (of scalar variables x, y, z) and the region D is a vector field.

Scalar point function and scalar field:

 If there exists a scalar ‘f’ given by a scalar function ‘f’ corresponding to each point P (with position vector r) in a region D of space, ‘f’ is called a scalar point function and D is called a scalar field.

Example: let P be a point whose position vector is

\[\overrightarrow{r}= 2\overrightarrow{i}+ \overrightarrow{j}+ 3\overrightarrow{k}\]

      Consider f= xyz + xy + z

                  Then the value of f at P is obtained by putting x = 2, y = I, z = 3

                    i.e., At P, f= 2.1.3 + 2.1 + 3 = II

                    Hence the scalar’ II ‘ is attached to the point P.

                     The function ‘f’ is a scalar point function (of scalar variables x, y, z), and D is a  

                   scalar field.

Note : There can be vector and scalar function of one or more scalar variables.

Vector differential operator

\[The\ vector\ differential\ operator\ ‘DEL’\ denoted\ as\ ‘\nabla’\ is\ defined\ by\]
\[\nabla = \frac{\partial}{\partial\ x}\overrightarrow{i} + \frac{\partial}{\partial\ y}\overrightarrow{j} + \frac{\partial}{\partial\ z}\overrightarrow{k}\]
\[\overrightarrow{i},\ \overrightarrow{j},\ \overrightarrow{k}\ are\ unit\ vectors\ in\ x,\ y,\ z\ directions\]
\[This\ operator\ \nabla\ is\ used\ in\ defining\ the\ gradient,\ divergence\ and\ curl.\]
\[Properties\ of\ \nabla\ are\ similar\ to\ those\ of\ vectors\]

The operator is applied to both vector and scalar functions.

Gradient

\[If\ \phi(x,y,z)\ is\ a\ scalar\ function\]

defined at each point (x, y, z) in a certain region of space and is differentiable

\[the\ gradient\ of\ \phi\ (shortly\ written\ as\ grad\Phi)\ is\ defined\ as\]
\[\nabla \phi= (\frac{\partial}{\partial\ x}\overrightarrow{i} + \frac{\partial}{\partial\ y}\overrightarrow{j} + \frac{\partial}{\partial\ z}\overrightarrow{k})\phi\]
\[= (\frac{\partial\phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\phi}{\partial\ z})\overrightarrow{k}\]

Basic properties of the Gradient

\[If\ \phi\ and\ \psi\ are\ two\ scalar\ functions\]
\[1)\ grad(\phi + \psi) = grad\ \phi + grad\ \psi \ or\ \nabla( \phi + \psi) = \nabla\phi + \nabla\psi\]
\[2)\ grad(\phi \psi) = \phi\ grad\phi +\psi\ grad\psi \ or\ \nabla( \phi + \psi) = \nabla\phi + \nabla\psi\]
\[3)\ \phi(x,y,z)\ = c\ (being\ constant)\ represents\ a\ surface\]
\[then\ unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (x,y,z)\ is\ \frac{\nabla\phi}{|\nabla\phi|}\]
\[4)\ angle\ between\ surfaces= angle\ between\ normals\]
\[\theta = \cos ^-1 ( \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}})\]

Example: If  f = x2yz,  find grad f at the point (1,-2,1)

Soln: f = x2yz

\[\nabla\ f=(\frac{\partial\ f}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ f}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ f}{\partial\ z})\overrightarrow{k}————(1)\]
\[\frac{\partial\ f}{\partial\ x} = yz\ \frac{\partial}{\partial\ x}({x^2})\]

= yz ( 2x )

\[\frac{\partial\ f}{\partial\ x} = 2xyz\]
\[\frac{\partial\ f}{\partial\ y} = x^2z\ \frac{\partial}{\partial\ y}(y)\]

= x2z ( 1 )

\[\frac{\partial\ f}{\partial\ y} = x^2z\]
\[\frac{\partial\ f}{\partial\ z} = x^2y\ \frac{\partial}{\partial\ z}(z)\]

= x2y ( 1 )

\[\frac{\partial\ f}{\partial\ z} = x^2y\]

