Learning Objectives:

• Identify types of matrices such as Square, Diagonal, Scalar, Row, Column, Identity.

Row Matrix:

A matrix which has only one row.

Column Matrix:

A matrix which has only one column.

Square Matrix:

A matrix having same no. of rows and columns.

Diagonal Matrix:

A square matrix where only diagonal elements are present, non-diagonal elements are zero.

Scalar Matrix:

Diagonal matrix where all diagonal elements are identical.

Identity Matrix:

Scalar Matrix where the diagonal elements are one.

Singular Matrix:

A Matrix whose determinant is equal to zero.

Solved Example: 4-1-01

For which value of x will the matrix given below become singular? $\begin{bmatrix} 8 & x & 0\\ 4 & 0 & 2\\ 12 & 6 & 0 \end{bmatrix}$

A. 4

B. 6

C. 8

D. 12

For singularity of the matrix, |A| = 0 \[\begin{vmatrix} 8 & x & 0 \\ 4 & 0 & 2\\ 12 & 6 & 0 \end{vmatrix} = 0\] \begin{align*} 8[0 - 2 \times 6] - x[0 - 24] + 0[24 - 0] &= 0\\ 8 \times(- 12) + 24x &= 0\\ -96 + 24x &= 0 \\ x &= \dfrac{96}{24} = 4 \end{align*}

Correct Answer: A

Solved Example: 4-1-02

If A is a square matrix then orthogonality property mandates:

A. $AA^T = A^{-1}$

B. $AA^T = 0$

C. $AA^T = I$

D. $AA^T = A^2$

If $AA^T = I$ or $A^{-1} = A^T$, the matrix is called Orthogonal.

Correct Answer: C

Solved Example: 4-1-03

The matrix \(\left[\begin{array}{ccc}0 & 5 & 8 \\ 5 & 0 & 12 \\ 8 & 12 & 0\end{array}\right]\) is a

A. Diagonal matrix

B. Symmetric matrix

C. Skew symmetric matrix

D. None of the above

Correct Answer: B

Learning Objectives:

• Find determinant of a 2X2 and a 3X3 matrix.

Only square matrices have determinant. (i.e. have the same number of rows as columns).

Determinant of a 2 X 2 Matrix:

For a 2 X 2 matrix, the determinant can be calculated as: \[\begin{vmatrix} 1 & 2\\ 3& 4 \end{vmatrix} = 1\times 4 - 2 \times 3 = -2\]

Determinant of a 3 X 3 Matrix:

For a 3 X 3 matrix, the determinant can be calculated as: \[\begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i \end{vmatrix}\]

• Multiply a by the determinant of the 2 \(\times\) 2 matrix that is not in a’s row or column.
• Likewise for b, and for c
• Add them up, making sure that b has a negative sign.

\[\begin{vmatrix} A \end{vmatrix} = a \begin{vmatrix} e & f\\ h & i \end{vmatrix} - b \begin{vmatrix} d & f\\ g & i \end{vmatrix} + c \begin{vmatrix} d & e\\ g & h \end{vmatrix}\]

Vassia Atanassova - Spiritia, Public domain, via Wikimedia Commons

Solved Example: 324234

The determinant of a matrix is zero. What can be inferred about the matrix?

A. It is invertible.

B. Its rows are linearly independent.

C. It is singular.

D. It has full rank.

Correct Answer: C

Solved Example: 4-2-01

If A and B are square matrices of size n x n, then which of the following statement is not true?

A. det (AB) = det (A) det (B)

B. det (kA) = k$^n$ det (A)

C. det (A + B) = det (A) + det (B)

D. det (A$^T$) =1/det (A$^{-1}$)

det (A + B) = det (A) + det (B) is not true, all other identities are true.

Correct Answer: C

Solved Example: 4-2-02

The determinant of the following matrix: \[\begin{vmatrix} 1 & 0 & 5\\ 2 & 2 & -3 \\ 1 & -1 & 4 \end{vmatrix}\] is:

A. -15

B. 15

C. 7

D. 23

Method I:The deteminant of a typical 3 X 3 matrix can be calculated by hand or by a scientific calculator.

Method II: Varies based on the model of calculator you are using. However, irrespective of your calculator model, the steps will be as follows:
Step I : Switch your calculator in Matrix or MAT mode.
Step II: Define the size of matrix in terms of rows and columns. In this case, it will be 3 rows and 3 columns.
Step III: Manually enter all 9 elements in respective positions.
Step IV: Calculate the determinant using DET(A) or similar command.

For exact procedure, note down the calculator model and on Youtube, search for determinant of a matrix using XYZ calculator.

Correct Answer: A

Solved Example: 4-2-03

If the matrix \[\left[ {\begin{array}{*{20}{c}} 1&3&{\lambda + 2}\\ 2&4&8\\ 3&5&{10} \end{array}} \right]\] is singular, then $\lambda$ equals:

A. -2

B. 2

C. 4

D. -4

If a Matrix is singular then its determinant is zero. \begin{align*} \left| A \right| &= 0\\ \left[ {\begin{array}{*{20}{c}} 1&3&{\lambda + 2}\\ 2&4&8\\ 3&5&{10} \end{array}} \right] &= 0\\ 1(40 - 40) -3(20 -24) + (\lambda + 2)(10-12) &= 0\\ -3(-4) + (\lambda + 2)(-2) &= 0\\ 12 - 2\lambda - 4 &=0\\ 8 &= 2 \lambda\\ \lambda &= 4 \end{align*}

Correct Answer: C

Solved Example: 456456

If a matrix has a row of zeros, what is the value of its determinant?

A. 0

B. 1

C. Undefined

D. It depends on other rows

Correct Answer: A

Learning Objectives:

• Add, subtract and multiply two matrices.
• Find transpose of a matrix.
• Find inverse of a matrix.

Scalar Multiplication of a Matrix:

If a matrix is multiplied by scalar, then ALL the elements get multiplied by that scalar.

Transpose of a Matrix:

If rows of a matrix are interchanged as column of the matrix, then the resulting matrix is the transpose of the original one.

Multiplication of Two Matrices:

A is m X n matrix (means having m rows and n columns), B is n X s matrix (means having n rows and s columns), then they can be multiplied and the result matrix C will be a m X s matrix.

