Engineering Mathematics Test 4

Given that,

\(\vec a = \hat i + 2\hat j + 5\hat k\)

\(\vec b = \hat i + 2\hat j - \hat k\)

\(\vec a,\vec b,\vec c\) are three orthogonal vectors. The dot product of any two vectors should be equal to zero.

\(\Rightarrow \vec a.\vec b = \vec b.\vec c = \vec a.\vec c = 0\)

Let the required \(\vec c\) is \(x̂ i + ŷ j + ẑ k\)

\(\vec b.\vec c = 0\)

\(\Rightarrow \left( {\hat i + 2\hat j - \hat k} \right).\left( {x̂ i + ŷ j + ẑ k} \right) = 0\)

\(\Rightarrow x + 2y - z = 0\)       ----(1)

\(\vec a.\vec c = 0\)

⇒ (î  + 2ĵ + 5k̂) . (xî + yĵ + zk̂) = 0

⇒ x + 2y + z = 0       ----(2)

Form (1) and (2),

z = 0

⇒ x + 2y = 0

⇒ x = - 2y

\(c=x̂ i + ŷ j \)

From the options

Concept:

For a vector field, V defined as Vx î + Vy ĵ + Vz k̂, the divergence (∇.V) is given by:

\(∇.V= \left( {\frac{\partial }{{\partial x}}\hat i + \frac{\partial }{{\partial y}}\hat j + \frac{\partial }{{\partial z}} \hat k} \right).\left( {V_x \hat i + V_y \hat j + V_z \hat k} \right)\)

\(∇.V=\frac{\partial V_x }{\partial x}+\frac{\partial V_y }{\partial y}+\frac{\partial V_z }{\partial z}\)

Also, for a Scalar function A, the gradient is defined as:

\(\nabla A = \frac{{\partial {A_x}}}{{\partial x}}{a_x} + \frac{{\partial {A_y}}}{{\partial y}}{a_y} + \frac{{\partial {A_z}}}{{\partial z}}{a_z}\)

Calculation:

For the given vector field \(\vec F\):

\(div\;\vec F = \nabla \cdot \vec F = 2xy - 2 + {x^2}\)

\(grad\left( {div\;\vec F} \right) = \left( {2x + 2y} \right)\hat i + \left( {2x} \right)\hat j\)

At given point (1, 2, 3)

\(grad\left( {div\;\vec F} \right) = 6\hat i + 2\hat j\)

We have \(\nabla \phi = {\rm{I}}\frac{{\partial \phi }}{{\partial x}} + J\frac{{\partial \phi }}{{\partial y}} + k\frac{{\partial \phi }}{{\partial z}}\)

= (10xy + 2.5z²)I + (5x² – 10yz)J + (-5y² + 5zx)k

= 12.5 I – 5J at P (1, 1, 1)

Equation of line is given as 

\(\frac{{{\rm{x}} - 1}}{2} = \frac{{{\rm{y}} - 3}}{{ - 2}} = {\frac{z}{1}\rm{}}\)

The given line is parallel to vector 2i -2j + k (Take the denominator terms)

the unit vector in direction of the given line is,

\(\hat A = \frac{{2I - 2J + k}}{3}\)

Required directional derivative = ∇ϕ.Â

\(\begin{array}{l} = \left( {12.5{\rm{I}} - 5{\rm{J}}} \right) - \frac{{\left( {2{\rm{I}} - 2{\rm{J}} + {\rm{k}}} \right)}}{3}\\ = \frac{{\left( {25 + 10} \right)}}{3} = \frac{{35}}{3} \end{array}\)

Concept:

A vector F is said to be solenoidal when ∇. F = 0

A vector F is said to be irrotational when ∇ × F = 0

Calculation:

\(\vec V = \left( {x + y + az} \right)i + \left( {bx + 2y - z} \right)j + + \left( { - x + cy + 2z} \right)k\)