Equation (1) becomes

\[\nabla\ f=(2xyz)\overrightarrow{i} + (x^2z)\overrightarrow{j} + (x^2y)\overrightarrow{k}\]

At the point (1,-2, 1),

\[\nabla\ f=2(1)(-2)(1)\overrightarrow{i} + (1)^2(1)\overrightarrow{j} + (1)^2(-2))\overrightarrow{k}\]
\[\nabla\ f=-4\overrightarrow{i} + \overrightarrow{j} – 2\overrightarrow{k}\]

Example: Find the unit normal to the surface xy +yz + zx= 3 at the point (1, 1, 1).

Soln:

\[\phi(x,y,z)\ = c\ (being\ constant)\ represents\ a\ surface\]
\[then\ unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (x,y,z)\ is\ \frac{\nabla\phi}{|\nabla\phi|}\]
\[Here\ \phi = xy +yz + zx\]
\[\nabla\ \phi=(\frac{\partial\ \phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ \phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ \phi}{\partial\ z})\overrightarrow{k}————(1)\]
\[\frac{\partial\ \phi}{\partial\ x} = y \frac{\partial}{\partial\ x}(x) + 0 + z \frac{\partial}{\partial\ x}(x)\]

= y ( 1 )  + z ( 1 )

\[\frac{\partial\ \phi}{\partial\ x} = y + z\]
\[\frac{\partial\ \phi}{\partial\ y} = x \frac{\partial}{\partial\ y}(y) + z \frac{\partial}{\partial\ y}(y) + 0\]

= x ( 1 )  + z ( 1 )

\[\frac{\partial\ \phi}{\partial\ y} = x + z\]
\[\frac{\partial\ \phi}{\partial\ z} = 0 + y \frac{\partial}{\partial\ z}(z) + x \frac{\partial}{\partial\ z}(z)\]

= y ( 1 )  + x ( 1 )

\[\frac{\partial\ \phi}{\partial\ z} = y + x\]

Equation (1) becomes

\[\nabla\ \phi=(y + z)\overrightarrow{i} + (x + z)\overrightarrow{j} + (y + x)\overrightarrow{k}\]

At the point (1,1, 1),

\[\nabla\ \phi=(1 + 1)\overrightarrow{i} + (1 + 1)\overrightarrow{j} + (1 + 1)\overrightarrow{k}\]
\[\nabla\ \phi= 2\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}\]
\[|\nabla\phi| = \sqrt{(2)^2 + (2)^2 + (2)^2 }\]
\[= \sqrt{(4 + 4 + 4) }\]
\[= 2\sqrt{3}\]
\[unit\ normal\ to\ the\ surface\ \phi\ at\ the\ point\ (1,1,1)\ =\ \frac{\nabla\phi}{|\nabla\phi|}\]
\[ = \frac{2\overrightarrow{i}+ 2\overrightarrow{j}+ 2\overrightarrow{k}}{2\sqrt{3}}\]

Example:  Find the acute angle between the surface xy2z = 4 and  x2 + y2 + z2 =6  at the point (2, 1, 1).

Soln:

Let f = xy2z = 4 be the surface  —— (1)

                      Normal vector to (1) at (2,1,1)

\[\nabla\ f=(\frac{\partial\ f}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ f}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ f}{\partial\ z})\overrightarrow{k}————(A)\]
\[\frac{\partial\ f}{\partial\ x} = y^2 z \frac{\partial}{\partial\ x}(x)\]

= y2z ( 1)

\[\frac{\partial\ f}{\partial\ x} = y^2z\]
\[\frac{\partial\ f}{\partial\ y} = x z \frac{\partial}{\partial\ y}(y^2)\]

= x z  (2y)

\[\frac{\partial\ f}{\partial\ y} = 2xyz\]
\[\frac{\partial\ f}{\partial\ z} = x y^2\ \frac{\partial}{\partial\ z}(z)\]

= xy2  (1)

\[\frac{\partial\ f}{\partial\ z} = xy^2\]
\[\nabla\ f = y^2z \overrightarrow{i} + 2xyz\overrightarrow{j} + x y^2\overrightarrow{k}\]