In other words, columns of first matrix should match row of second matrix. Also, the common dimension gets eliminated and the product matrix does not have common dimension, neither as row nor as column. e.g. A 3 X 4 matrix ( 3 rows and 4 columns) can be multiplied with a 4 X 6 matrix, because the ’inner’ dimensions (4 and 4) are matching, which gets ’eliminated’ and the answer will be 3 X 6 matrix.

\[\begin{bmatrix} 3 & 4 & 2\\ 6&5 & 0\\ 1& 2 & -2 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 2\\ 2&-2 & 0\\ 3& 2 & -1 \end{bmatrix}\] \[ = \begin{bmatrix} 3+8+6 & 0-8+4 & 6+0-2\\ 6+10+0&0-10+0 & 12+0+0\\ 1+4-6& 0-4-4 & 2+0+2 \end{bmatrix} = \begin{bmatrix} 17 & -4 & 4\\ 16&-10 & 12\\ -1& -8 & 4 \end{bmatrix}\]

In matrix multiplication, the order in which the matrices are multiplied is important. (In other words, matrix multiplication is NOT commutative.) \[A \times B \neq B \times A\] You may take the same matrices mentioned above and verify that the answer is NOT the same.

Quartl, CC BY-SA 3.0, via Wikimedia Commons

Inverse of a Matrix:

For a square matrix A, the inverse is written \(A^{-1}\). When A is multiplied by \(A^{-1}\) the result is the identity matrix I. Non-square matrices do not have inverses.

Note: Not all square matrices have inverses. A square matrix which has an inverse is called invertible or non singular, and a square matrix without an inverse is called non-invertible or singular. A square matrix whose determinant is zero, does not have inverse.

For a 2 X 2 matrix, the inverse can be easily found out by the following formula: If \[A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\]

Then \(\displaystyle A^{-1} = \dfrac{1}{\mathrm{det}\ A}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}\) \(\displaystyle A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}\)
These matrix operations (transpose, determinant, multiplication and inverse) can be performed using a scientific calculator, so you must be comfortable with the MATRIX or MAT mode of your calculator.

Solved Example: 4-3-01

Multiplication of matrices E and F is G. matrices E and G are: \[E = \begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}, \quad G= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\] What is the matrix F ?

A. $\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$

B. $\begin{bmatrix} \cos \theta & \cos \theta\\ -\cos \theta & \sin \theta \end{bmatrix}$

C. $\begin{bmatrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta \end{bmatrix}$

D. $\begin{bmatrix} \sin \theta & -\cos \theta\\ \cos \theta & \sin \theta \end{bmatrix}$

\begin{align*} E \times F &= G\\ E^{-1} \times E \times F &= E^{-1} \times G\\ F &= E^{-1} \times G \end{align*} Since G is a unit matrix \[F = E^{-1}\] For a 2 X 2 matrix, the inverse can be easily found out by the following formula: If \[A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\] Then: \begin{align*} A^{-1} &= \dfrac{1}{\mathrm{det}\ A}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}\\ &= \dfrac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix} \end{align*} Here the determinant is 1 because, \[\cos^2 \theta + \sin^2 \theta = 1\] \[F = E^{-1} = \begin{bmatrix} \cos \theta & \sin \theta\\ -\sin \theta & \cos \theta\end{bmatrix}\]

Correct Answer: C

Solved Example: 4-3-02

Which of the following is the inverse of the matrix? \[\begin{bmatrix} -5 & 7\\ 3 & -4 \end{bmatrix}\]

A. $\begin{bmatrix} -4 & -7\\ -3 & -5 \end{bmatrix}$

B. $\begin{bmatrix} 4 & 7\\ 3 & 5 \end{bmatrix}$

C. $\begin{bmatrix} 5 & -7\\ -3 & 4 \end{bmatrix}$

D. The inverse does not exist.

Use the formula $\displaystyle A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$

Correct Answer: B

Solved Example: 4-3-03

Compute the product AB of the two matrices $\begin{bmatrix} 3 & 1\\ 2 & 4\\ 6 & 5\\ 1 & 2 \end{bmatrix} *\begin{bmatrix} -3 & 1 \\ 4 & 2 \end{bmatrix}$

A. $\begin{bmatrix} -12 &1 \\ 10& 10\\ 2 & 16\\ 5 &5 \end{bmatrix}$

B. $\begin{bmatrix} -5 &5 \\ 10& 10\\ 2 & 16\\ 5 &5 \end{bmatrix}$

C. $\begin{bmatrix} -5 &5 \\ 10& 10\\ 2 & 16\\ -5 &5 \end{bmatrix}$

D. $\begin{bmatrix} -1 &12 \\ 10& 10\\ 2 & 16\\ 5 &5 \end{bmatrix}$

\[\begin{bmatrix} 3 & 1\\ 2 & 4\\ 6 & 5\\ 1 & 2 \end{bmatrix} *\begin{bmatrix} -3 & 1 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -9+4 & 3+2\\ -6+16 & 2+8\\ -18+20 & 6+10\\ -3+8 & 1+4 \end{bmatrix} = \begin{bmatrix} -5 & 5\\ 10 & 10\\ 2 & 16\\ 5 & 5 \end{bmatrix}\]

Correct Answer: B

Solved Example: 4-3-04

If A = $\begin{bmatrix} 4 & -1\\ 5& -2 \end{bmatrix}$ and B = $\begin{bmatrix} 1 & 2\\ -2& 1 \end{bmatrix}, \mathrm{\ then\ } (AB)^T = ?$

A. $\begin{bmatrix} 6 & 7\\ 9& 8 \end{bmatrix}$

B. $\begin{bmatrix} 6 & 9\\ 7& 8 \end{bmatrix}$

C. $\begin{bmatrix} 7 & 6\\ 8& 9 \end{bmatrix}$

D. $\begin{bmatrix} 9 & 8\\ 6& 7 \end{bmatrix}$

\[AB = \begin{bmatrix} 4+2 & 8-1\\ 5+4 & 10-2 \end{bmatrix} = \begin{bmatrix} 6 & 7\\ 9 & 8 \end{bmatrix},\] To calculate the transpose of the matrix, just interchange rows into columns. \[(AB)^T = \begin{bmatrix} 6 & 9\\ 7 & 8 \end{bmatrix}\]

Correct Answer: B

Solved Example: 4-3-05

If \[A = \begin{bmatrix} 4 & -1\\ 5& -2 \end{bmatrix} \mathrm{\ and\ } (AB)^T = \begin{bmatrix} 6 & 9\\ 7& 8 \end{bmatrix}\] then B = ??

