Vector Calculus MCQ Quiz in தமிழ் - Objective Question with Answer for Vector Calculus - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Dot product:

The dot product of two vectors is given by:

$\overrightarrow{A}\overrightarrow{B}=|A||B|cos\theta$

where, cos θ = Angle between vectors A and B

|A| = Magnitude of vector A

|B| = Magnitude of vector B

$\overrightarrow{B}=|A|cos\theta$

Hence, the dot product is also the projection of one vector on another vector.

Concept:

$\mathrm{d}\mathrm{i}\mathrm{v}\overline{\mathrm{r}}=\nabla .\overrightarrow{\mathrm{r}}$

Where ∇ = $\frac{\partial }{\partial \mathrm{x}}\hat{\mathrm{i}}+\frac{\partial }{\partial \mathrm{y}}\hat{\mathrm{j}}+\frac{\partial }{\partial \mathrm{z}}\hat{\mathrm{k}}$

Calculation:

Given:  $\overrightarrow{\mathrm{v}}$= 3xzî + 2xyĵ - yz2 k̂

$\text{div}\overrightarrow{\mathrm{v}}=\nabla .\overrightarrow{\mathrm{v}}$

$=(\frac{\partial }{\partial \mathrm{x}}\hat{\mathrm{i}}+\frac{\partial }{\partial \mathrm{y}}\hat{\mathrm{j}}+\frac{\partial }{\partial \mathrm{z}}\hat{\mathrm{k}})\cdot (3\mathrm{x}\mathrm{z}\hat{\mathrm{i}}+2\mathrm{x}\mathrm{y}\hat{\mathrm{j}}-\mathrm{y}{\mathrm{z}}^2\hat{\mathrm{k}})$

$$\begin{gathered} =\frac{\partial (3\mathrm{x}\mathrm{z})}{\partial \mathrm{x}}+\frac{\partial (2\mathrm{x}\mathrm{y})}{\partial \mathrm{y}}+\frac{\partial (-\mathrm{y}{\mathrm{z}}^2)}{\partial \mathrm{z}} \\ =3\mathrm{z}\frac{\partial (\mathrm{x})}{\partial \mathrm{x}}+2\mathrm{y}\frac{\partial (\mathrm{x})}{\partial \mathrm{y}}-\mathrm{y}\frac{\partial ({\mathrm{z}}^2)}{\partial \mathrm{z}} \end{gathered}$$

= 3z + 2x - 2yz

Concept:

Laplacian operator in the Cartesian system is:

${\mathrm{\nabla }}^2=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$

Laplacian operator in Cylindrical system is:

${\mathrm{\nabla }}^2=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{{\partial }^2}{\partial z^2}$

Laplacian operator in Spherical system is:

${\mathrm{\nabla }}^2=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\varphi }\frac{{\partial }^2}{\partial {\varphi }^2}$

Calculation:

The Laplacian of a scalar field V(x,y,z) = x² + y² + z²

${\mathrm{\nabla }}^2\mathrm{V}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}$

${\mathrm{\nabla }}^2\mathrm{V}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\frac{\partial }{\partial x}\left(2x\right)+\frac{\partial }{\partial y}\left(2y\right)+\frac{\partial }{\partial z}\left(2z\right)$

= 2 + 2 + 2 = 6

CONCEPT: 

The Resultant of two vectors is given by:

R2 = (A2 + B2 + 2AB cosθ) 

where R is the resultant vector, A and B are the two vectors, and θ is the angle between two vectors.​

Maximum value of R = A + B when θ = 0° 

Minimum value of R = A - B when θ = 180° 

CALCULATION:

Given that Two vectors are 15 units and 10 units

The maximum value of the resultant R = A + B = 15 + 10 = 25 units

The minimum value of the resultant R = A - B = 15 - 10 = 5 units

• The resultant vector can be between 5 to 25 units.
• So the result could not be 3 units.
• Hence the correct answer is option 1.

Concept:

a = 3i + j + k; b = 2i – 2j + k

$\begin{array}{l}\cos \theta =\frac{a.b}{\left|a\right|\left|b\right|}\\ =\frac{6-2+1}{\sqrt{9+1+1}.\sqrt{4+4+1}}\end{array}$

$=\frac{5}{3\sqrt{11}}=\frac{5}{\sqrt{99}}$

∴ sin² θ +cos² θ = 1

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{25}{99}}$

Explanation:

Laplace’s equation states that the sum of the second-order partial derivatives of U (function) with respect to the coordinates, equals zero.

∇2 is called the laplacian or Laplace operator.

In Cartesian coordinates:

${\nabla }^{2}U=\frac{{\partial }^{2}U}{\partial x^2}+\frac{{\partial }^{2}U}{\partial y^2}+\frac{{\partial }^{2}U}{\partial z^2}=0$

In Cylindrical coordinates:

${\nabla }^{2}U=\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial U}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^{2}U}{\partial {\mathrm{\Theta }}^2}+\frac{{\partial }^{2}U}{\partial z^2}=0$

Concept-

Length of vector $a\hat{i}+b\hat{j}+c\hat{k}=\sqrt{(a{)}^2+(b{)}^2+(c{)}^2}$

Let Initial point A(x₁, y₁, z₁) and terminal point B(x₂, y2, z2) of a vector than 

$|\overrightarrow{AB}|=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{i}+(z_2-z_1)\hat{i}$

Calculation-

Initial point is P(2,-3,4) and terminal point Q(-2, 1, 1)

Vector $|\overrightarrow{PQ}|$ = (-2 - 2)$\hat{i}$ + (1 - (-3))$\hat{j}$+ (1 - 4)$\hat{k}$

Vector $|\overrightarrow{PQ}|$ = -4$\hat{i}$ + 4$\hat{j}$ - 3$\hat{k}$

Now the length of a vector  $|\overrightarrow{PQ}|$ is given by 

$|\overrightarrow{PQ}|=\sqrt{(-4{)}^2+(4{)}^2+(-3{)}^2}$

$|\overrightarrow{PQ}|=\sqrt{16+16+9}$

$|\overrightarrow{PQ}|=\sqrt{41}$

∴ length of vector $|\overrightarrow{PQ}|is\sqrt{41}$

The coordinates (5, 30°, 2) are in the form (r, ϕ, z). 