At the point (2,1, 1),

\[\nabla\ f = (1)^2 (1) \overrightarrow{i} + (2) (1) (1)\overrightarrow{j} + (2)(1)^2\overrightarrow{k}\]
\[\nabla\ f = \overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k} = a (say)\]

                   Let g = x2 + y2 + z2 =6  be the surface  —— (2)

                      Normal vector to (2) at (2,1,1)

\[\nabla\ g = (\frac{\partial\ g}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ g}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ g}{\partial\ z})\overrightarrow{k}————(B)\]
\[\frac{\partial\ g}{\partial\ x} = 2x\]
\[\frac{\partial\ g}{\partial\ y} = 2y\]
\[\frac{\partial\ g}{\partial\ z} = 2z\]
\[\nabla\ g = 2x \overrightarrow{i} + 2y\overrightarrow{j} + 2z\overrightarrow{k}\]

At the point (2,1, 1),

\[\nabla\ g = 2 (2) \overrightarrow{i} + (2) (1) \overrightarrow{j} + (2)(1)\overrightarrow{k}\]
\[\nabla\ g = 4\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k} = b (say)\]

Angle between the surfaces = Angle between the normal to them

                                                   = Angle between a and b

\[= \cos ^-1 ( \frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}})\]
\[= \cos ^-1 ( \frac{(\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}).(4\overrightarrow{i} + 2\overrightarrow{j} + 2\overrightarrow{k}}{\sqrt{(1)^2 + (4)^2 + (2)^2 }\sqrt{(4)^2 + (2)^2 + (2)^2 }})\]
\[= \cos ^-1 ( \frac{4 + 8 + 4}{\sqrt{1 + 16 + 4 }\sqrt{16 + 4 + 4}})\]
\[= \cos ^-1 ( \frac{16}{\sqrt{21}\sqrt{24}})\]
\[= \cos ^-1 ( \frac{16}{\sqrt{(3) (7)}\ 2\sqrt{(3) ( 2 )}})\]
\[= \cos ^-1 ( \frac{8}{3\sqrt{14}})\]

3.1 PRODUCT OF THREE AND FOUR VECTORS

Scalar triple Product

\[Let\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} be\ any\ three\ vectors,\]\[their\ scalar\ triple\ product\ is\ denoted\ by\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\]

Properties of Scalar triple Product:

\[1.\ Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ ,\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k} and\ \overrightarrow{c}= c_1\overrightarrow{i}\ + c_2\overrightarrow{j}+ c_3\overrightarrow{k}\]
\[Then\ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}\]
\[2.\ The\ three\ vectors\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ if\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[3.\ The\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}, \overrightarrow{OC}\ and\ \overrightarrow{OD}\ are\ coplanar\ if\ [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}] = 0\]

Example  :

\[Evaluate:\ [\overrightarrow{i} + \overrightarrow{j}\ \overrightarrow{j} + \overrightarrow{k}\ \overrightarrow{k} + \overrightarrow{i}]\]

Soln:

\[\overrightarrow{a}= \overrightarrow{i}\ + \overrightarrow{j}\]
\[\overrightarrow{b}= \overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{k} + \overrightarrow{i}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ \end{vmatrix}\]

=    1  ( 1 – 0 )  – 1 ( 0 – 1 )  + 0 (  0 – 1 )

=   1 ( 1 )  -1 (- 1 ) +  0

=   1 + 1

=   2

Example  :

\[Prove\ that\ the\ vectors\ 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k},\ 3\overrightarrow{i} + 4\overrightarrow{j}+ \overrightarrow{k}\ and\ \overrightarrow{i} – 2\overrightarrow{j}+ \overrightarrow{k}\ are\ coplanar. \]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= 3\overrightarrow{i}+ 4\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix} 2 & 1 & 1\\ 3 & 4 & 1\\ 1 & -2 & 1\\ \end{vmatrix}\]

=    2  ( 4 + 2 )  – 1 (3 – 1 )  + 1 ( -6 – 4 )

=   2 ( 6 )  – 1 (2 ) +  1 ( – 10 )

=   12 – 2  –  10   =    10  –  10

\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] = 0\]

The given points are coplanar.