A. $\begin{bmatrix} 1 & 2\\ -2& 1 \end{bmatrix}$

B. $\begin{bmatrix} -1 & 2\\ -2& 1 \end{bmatrix}$

C. $\begin{bmatrix} -1 & 2\\ -2& 1 \end{bmatrix}$

D. $\begin{bmatrix} 1 & 2\\ 2& 1 \end{bmatrix}$

\[(AB)^T = \begin{bmatrix} 6 & 9\\ 7& 8 \end{bmatrix},\ (AB) = \begin{bmatrix} 6 & 7\\ 9& 8 \end{bmatrix}\]

Also, A * B = (AB)

Premultiplying with $A^{-1}$, we get

Here, $A^{-1} = \dfrac{1}{3} * \begin{bmatrix} 2 & -1\\ 5& -4 \end{bmatrix} $

Substituting, \[ B = \dfrac{1}{3} * \begin{bmatrix} 2 & -1\\ 5& -4 \end{bmatrix} \times \begin{bmatrix} 6 & 7\\ 9& 8 \end{bmatrix} = \begin{bmatrix} 1 & 2\\ -2& 1 \end{bmatrix}\]

Correct Answer: A

Solved Example: 4-3-06

What is the value of matrix element c(2,3), given C = A + B + A, where: \[A = \begin{bmatrix} 0& 2 & -1\\ 1& 3 & -1\\ 2 & 2 & 1 \end{bmatrix}, \quad B= \begin{bmatrix} 2& -1 & -1\\ 1& 0 & 5\\ -2 & 1 & 1 \end{bmatrix}\]

A. 9

B. 6

C. 3

D. 15

We do NOT have to evaluate complete addition C matrix. Only the requested element needs to be calculated.

c(2,3) means 2$^{\mathrm{nd}}$ row and 3$^{\mathrm{rd}}$ column of the matrix C.

c(2,3) = (-1) + 5 + (-1) = 3

Correct Answer: C

Solved Example: 4-3-07

The solution to the system of equations $\begin{bmatrix} 2 & 5\\ -4 & 3 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 2\\ -30 \end{bmatrix}$ is:

A. 6,2

B. -6,2

C. -6,-2

D. 6,-2

This is a system of equations \[2x + 5y = 2\] \[-4x +3y =-30\] The quick way is substitute the values of x and y to check whether the right hand side matches or not. Or the correct technical way is, multiply the first question by 2, then it becomes \[4x + 10y = 4\] Adding it with the second equation, \[13y = -26, \quad y = -2\] Substitute in any equation to get x = 6

Correct Answer: D

Solved Example: 4-3-08

If A and B be real symmetric matrices of size n x n, then:

A. AA$^T$ = 1

B. A = A$^{-1}$

C. AB = BA

D. (AB)$^T$ = BA

A symmetric matrix is a square matrix that is equal to its transpose.

Correct Answer: D

Solved Example: 4-3-08

If the order of the matrix is m $\times$ n, then how many elements will there be in the transpose of that matrix?

A. mn

B. m$^2$ n$^2$

C. mn$^2$

D. 2mn

Correct Answer: A

Solved Example: 4-3-09

Regarding matrix operations, select the correct alternative.

A. Matrix addition and matrix multiplication both are commutative.

B. Matrix addition is commutative but matrix multiplication is not commutative.

C. Matrix addition is not commutative but matrix multiplication is commutative.

D. Neither matrix addition nor matrix multiplication is commutative.

A + B = B + A, hence matrix addition is commutative.
But AB $\neq$ BA, hence matrix multiplication is not commutative.

Correct Answer: B

Learning Objectives:

• To be able to find length of a vector as well as unit normal vector.

Length (Magnitude) of a Vector:

Imagine a box with length, width and height equal to the i,j,k components of a vector. Then the length of the vector will be diagonal of this box, and it can be calculated using Pythagorean theorem.

If \( \vec{a} = a_{1}\bar{i} + a_{2}\bar{j} + a_{3}\bar{k}\) then length of the vector = \[\left | \vec{a} \right | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}\]

e.g. if \(\bar{a}\) = 2i + 3j + 4k then, \[\left | \bar{a} \right | = \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}\]

Calculating Unit Vector:

Unit vector has a magnitude (length) = 1 and is given by: \[ \hat{a} = \pm \dfrac{\bar{a}}{\left | \bar{a} \right |}\] e.g. if \(\bar{a}\) = 2i + 3j + 4k then, \[\left | \bar{a} \right | = \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}\] \[ \hat{a} = \pm \dfrac{2i + 3j + 4k}{\sqrt{29}}\] The \(\pm\) sign is because there can be two unit vectors opposite to each other.

Maxmath12, CC0, via Wikimedia Commons

Addition and Subtraction of Vectors:

Add/subtract corresponding i, j and k components with similar components of another vector.

e.g. if \(\bar{a}\) = 2i + 3j + 4k and
\(\bar{b}\) = 3i - 8j + k then,
\(\bar{a}\) + \(\bar{b}\) = (2+3)i + (3+(-8))j + (4+1)k = 5i -5j +5k

Similarly,

\(\bar{a}\) - \(\bar{b}\) = (2-3)i + (3-(-8))j + (4-1)k = -i +11j +3k

Sonja Weiland, CC BY-SA 4.0, via Wikimedia Commons

Solved Example: 23423

What is a unit vector?

A. A vector with a length of 1

B. A vector with a length of 0

C. A vector that points in the opposite direction

D. A vector that represents a point in space

Correct Answer: A

Solved Example: 23452345

If $\bar{b}$=(4,−3), which of the following is true?

A. It is a unit vector.

B. Its magnitude is 5.

C. It points in the positive direction.

D. Its direction cannot be determined.

Correct Answer: A

Solved Example: 4-4-01

Determine the magnitude of the force vector $\overline {F}$ = 20$\overline {i}$ + 60$\overline {j}$ - 90$\overline {k}$ (N).

A. 130 N

B. 120 N

C. 100 N

D. 110 N

\[\overline {F} =20\overline {i}+60\overline {j}-90\overline {k}\] \begin{align*} \left| \overline {F}\right|&=\sqrt {20^{2}+60^{2}+\left( -90\right) ^{2}}\\ &=\sqrt {400+3600+8100}\\ &=\sqrt {12100}\\ &= 110 \ N \end{align*}

Correct Answer: D

Solved Example: 4-4-02

For the spherical surface $x^2+ y^2+ z^2 = 1$, the unit outward normal vector at the point $(\dfrac {1}{\sqrt {2}}, \dfrac {1}{\sqrt {2}}, 0)$ is given by:

A. $\dfrac {1}{\sqrt {2}}i +\dfrac {1}{\sqrt {2}}j$

B. $\dfrac {1}{\sqrt {2}}i -\dfrac {1}{\sqrt {2}}j$

C. $\dfrac{1}{\sqrt {2}}i +\dfrac {1}{\sqrt {2}}j + \dfrac {1}{\sqrt {2}}k$

D. k

Given : $x^2+ y^2+ z^2 = 1$

Correct Answer: A

Solved Example: 4-4-03

Find the value of p if vector \[\overrightarrow{A}=p\widehat{i}+0.25\widehat{j}-0.1\widehat{k}\] is a unit vector.