\(\nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {\left( {x + y + az} \right)}&{\left( {bx + 2y - z} \right)}&{\left( { - x + cy + 2z} \right)} \end{array}} \right| = 0\)

⇒ i (c + 1) – j (-1 – a) + k (b – 1) = 0

⇒ a = -1, b = 1, and c = -1

\(\vec V = \left( {x + y - z} \right)i + \left( {x + 2y - z} \right)j + \left( { - x - y + 2z} \right)k\)

Divergence of the given vector is

\(div\;\vec V = \nabla .\vec V = \frac{\partial }{{\partial x}}\left( {x + y - z} \right) + \frac{\partial }{{\partial y}}\left( {x + 2y - z} \right) + \frac{\partial }{{\partial z}}\left( { - x - y + 2z} \right)\)

\(\vec \nabla = \hat i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}\)

\(\vec u = y\hat i - x\hat j\)

\(\vec \omega = \vec \nabla \times \vec u = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ y&{ - x}&0 \end{array}} \right|\)

\(\vec \omega = o\hat i + o\hat j - 2\hat k = - 2\hat k\)

\(\left( {\vec \omega .\vec \nabla } \right) = \left( { - 2\hat k} \right).\left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}} \right) = - \frac{\partial }{{\partial z}}\left( { - 2} \right) = 0\)

\(\left( {\vec \omega .\vec \nabla } \right)\vec u = 0\)

Some of the important identities: (U: scalar, a: vector)

1. curl grad U = ∇ × ∇U = 0

2. div curl a = ∇.∇ × a = 0

3. ∇. (Ua) = U diva + (grad U).a

4. ∇ × (Ua) = U∇ × a + (∇U) × a

5. div (a × b) = curl a.b – a. curl b

If ϕ(x, y, z) is a potential function, then the vector field is given by

\(\vec F = grad\;\phi = \frac{{\partial \phi }}{{\partial x}}\hat i + \frac{{\partial \phi }}{{\partial y}}\hat j + \frac{{\partial \phi }}{{\partial z}}\hat k\) 

Given vector field is,

\(\vec F = \left( {4xy + {y^2}z} \right)\hat i + \left( {2{x^2} + 2xyz + {z^2}} \right)\hat j + \left( {x{y^2} + 2yz} \right)\hat k\) 

\(\frac{{\partial \phi }}{{\partial x}} = 4xy + {y^2}z,\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + {z^2},\frac{\partial }{{\partial z}} = x{y^2} + 2yz\) 

Consider \(\frac{{\partial \phi }}{{\partial x}} = 4xy + {y^2}z\) 

By integrating w.r.t ‘x’

\(\Rightarrow \phi = \frac{{4{x^2}y}}{2} + x{y^2}z + u\left( {y,z} \right)\) 

⇒ ϕ (x, y, z) = 2x²y + xy²z + u(y, z)

By differentiating w.r.t ‘y’

\(\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + \frac{{\partial u}}{{\partial y}}\) 

We know that,

\(\frac{{\partial \phi }}{{\partial y}} = 2{x^2} + 2xyz + {z^2}\) 

\(\Rightarrow \frac{{\partial u}}{{\partial y}}\left( {y,z} \right) = {z^2}\) 

By integrating w.r.t ‘y’

⇒ u = z²y + v(z)

Now, ϕ (x, y, z) = 2x²y + xy²z + z²y + v(x)

By differentiating w.r.t ‘z’

\(\frac{{\partial \phi }}{{\partial z}} = x{y^2} + 2yz + \frac{{\partial v}}{{\partial z}}\) 

We know that,

\(\frac{{\partial \phi }}{{\partial z}} = x{y^2} + 2yz\) 

\( \Rightarrow \frac{{\partial v}}{{\partial z}} = 0\) 

⇒ v = constant

Now, the potential function becomes

ϕ (x, y, z) = 2x²y + xy²z + z²y + c

We know that, \(\nabla f = f'\left( r \right)\nabla r = f'\left( r \right)\frac{{\vec r}}{r}\)