Example  :

\[Find\ the\ value\ of\ m\ if\ the\ vectors\ 2\overrightarrow{i}-\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + 2\overrightarrow{j}- 3\overrightarrow{k}\ and\ 3\overrightarrow{i} + m\overrightarrow{j}+ 5\overrightarrow{k}\ are\ coplanar. \]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{c}= 3\overrightarrow{i}+ m\overrightarrow{j}+ 5\overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ \implies [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix} 2 &- 1 & 1\\ 1 & 2 & -3\\ 3 & m & 5\\ \end{vmatrix}=0\]

2  ( 10 + 3m )  + 1 ( 5 + 9 )  + 1 ( m – 6 ) = 0

20 + 6m + 14 + m – 6 =  0

7m + 28 = 0

7m   =   -28

m =  – 4

Example  :

\[Show\ that\ the\ points\ whose\ position\ vectors\ 4\overrightarrow{i}+ 5\overrightarrow{j}+ \overrightarrow{k},\ – \overrightarrow{j}- \overrightarrow{k},\ 3\overrightarrow{i} + 9\overrightarrow{j}+ 4\overrightarrow{k}\]\[and\ -4\overrightarrow{i} + 4\overrightarrow{j}+ 4\overrightarrow{k}\ lie\ on\ the\ same\ plane.\ (or)\ Coplanar.\]

Soln:

\[\overrightarrow{OA}= 4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{OB}= -\overrightarrow{j} – \overrightarrow{k}\]
\[\overrightarrow{OC}= 3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}\]
\[\overrightarrow{OD}= -4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=-\overrightarrow{j} – \overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})\]
\[=-\overrightarrow{j} – \overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j} – \overrightarrow{k}\]
\[\overrightarrow{AB}= -4\overrightarrow{i} -6\overrightarrow{j} -2\overrightarrow{k}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})\]
\[=3\overrightarrow{i}\ + 9\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{AC}= -\overrightarrow{i} +4\overrightarrow{j} +3\overrightarrow{k}\]
\[\overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA}\]
\[=-4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}- (4\overrightarrow{i}\ + 5\overrightarrow{j}+ \overrightarrow{k})\]
\[=-4\overrightarrow{i}\ + 4\overrightarrow{j}+ 4\overrightarrow{k}- 4\overrightarrow{i}\ – 5\overrightarrow{j}-\overrightarrow{k})\]
\[\overrightarrow{AD}= -8\overrightarrow{i} -\overrightarrow{j} +3\overrightarrow{k}\]
\[ [\overrightarrow{AB}\ \overrightarrow{AC} \overrightarrow{AD}] =\begin{vmatrix} -4 & -6 & -2\\ -1 & 4 & 3\\ -8 & -1 & 3\\ \end{vmatrix}\]

=    -4 ( 12 + 3)  +  6 ( -3 + 24 )  – 2 ( 1 + 32 )

=   -4 (15)  + 6 (21) – 2 ( 33 )

=   -60 + 126 – 66

=   -126 + 126

\[ [\overrightarrow{AB}\ \overrightarrow{AC} \overrightarrow{AD}] =0\]

The given points lie in the same plane.

Vector Triple Product

\[Let\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} be\ any\ three\ vectors,\] \[then\ the\ product \overrightarrow{a}×( \overrightarrow{b}×\overrightarrow{c})\ and\ ( \overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c} \]\[are\ called\ vector\ triple\ product\ of\ \overrightarrow{a},\overrightarrow{b},\ \overrightarrow{c}\]

Note:

\[\overrightarrow{a} × (\overrightarrow{b} × \overrightarrow{c}) = (\overrightarrow{a}. \overrightarrow{c}) \overrightarrow{b} – (\overrightarrow{a}.\overrightarrow{b}) \overrightarrow{c}\]