A. $\sqrt{0.5}$

B. $\sqrt{0.75}$

C. $\sqrt{1.9275}$

D. $\sqrt{0.9275}$

Since the given vector is a unit vector, its length will be = 1.

Correct Answer: D

Solved Example: 4-4-04

Unit vector can be found out by:

A. Taking Square roots of the components of a vector

B. Squaring the components of a vector

C. Multiplying a vector by its magnitude

D. Dividing a vector by its magnitude

Correct Answer: D

Solved Example: 4-4-05

Unit vector along z-axis is:

A. $\bar{i}$

B. $\bar{j}$

C. $\bar{k}$

D. $\bar{i} + \bar{j} + \bar{k}$

Correct Answer: C

Solved Example: 4-4-06

Let $\bar{x}$ be a vector. Find |x|, if for an unit vector $\hat{a}$ its known that \[(\bar{x} - \hat{a}) \cdot (\bar{x} + \hat{a}) = 48\]

A. 7

B. 8

C. $\sqrt{48}$

D. $3\sqrt{3}$

\begin{align*} (\bar{x} - \hat{a}) \cdot (\bar{x} + \hat{a}) &= 48\\ (\bar{x} \cdot \bar{x}) - (\hat{a} \cdot \hat{a}) &= 48\\ |x|^2 - |a|^2 &= 48\\ |x|^2 - 1 &= 48\\ |x|^2 &= 49\\ |x| &= 7\\ \end{align*}

Correct Answer: A

Solved Example: 4-5-01

Given the vectors $\bar{A}$ = i - 2j + 4k and $\bar{B}$ = 3i + j - 2k, find $\bar{R}$ = $\bar{A}$ + $\bar{B}$.

A. 4i - j + 2k

B. 4i + j + 2k

C. 4i + j - 2k

D. -4i - j + 2k

The components of the given vectors are: \[A_x = 1, A_y = -2, A_z = 4, \] \[B_x = 3, B_y = 1, B_z = -2.\] \[R_x = A_x + B_x = 4, \quad R_y = A_y+ B_y = -1, \quad R_z = A_z + B_z = 2,\] \[\bar{R} = \bar{A} + \bar{B} = 4i - j + 2k\]

Correct Answer: A

Solved Example: 4-5-03

Addition of two vectors can be found out by:

A. Triangle Law

B. Hexagonal Law

C. Circular Law

D. Lami's theorem

Correct Answer: A

Solved Example: 4-5-04

If $|\bar{A}|$ = 3, $|\bar{B}|$ = 4 and $|\bar{A} + \bar{B}|$ = 5, then:

A. $\bar{A}$ and $\bar{B}$ are collinear acting in the same direction.

B. $\bar{A}$ and $\bar{B}$ are collinear acting in the opposite direction.

C. $\bar{A}$ and $\bar{B}$ are perpendicular to each other.

D. Cannot be predicted unless some additional information is given.

Correct Answer: C

Solved Example: 4-5-05

Which of the following vectors is a unit vector?

A. $\dfrac{\bar{i}}{\sqrt{2}}$

B. $\bar{j} + \bar{k}$

C. $\dfrac{\bar{i} + \bar{k}}{\sqrt{2}}$

D. $\dfrac{\bar{j}}{\sqrt{2}}$

Correct Answer: C

Solved Example: 4-5-06

If $\vec{A} = 2\hat{i}+5\hat{j}$ and $\vec{B} = 2\hat{i}-\hat{j}$ then the unit vector along the $|\vec{A}+\vec{B}|$ is:

A. $\dfrac{1}{\sqrt 2}(\hat {i}- \hat {j})$

B. $\dfrac{1}{\sqrt {29}}(\hat {2i}- \hat {5j})$

C. $\dfrac{1}{\sqrt {5}}(\hat {2i}- \hat {j})$

D. $\dfrac{1}{\sqrt 2}(\hat {i}+ \hat {j})$

\begin{align*} \vec{A} &= 2\hat{i}+5\hat{j}\\ \vec{B} &= 2\hat{i}-\hat{j} \end{align*} \[\vec{A} + \vec{B} = 4\hat{i}+4\hat{j}\] \[|\vec{A} + \vec{B}| = \sqrt{4^2 + 4^2} = 4\sqrt{2}\] \[\mathrm{Unit\ vector} = \dfrac{\mathrm{vector}}{\mathrm{magnitude}} = \dfrac{4\hat{i}+4\hat{j}}{4\sqrt{2}} = \dfrac{1}{\sqrt 2}(\hat {i}+ \hat {j})\]

Correct Answer: D

Learning Objectives:

• Evaluate dot and products.
• Understand physical interpretation of dot and cross products and their applications.
• Understand properties and identities related to dot and cross products.

Calculation of Dot Product:

Let us take the same example vectors again.
e.g. if \(\bar{a}\) = 2i + 3j + 4k and
if \(\bar{b}\) = 3i - 8j + k then, \[\overline{\mathbf{a}}. \overline{\mathbf{b}} = 2 \times 3 + 3 \times (-8) + 4 \times 1 = -14\] Dot product can be positive or negative but since it is a scalar it will not have i, j and k components.

Svjo, CC BY-SA 3.0, via Wikimedia Commons

Application of Dot Product:

Dot product is used to calculate scalar results such as:

• Work done by a force.
• Projection of a vector on another vector.
• Angle between two vectors.

where \(\left | \bar{a} \right |\) and \(\left | \bar{b} \right |\) represent the magnitudes of individual vectors and \(\theta\) is the angle between the two vectors.
The order of vectors while taking dot product is not important.

Now, consider the dot product of same two unit vectors,

This is because the angle between two same unit vectors is zero, hence the \(\cos \theta\) becomes one and also the magnitudes of unit vectors is one.
Now, consider the dot product of dissimilar unit vectors.

Again, here the magnitudes of unit vectors are one, but since the dissimilar unit vectors i,j and k are always perpendicular to each other, the \(\cos \theta\) becomes zero and hence, the dot products will become zero.