Given that, \(\nabla f = \frac{{\vec r}}{{{r^5}}}\)

\(\Rightarrow f'\left( r \right)\frac{{\vec r}}{r} = \frac{{\vec r}}{{{r^5}}}\)

\(\Rightarrow f'\left( r \right) = \frac{1}{{{r^4}}}\)

On integrating we get,

\(\Rightarrow f\left( r \right) = - \frac{1}{{3{r^3}}} + C\)

f(1) = 0

\(\Rightarrow 0 = - \frac{1}{3} + C \Rightarrow C = \frac{1}{3}\)

\(\Rightarrow f\left( r \right) = - \frac{1}{{3{r^3}}} + \frac{1}{3} = \frac{1}{3}\left( {1 - \frac{1}{{{r^3}}}} \right)\)

Concept:

If \(\vec a\) and \(\vec b\) are two vectors given in the component form as \(\vec a = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k,\;and\;\vec b = {a_2}\hat i + {b_2}\hat j + {c_2}\hat k\). Then, their cross or vector product is,

\(\vec a \times {\rm{\;}}\vec b = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right|\)

If \(\vec a\) and \(\vec b\) are two vectors given in the component form as \(\vec a = {a_1}\hat i + {b_1}\hat j + {c_1}\hat k,\;and\;\vec b = {a_2}\hat i + {b_2}\hat j + {c_2}\hat k\). Then, both are perpendicular if their dot product is zero.

Calculation:

\(\vec a = \lambda \hat i - 9\hat j - \hat k,\;\vec b = 3\hat i + 3\hat j + \hat k\;and\;\vec c = 4\hat i + 2\hat j + \hat k\)

\(\vec b \times {\rm{\;}}\vec c = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\3&3&1\\4&2&1\end{array}} \right|\)

\(= \hat i\left( {3 - 2} \right) - \hat j\left( {3 - 4} \right) + \hat k\left( {6 - 12} \right)\)

\(= \hat i + \hat j - 6\hat k\)

vector \(\vec a\) is perpendicular to \(\vec b \times {\rm{\;}}\vec c\)

\(\Rightarrow \vec a.\left( {\vec b \times {\rm{\;}}\vec c} \right) = 0\)

\(\Rightarrow \left( {\lambda \hat i - 9\hat j - \hat k} \right).\left( {\hat i + \hat j - 6\hat k} \right) = 0\)

⇒ λ – 9 + 6 = 0

Concept:

If ϕ(X, Y, Z) is a differential scalar function then the rate of change of ϕ at a point P in the direction of a given vector \(\bar a=\left( {x\hat i + y\hat j - z\hat k} \right)\) is called a D.D of ϕ. It is given by:

\({\rm{D}}.{\rm{D}} = {\rm{∇ }}ϕ .\frac{{\rm{\bar a}}}{{\left| {{\rm{\bar a}}} \right|}}\)

Where,

∇ϕ = derivative of function ϕ(X, Y, Z)  

\(\left| \bar a \right|\; = \sqrt {{x^2} + {y^2} + {z^2}} \)

Analysis:

r̅ = xi + yj + zk

r̅ = 3t²i + (t³ - 3t²)j + (4t - 6)k

Since the velocity is the time derivative of the distance, we can write:

\(\bar v = \frac{{d\bar r}}{{dt}} = 6ti + \left( {3{t^2} - 6t} \right)j + 4k\)

At t = 2, V̅ = 12i + 4k

Component of velocity in the direction of (2i + 3j - 5k) will be nothing but the directional derivative as shown:

\(= \frac{{\left( {12i + 4k} \right)\left( {2i + 3j - 5k} \right)}}{{\sqrt {4 + 9 + 25} }} = \frac{{24 - 20}}{{\sqrt {38} }} = 0.648\)