Product of Four vectors

\[Let\ \overrightarrow{a},\overrightarrow{b} ,\overrightarrow{c}, \overrightarrow{d} be\ any\ four\ vectors,\] \[then\ the\]\[ scalar\ product\ of\ these\ four\ vectors\ is\ defined\ as\ ( \overrightarrow{a} × \overrightarrow{ b} ) . ( \overrightarrow{c} × \overrightarrow{ d} )\]

Example  :

\[If\ \overrightarrow{a} = 2\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} -2\overrightarrow{j}+ 3\overrightarrow{k}\ and\ \overrightarrow{c} = 3\overrightarrow{j}-\overrightarrow{j} +5\overrightarrow{k}\] \[find\ (\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i}+ 3\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i} -2\overrightarrow{j}+ 3\overrightarrow{k}\]
\[\overrightarrow{c} = 3\overrightarrow{j}-\overrightarrow{j} +5\overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\ 2 & 3 & 1\\ 1 & -2 & 3\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( 9+ 2) -\overrightarrow{j}(6-1)+\overrightarrow{k}(-4-3)\]
\[ = \overrightarrow{i}(11) -\overrightarrow{j}(5)+\overrightarrow{k}(-7)\]
\[ \overrightarrow{a}× \overrightarrow{b}= 11\overrightarrow{i}-5\overrightarrow{j}-7\overrightarrow{k}\]
\[(\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c} = \begin{vmatrix} \overrightarrow{i} &\overrightarrow{j} & \overrightarrow{k}\\ 11 & -5 & -7\\ 3 & -1 & 5\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -25 – 7) -\overrightarrow{j}(55 +21)+\overrightarrow{k}(-11+15)\]
\[ = \overrightarrow{i}(-32) -\overrightarrow{j}(76)+\overrightarrow{k}(4)\]
\[ (\overrightarrow{a} × \overrightarrow{ b} ) × \overrightarrow{c}= 11\overrightarrow{i}-5\overrightarrow{j}-7\overrightarrow{k}\]

Example  :

\[If\ \overrightarrow{a} = 2\overrightarrow{i} – \overrightarrow{j}+ 2\overrightarrow{k},\ \overrightarrow{b} = \overrightarrow{i} +\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}\}\ and\ \overrightarrow{d} = \overrightarrow{i}-\overrightarrow{j} – \overrightarrow{k}\]\[find\ (\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} )\]

Soln:

\[\overrightarrow{a}= 2\overrightarrow{i} – \overrightarrow{j}+ 2\overrightarrow{k}\]
\[\overrightarrow{b} = \overrightarrow{i} +\overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{c} = \overrightarrow{i}+2\overrightarrow{j} +3\overrightarrow{k}\]
\[\overrightarrow{d} = \overrightarrow{i}-\overrightarrow{j} – \overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 2 & -1 & 2\\ 1 & 1 & 1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -1 – 2) -\overrightarrow{j}(2-2)+\overrightarrow{k}(2 + 1)\]
\[ = \overrightarrow{i}(-3) -\overrightarrow{j}(0)+\overrightarrow{k}(3)\]
\[ \overrightarrow{a}× \overrightarrow{b}= -3\overrightarrow{i}+3\overrightarrow{k}\]
\[\overrightarrow{c}×\overrightarrow{d} =\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ 1 & 2 & 3\\ 1 & -1 & -1\\ \end{vmatrix}\]
\[ = \overrightarrow{i}( -2 + 3) -\overrightarrow{j}(-1-3)+\overrightarrow{k}(-1 -2)\]
\[ = \overrightarrow{i}(1) -\overrightarrow{j}(-4)+\overrightarrow{k}(-3)\]
\[ \overrightarrow{c}× \overrightarrow{d}= \overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k}\]
\[(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = (-3\overrightarrow{i}+3\overrightarrow{k}) .(\overrightarrow{i}+4\overrightarrow{j}-3\overrightarrow{k})\]

  =   -3 ( 1 ) + 3 ( – 3 )

=    -3 -9

=   – 12

\[(\overrightarrow{a} × \overrightarrow{ b} ) . (\overrightarrow{c} × \overrightarrow{ d} ) = -12\]