Calculation of Cross Product:

When two vectors are multiplied in such a way that their multiplication is also a vector, then it is referred as cross product.
Cross products are mostly useful when we are studying phenomenons involving rotational effects, such as moment of a force.
In order to evaluate cross product, create a 3X3 determinant with first row as i,j,k and then write down i,j,k component values of individual vectors.

$\vec{a} \times \vec{b} = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ a_{x} & a_{y} & a_{z} \\ b_{x}&b_{y} & b_{z} \end{vmatrix}$       Page 59

Let us take the same example vectors again,
e.g. if \(\bar{a}\) = 2\(\overline{i}\) + 3\(\overline{j}\) + 4\(\overline{k}\) and
\(\bar{b}\) = 3\(\overline{i}\) - 8\(\overline{j}\) + \(\overline{k}\) then,
\[\vec{a} \times \vec{b} = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ 2 & 3 & 4 \\ 3& -8 & 1 \end{vmatrix}\] Then, \(\vec{a} \times \vec{b}\) = 3-(-32)\(\overline{i}\) - (2-12)\(\overline{j}\) + (-16-9)\(\overline{k}\) = 35\(\overline{i}\) +10\(\overline{j}\) -25\(\overline{k}\)

MikeRun, CC BY-SA 4.0, via Wikimedia Commons

Properties of Cross Product:

\[\left|\vec{a} \times \vec{b}\right| = \left|\bar{a}\right| \left|\bar{b}\right|\sin \theta\]

The order of vectors while taking cross product is important.

\[ \displaystyle \overline{\mathbf{a}}\times \overline{\mathbf{b}} = - (\overline{\mathbf{b}}\times \overline{\mathbf{a}})\] Now consider the cross product of two similar unit vectors,

Solved Example: 234234

The dot product of two vectors $\bar{a}$ and $\bar{b}$ is defined as;

A. absin⁡θ

B. abcos⁡θ

C. ∣\bar{a}+\bar{b}∣

D. None of the above

Correct Answer: B

Solved Example: 34534

The result of the cross product a×b is:

A. A scalar

B. A vector

C. A matrix

D. None of the above

Correct Answer: B

Solved Example: 345345

Which of the following properties is true for the dot product?

A. Commutative

B. Associative

C. Distributive

D. All of the above

Correct Answer: D

Solved Example: 4-6-01

The angle between two unit-magnitude coplanar vectors P(0.866, 0.500, 0) and Q(0.259, 0.966, 0) will be:

A. 0$^\circ$

B. 30$^\circ$

C. 45$^\circ$

D. 60$^\circ$

\[\overline{P}\cdot \overline{Q}=\left|\overline{P}\right| \left|\overline{Q}\right| \cos \theta\] \[\left|\overline{P}\right| =\sqrt {0.866^{2}+0.5^{2}}=1\] \[\left|\overline{Q}\right| =\sqrt {0.259^{2}+0.966^{2}}=1\] \[\cos \theta =\dfrac {0.866\times 0.259+0.5\times 0.966}{1\times 1}=0.707\] \[\theta =\cos^{-1}\left( 0.707\right) =45^\circ\]

Correct Answer: C

Solved Example: 4-6-02

Given the 3-dimensional vectors: $\bar{A}$ = i(xy) + j(2yz) + k(3zx) and $\bar{B}$ = i(yz) + j(2zx) + k(3xy). Determine the scalar product at the point (1,2,3).

A. 144

B. 138

C. 132

D. 126

\begin{align*} \bar{A} &= i(xy) + j(2yz) + k(3zx)\\ \bar{B} &= i(yz) + j(2zx) + k(3xy)\\ \bar{A} \cdot \bar{B} &= xy^2z + 4xyz^2 + 9x^2yz \end{align*} At (1,2,3): \begin{align*} \bar{A} \cdot \bar{B} &= 1.(2^2).3 + 4.1.2.(3^2) + 9.(1^2).2.3\\ &= 12 + 72 + 54\\ &= 138 \end{align*}

Correct Answer: B

Solved Example: 4-6-03

What is the angle between two vectors $\overline {A}$ and $\overline {B}$ if $\overline {A}$ = 4$\overline {i}$ - 12$\overline {j}$ + 6$\overline {k}$ and $\overline {B}$ = 24$\overline {i}$ - 8$\overline {j}$ + 6$\overline {k}$?

A. 168.45$^\circ$

B. 51.22$^\circ$

C. 86.32$^\circ$

D. -84.64$^\circ$

\[\overline {A}=4\overline {i}-12\overline {j}+6\overline {k}\] \[\overline {B}=24\overline {i}-8\overline {j}+6\overline {k}\] \[\overline {A}\cdot \overline {B}=\left( 4\right) \left( 24\right) +\left( -12\right) \left( -8\right) +\left( 6\right) \left( 6\right)\] \[=96+96+36= 228\] \[\left|\overline {A} \right|=\sqrt {4^{2}+\left( -12\right) ^{2}+\left( 6\right) ^{2}} =14\] \[\left| \overline {B}\right| =\sqrt {\left( 24\right) ^{2}+\left( -8\right) ^{2}+\left( 6\right) ^{2}} = 26\] \[\cos \theta =\dfrac {\overline {A} \cdot \overline {B}}{\left| \overline {A}\right| \left| \overline {B}\right| } = \dfrac{228}{14 \times 26} = 0.621 \] \[\theta = \cos^{-1} (0.626) = 51.22 ^\circ\]

Correct Answer: B

Solved Example: 4-6-04

Determine the dot product of the two vectors U = 8$\overline {i}$ - 6$\overline {j}$ + 4$\overline {k}$ and V = 3$\overline {i}$ + 7$\overline {j}$ + 9$\overline{k}$.

A. 18

B. 16

C. 14

D. 12

\[\overline {U}=8\overline {i}-6\overline {j}+4\overline{k}\] \[\overline {V}=3\overline {i}+7\overline {j}+9\overline {k}\] \begin{align*} \overline {U}\cdot \overline {V} &= \left( 8\right) \left( 3\right) +\left( -6\right) \left( 7\right) +\left( 4\right) \left( 9\right)\\ &= 24-42+36\\ &=18 \end{align*}

Correct Answer: A

Solved Example: 4-6-05

Two perpendicular vectors are given in terms of their components by U = $U_x$ $\overline{i}$ - 4$\overline{j}$ + 6$\overline{k}$ and V = 3$\overline{i}$ + 2$\overline{j}$ - 3$\overline{k}$. Determine the component $U_x$.