Hence, it is a cylindrical coordinates system.

Representation of coordinate system:

1.) Cartesian coordinate system:

• It is represented by (x, y, z).
• Its range is (-∞ < x < ∞), (-∞ < y < ∞) and (-∞ < z < ∞). 

​2.) Cylindrical coordinate system:

• It is represented by (r, ϕ, z).
• Its range is (0 < r < ∞), (0 < y < 2π) and (-∞ < z < ∞). 

Conversion of cylindrical to the cartesian system:

• x = r cosϕ
• y = r sinϕ 
• z = z 

​​3.) Spherical coordinate system:

• It is represented by (ρ, θ, ϕ).
• Its range is (0 < ρ < ∞), (0 < ϕ < π) and (0 < θ < 2π). 

Conversion of spherical to the cartesian system:

• x = ρ sin ϕ cos θ 
• y = ρ sin ϕ sin θ 
• z = ρ cos ϕ

Calculation:

At Point P: x = -2, y = 6, z = 3. Hence,

$\mathrm{r}\mathrm{h}\mathrm{o}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=\sqrt{4+36}=6.32$

$\varphi =\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\frac{\mathrm{y}}{\mathrm{x}}={\tan }^{-1}\frac{6}{-2}={108.43}^{\circ }$

z = 3

$\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}=\sqrt{4+36+9}=7$

$\theta ={\tan }^{-1}\frac{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}{\mathrm{z}}={\tan }^{-1}\frac{\sqrt{40}}{3}={64.62}^{\circ }$

Thus,

P(-2, 6, 3) = P(6.32, 108.43°, 3) = P(7, 64.62°, 108.43°)

In the Cartesian system, A at P is

A = 6ax + ay

For vector A, Ax = y, Ay = x + z, Az = 0. Hence, in the cylindrical system.

$\left[\begin{array}{c}{A}_{\rho }\\ A_{\varphi }\\ A_z\end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0\\ -\sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}y\\ x+z\\ 0\end{array}\right]$

or

Aρ = y cos ϕ + (x + z) sin ϕ

Aϕ = -y sin ϕ + (x + z) cos ϕ

Az = 0

But x = ρ cos ϕ, y = sin ϕ, and substituting these yields

A = (Aρ, Aϕ, Az) = [ρ cos ϕ sin ϕ + (ρ cos ϕ + z) sin ϕ]aρ + [-ρ sin2 ϕ + (ρ cos ϕ + z)cos ϕ]aϕ

At P

ρ = $\sqrt{40},\tan \varphi =\frac{6}{-2}$

Hence, 

$\mathrm{c}\mathrm{o}\mathrm{s}\varphi =\frac{-2}{\sqrt{40}},\sin \varphi =\frac{6}{\sqrt{40}}$

$\mathrm{A}=[\sqrt{40}.\frac{-2}{\sqrt{40}}.\frac{6}{\sqrt{40}}+(\sqrt{40}.\frac{-2}{\sqrt{40}}+3).\frac{6}{\sqrt{40}}]{\mathrm{a}}_{\rho }+[-\sqrt{40}.\frac{36}{40}+(\sqrt{40}.\frac{-2}{\sqrt{40}}+3).\frac{-2}{\sqrt{40}}]{\mathrm{a}}_{\varphi }$

$\frac{-6}{\sqrt{40}}{\mathrm{a}}_{\rho }-\frac{38}{\sqrt{40}}{\mathrm{a}}_{\varphi }=-0.9487{\mathrm{a}}_{\rho }-6.008{\mathrm{a}}_{\varphi }$

Gradient:

• The gradient of a scalar field is a vector field and whose magnitude is the rate of change and which points in the direction of the greatest rate of increase of the scalar field.
• Therefore, The gradient is a vector, that is defined for every point in a scalar field, whose magnitude gives the rate of change magnitude, and whose direction points in the direction of the greatest rate of increase of the scalar field
• If the vector is resolved, its components represent the rate of change of the scalar field with respect to each directional component.

For a two-dimensional scalar field ∅ (x,y), the gradient at any point is given by:

$grad\varphi \left(x,y\right)=\mathrm{\nabla }\varphi \left(x,y\right)=\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y}\right)\varphi =\left(\frac{\partial \varphi }{\partial x},\frac{\partial \varphi }{\partial y}\right)$

And for a three-dimensional scalar field ∅ (x, y, z), the gradient is defined as:

$grad\varphi \left(x,y,z\right)=\mathrm{\nabla }\varphi \left(x,y,z\right)=\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\right)\varphi =\left(\frac{\partial \varphi }{\partial x},\frac{\partial \varphi }{\partial y},\frac{\partial \varphi }{\partial z}\right)$

Key Points

• The gradient of a scalar field is the derivative of f in each direction. Note that the gradient of a scalar field is a vector field. An alternative notation is to use the del or nabla operator, ∇f = grad f.
• For a three dimensional scalar, its gradient is given by $grad\left(V\right)=\mathrm{\nabla }V$ 
• Gradient is a vector that represents both the magnitude and the direction of the maximum space rate of increase of a scalar.