A. 5.67

B. 6.67

C. 7.67

D. 8.67

\[\overline {U}=U_{x}\overline {i}-4\overline {j}+6\overline{k}\] \[\overline {V}=3\overline {i}+2\overline {j}-3\overline{k}\] Since, the Vectors are perpendicular, the \begin{equation*} \label{eq012} \begin{split} \overline {U}\cdot \overline {V} & =0 \\ U_x\left( 3\right) +\left( -4\right) \left( 2\right) +\left( 6\right) \left( -3\right) & =0 \\ 3U_x -8-18 & =0 \\ 3U_{x} & =26,\ \mathrm{Or,\ } U_{x} =8.67 \end{split} \end{equation*}

Correct Answer: D

Solved Example: 4-6-06

A force $\overrightarrow{F}= -K(y\hat{i}+x\hat{j})$ (where K > 0) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x- axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the forces $\overrightarrow{F}$ on the particle is:

A. $-2\ K{{a}^{2}}$

B. $2\ K{{a}^{2}}$

C. $-\ K{{a}^{2}}$

D. $K{{a}^{2}}$

Recall that: The dot product of same unit vectors,$i.i = j.j = k.k =1$

This is because the angle between two same unit vectors = 0, the $\cos \theta$ becomes one and also the magnitudes of unit vectors is one.

Here, in this problem, the motion is made up by two separate paths as shown in the figure.

$W = \bar{F}.\bar{s}= -K(y\hat{i}+x\hat{j}). a\hat{i}= -K (y\hat{i}). a\hat{i}= -K a y (1)= 0$

So, the total work is = -K a$^2$

Correct Answer: C

Solved Example: 4-6-07

If three vectors $\bar{a}$, $\bar{b}$ and $\bar{c}$ are with magnitude 3, 4 and 5 respectively and $\bar{a}$+$\bar{b}$+$\bar{c}$ = $\bar{0}$, then the value of $\bar{a}$.$\bar{b}$ + $\bar{b}$.$\bar{c}$ + $\bar{c}$.$\bar{a}$ is:

A. -23

B. -25

C. 30

D. 26

First Method: Given, \(\bar{a} + \bar{b} + \bar{c} = \bar{0}\)
Taking dot product with itself, we get, \((\bar{a} + \bar{b} + \bar{c}) \cdot (\bar{a} + \bar{b} + \bar{c}) = \bar{0} \cdot \bar{0}\)

Second Method: Given, \(\bar{a} + \bar{b} + \bar{c} = \bar{0}\)
Then, they form a cyclic closed loop as shown in the figure.

Since 3,4,5 is a Pythagorean triplet, the triangle formed is a right angled triangle as shown in the figure.
Since \(\bar{a}\) is \(\perp\) with \(\bar{b}\) \[\bar{a}. \bar{b} = 0\] \[\bar{b}. \bar{c} = bc\ \cos (180 - \alpha) = - bc \cos \alpha = - 4 \times 5 \times \left( \dfrac{4}{5} \right) = -16\] \[\bar{c}. \bar{a} = ca\ \cos (180 - \beta) = - ca \cos \beta = - 5 \times 3 \times \left( \dfrac{3}{5} \right) = -9\] Adding, \(\bar{a}\).\(\bar{b}\) + \(\bar{b}\).\(\bar{c}\) + \(\bar{c}\).\(\bar{a}\) = 0 - 16 - 9 = -25

Correct Answer: B

Solved Example: 4-7-01

Cross product of two unit vectors has a magnitude = 1. The $\angle$ between the vectors will be:

A. 60$^\circ$

B. 120$^\circ$

C. 45$^\circ$

D. 90$^\circ$

\[\left | \vec{a} \times \vec{b}\right | = a b \sin \theta\] \[1 = (1) (1) \sin \theta\] \[\theta = 90^\circ\]

Correct Answer: A

Solved Example: 4-7-02

What is the cross product $\overline{A}$ x $\overline{B}$ of the vectors, $\overline{A}$ = $\overline{i}$ + 4$\overline{j}$ + 6$\overline{k}$ and $\overline{B}$ = 2$\overline{i}$ + 3$\overline{j}$ + 5$\overline{k}$ ?

A. $\overline{i}$ - $\overline{j}$ -$\overline{k}$

B. –$\overline{i}$ + $\overline{j}$+$\overline{k}$

C. 2$\overline{i}$ + 7$\overline{j}$ - 5$\overline{k}$

D. 2$\overline{i}$ + 7$\overline{j}$ + 5$\overline{k}$

\[\overline {A}=\overline{i}+4\overline {j}+6\overline {k}\] \[\overline {B}=2\overline {i}+3\overline {j}+5\overline {k}\] \begin{align*} \overline {A} \times \overline {B} &= \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ 1 & 4 & 6 \\ 2& 3 & 5 \end{vmatrix}\\ &= \overline {i}\left( 20-18\right) -\overline {j}\left( 5-12\right) +\overline {k}\left( 3-8\right)\\ &=2\overline {i}+7\overline {j}-5\overline {k} \end{align*}

Correct Answer: C

Solved Example: 4-7-03

Find the magnitude of the following vector: $\bar{A} \times \bar{B}$ where, $\bar{A}$ = (-2, -5, 2) $\bar{B}$ = (-5, -2, -3)

A. 32.53

B. 21.56

C. 26.55

D. 24.03

$[(-2, -5, 2) \times (-5, -2, -3)] =(19, -16, -21)$
$\mathrm{Magnitude} = 23 \sqrt{2} = 32.5269$

Correct Answer: A

Solved Example: 4-7-04

Cross product $\bar{A} \times \bar{B}$ gives you:

A. A vector within the plane containing $\bar{A}$ and $\bar{B}$

B. A vector perpendicular to the plane containing $\bar{A}$ and $\bar{B}$

C. A unit vector perpendicular to the plane containing $\bar{A}$ and $\bar{B}$

D. A scalar equivalent of projection of $\bar{A}$ along $\bar{B}$ multiplied by magnitude of $\bar{B}$

Correct Answer: B

Solved Example: 4-7-05

A dot product $\bar{A} \cdot \bar{B}$ gives you:

A. A vector within the plane containing $\bar{A}$ and $\bar{B}$

B. A vector perpendicular to the plane containing $\bar{A}$ and $\bar{B}$

C. A unit vector perpendicular to the plane containing $\bar{A}$ and $\bar{B}$

D. A scalar equivalent of projection of $\bar{A}$ along $\bar{B}$ multiplied by magnitude of $\bar{B}$

Correct Answer: D

Learning Objectives:

• Define Del operator in terms of partial derivatives.
• Calculate gradient, divergence and curl.
• Understand physical interpretation and applications of gradient, divergence and curl.

The Del Operator:

The Del (\(\nabla\)) operator is defined as: \[\nabla = \left ( \dfrac{\partial}{\partial x} \mathbf{i} + \dfrac{\partial}{\partial y} \mathbf{\mathbf{j}} + \dfrac{\partial}{\partial z} \mathbf{k} \right )\] The Del operator itself is a vector. Depending upon whether it is operated on scalar or a vector, the following three possibilities will arise:

1. Operated on a scalar - Gradient
2. Operated on a vector with a dot product - Divergence
3. Operated on a vector with a cross product - Curl

Gradient:

Gradient is a vector with the magnitude and direction of the maximum change of the function in space. Gradient is calculated for scalars and the result of calculation (the gradient itself) is a vector.

Let us evaluate gradient of a scalar function \(\phi\) described below. \[\phi = x^3z\] Then, \[\dfrac{\partial \phi}{\partial x} = 3x^2z\] \[\dfrac{\partial \phi}{\partial y} = 0\] \[\dfrac{\partial \phi}{\partial z} = x^3\]

\[\nabla \phi = \left ( 3x^2z \mathbf{i} + x^3 \mathbf{k} \right )\]

If you want to calculate \(\nabla \phi\) at certain point, let’s say at (1,2,-1) then substitute these coordinates instead of x, y and z.
\[\nabla \phi = \left ( 3(1)^2(-1) \mathbf{i} + (1)^3 \mathbf{k} \right ) = - 3 \mathbf{i} + \mathbf{k}\]

Physical Interpretation of Gradient:

The gradient represents the direction of greatest change. The gradient points to the maximum of the function; follow the gradient, and you will reach the local maximum. It is a vector, so it points towards the direction of greatest change.

Divergence:

Divergence is calculated for a vector, but since it is a dot product between the del (\(\nabla\)) operator (which is a vector) and another vector, the final result is a scalar.

Let us take an example of a variable vector which keeps on changing depending upon the location (coordinates).
e.g. if \[\bar{a} = x^3zi + xy^2j + yzk\] \[\nabla . \mathbf{V} = \left ( \frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k} \right ) .(x^3z i+ xy^2 j+ yz k)\] \[\nabla . \mathbf{V} =(3x^2z + 2xy + z )\] If you want to calculate at some location, let’s say, at (1,2-1) then simply substitute the coordinates, \[\nabla . \mathbf{V} \mathrm{\ at\ } (1,2-1) =(3(1)^2(-1) + 2(1)(2) + (-1) ) = 0\]

Physical Interpretation of the Divergence:

• Consider a vector field \(\mathbf{F}\) that represents a fluid velocity: The divergence of \(\mathbf{F}\) at a point in a fluid is a measure of the rate at which the fluid is flowing away from or towards that point.
• A positive divergence is indicating a flow away from the point.
• Physically divergence means that either the fluid is expanding or that fluid is being supplied by a source external to the field.
• The lines of flow diverge from a source and converge to a sink.
• If there is no gain or loss of fluid anywhere then div \(\mathbf{F}\) = 0. Such a vector field is said to be solenoidal.

Curl:

Let us take same example of a variable vector.
e.g. if \(\bar{a} = x^3z\overline{i} + xy^2\overline{j} + yz\overline{k}\) \[\begin{split} \nabla \times \textbf{V} & = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ x^3z & xy^2 & yz \end{vmatrix}\\ \nabla \times \mathbf{V} & =((z-0) \overline{i} - (0-x^3) \overline{j} + (y^2-0) \overline{k}) \\ & = (z \overline{i} + x^3 \overline{j} + y^2 \overline{k}) \end{split}\]

If you want to calculate at some location, let’s say, at (1,2-1) then simply substitute the coordinates, \[\begin{aligned} \nabla \times \mathbf{V}\ \mathrm{at}\ (1,2,-1) &=(z \overline{i} + x^3 \overline{j} + y^2 \overline{k}) \\ &= (-1)( \overline{i} + (1)^3 \overline{j} + (2)^2 \overline{k}) \\ &= -\overline{i} + \overline{j} + 4\overline{k} \end{aligned}\]

Physical Interpretation of the Curl:

• Consider a vector field \(\mathbf{F}\) that represents a fluid velocity, then the curl of \(\mathbf{F}\) at a point in a fluid is a measure of the rotation of the fluid.
• If there is no rotation of fluid anywhere then \(\nabla \times F = 0\) Such a vector field is said to be irrotational or conservative.
• For a 2D flow with \(\overline{F}\) represents the fluid velocity, \(\nabla \times \overline{F}\) is perpendicular to the motion and represents the direction of axis of rotation.

Solved Example: 4-10-01

The divergence of vector $\textbf{i} = x\textbf{i} + y\textbf{j} + z\textbf{k}$ is:

A. $\textbf{i} + \textbf{j} + \textbf{k}$

B. 3

C. 0

D. 1

Divergence is defined as $\nabla$.r, where:

Correct Answer: B

Solved Example: 4-10-02

The divergence of the vector field $x^2y \mathbf{i}+ xy \mathbf{j} + z^2 \mathbf{k}$ at P (1, 1, 1) is:

A. 5

B. 1

C. 4

D. 2

\[\nabla . \mathbf{V} = \left ( \dfrac{\partial}{\partial x} \mathbf{i} + \dfrac{\partial}{\partial y} \mathbf{j} + \dfrac{\partial}{\partial z} \mathbf{k} \right ) .(x^2y\ i+ xy\ j+ z^2\ k)\] \[\nabla . \mathbf{V} = \dfrac{\partial}{\partial x} x^2y + \dfrac{\partial}{\partial y} xy + \dfrac{\partial}{\partial z} z^2 = 2xy + x + 2z\] Substitute x = 1, y = 1 and z = 1 (Given coordinates) $\nabla . \mathbf{V} = 5$

Correct Answer: A

Solved Example: 4-10-03

Determine the divergence of the vector: $\overline{V}$ = $(x^2) \overline{i} + (-xy) \overline{j} + (xyz) \overline{k}$ at the point (3,2,1).

A. 9

B. 11

C. 13

D. 7

\[\nabla =\dfrac {\partial}{\partial x} \overline{i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline{k}, \quad \overline {V}=\left(x^{2}\right) \overline {i}+\left( -xy\right) \overline {j}+ \left(xyz\right) \overline {k}\] \begin{align*} \nabla \cdot \overline {V} &= \left( \dfrac {\partial }{\partial x}\right) \left( x^{2}\right) +\dfrac {\partial }{\partial y}\left( -xy\right) +\dfrac {\partial }{\partial z}\left( xyz\right)\\ &= 2x-x+xy\\ &=x+xy \end{align*}

At x = 3, y = 2 and z = 1,

$\nabla \cdot \overline {V}$ = 3+6 = 9

Correct Answer: A

Solved Example: 4-10-04

The divergence of the vector field $(x - y) \textbf{i} + (y - x)\textbf{j} + (x + y + z)\textbf{k}$ is:

A. 0

B. 1

C. 2

D. 3

\begin{align*} \overline {V}&=\left( x-y\right) \overline {i}+\left( y-x\right) \overline {j}+\left( x+y+z\right) \overline {k},\\ \nabla &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \end{align*} \begin{align*} \nabla \cdot \overline {V} &=\dfrac {\partial }{\partial x}\left( x-y\right) +\dfrac {\partial }{\partial y}\left( y-x\right) +\dfrac {\partial }{\partial z}\left( x+y+z\right)\\ &=1+1+1\\ &=3 \end{align*}

Correct Answer: D

Solved Example: 4-10-05

The divergence of the vector field, $3xzi + 2xyj - yz^2k$ at a point (1,1,1) is equal to:

A. 7

B. 4

C. 3

D. 0

Let, $V = 3xzi + 2xyj - yz^2k$. We know divergence vector field of V is given by ($\nabla$ $\cdot$V)

$\left( \nabla \cdot V\right) =3\times 1+2\times 1-2\times 1\times 1 =3$

Correct Answer: C

Solved Example: 4-10-06

Let f be a scalar field, and let $\bar{F}$ be a vector field. Which of the following expressions is meaningful:

A. $\nabla f + \nabla . f$

B. $\nabla \bar{F} + \nabla . f$

C. $\nabla f + \nabla . \bar{F}$

D. $\nabla f + \nabla \times \bar{F}$

$\nabla$ is a vector operator. When it is applied to a scalar, you get a vector result. Hence $\nabla$f is a vector.

When it is applied to a vector as a dot product, you get a scalar result. Hence $\nabla$.$\bar{F}$ is a scalar.

When it is applied to a vector as a cross product, you get a vector result. Hence $\nabla$ $\times$ $\bar{F}$ is a vector.

You CANNOT add scalar with a vector.

Correct Answer: D

Solved Example: 4-10-07

A vector is said to be solenoidal when its:

A. Divergence = 0

B. Divergence = 1

C. Curl = 0

D. Curl = 1

When the divergence of a vector is zero, it is said to be solenoidal or divergent-free.

Correct Answer: A

Solved Example: 4-11-01

The vector field is F = xi - yj (where i and j are unit vector) is:

A. Divergence free, but not irrotational

B. Irrotational, but not divergence free

C. Divergence free and irrotational

D. Neither divergence free nor irrational

$F = xi - yj$ . First check divergency, for divergence, \begin{align*} \nabla \cdot F &=\left[ \dfrac {\partial }{0x}\overline {i}+\dfrac {\partial }{dy}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right] \left[ x\overline {i}-y\overline {j}\right]\\ &=1-1 =0 \end{align*} So we can say that F is divergence free. Now checking the irrotionality. For irrotation the curl F must be 0. \begin{align*} \nabla \times F &=\left[ \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right] \times \left( x\overline {i}-y\overline {j}\right)\\ &= \begin{vmatrix} i & j & k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\ x & -y & 0 \end{vmatrix}\\ &= 0 \end{align*} So, vector field is irrotational. We can say that the vector field is divergence free and irrotational.

Correct Answer: C

Solved Example: 4-11-02

A vector is said to be irrotational when its:

A. Divergence = 0

B. Divergence = 1

C. Curl = 0

D. Curl = 1

When the curl of a vector is zero, it is said to be irrotational.

Correct Answer: C

Solved Example: 4-11-03

Curl of vector V(x,y,z)  = 2x²i + 3z²j + y³k at x = y = z = 1 is:

A. 3i-6k

B. -3i

C. 3i

D. 3i-4j

\[\begin{split} \nabla \times \textbf{V} & = \begin{vmatrix} \overline{i} & \overline{j} & \overline{k}\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ 2x^2 & 3z^2 & y^3 \end{vmatrix}\\ \nabla \times \mathbf{V} & =((3y^2-6z) \overline{i} - (0-0) \overline{j} + (0-0) \overline{k}) \\ & = ((3y^2-6z) \overline{i})\\ & =(3-6)i \\ & =(-3)i \end{split}\]

Correct Answer: B

Solved Example: 4-8-01

The Del operator is called as:

A. Gradient

B. Curl

C. Divergence

D. Vector differential operator

The Del operator is used to replace the differential terms, thus called vector differential operator in electromagnetics.

Correct Answer: D

Solved Example: 4-9-01

The directional derivative of the scalar function $f (x,y,z) = x^2+ 2y^2+ z$ at the point P = (1, 1, 2) in the direction of the vector a = 3i - 4j is:

A. -4

B. -2

C. -1

D. 1

Direction derivative of a function f along a vector P is given by \[a=\nabla f\cdot \dfrac {a}{\left| a\right| }\] \[\nabla F=\left( \dfrac {\partial f}{\partial x}\overline {i}+\dfrac {\partial f}{\partial y}\overline {j}+\dfrac {\partial f}{\partial z}\overline {k}\right)\] \[f\left( x,y,z\right) =x^{2}+2y^{2}+z\] \[a=3\overline {i}-4\overline {j}\] \[a=\nabla \left( x^{2}+2y^{2}+z\right) .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {3^{2}+4^{2}}}\] At point P(1, 1, 2) the direction derivative is $=\left[ 2x\overline {i}+4y\overline {j}+\overline {k}\right] .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {25}}$ \[a=\dfrac {6\times 1-16\times 1}{5}=-\dfrac {10}{5}=-2\]

Correct Answer: B

Solved Example: 4-9-02

The vector which is normal to the surface $2xz^2-3xy-4x = 7$ at point (1,-1,2) is:

A. 2$\bar{i}$ - 3$\bar{j}$ + 8$\bar{k}$

B. 2$\bar{i}$ + 3$\bar{j}$ + 4$\bar{k}$

C. 7$\bar{i}$ - 3$\bar{j}$ + 8$\bar{k}$

D. 7$\bar{i}$ - 5$\bar{j}$ + 8$\bar{k}$

\[\mathrm{Let\ } \phi \left( x,y,z\right) =2xz^{2}-3xy-4x=0\] Vector normal to a surface $\phi$ is given by gradient of $\phi$. \begin{align*} \nabla \phi &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \left( 2xz^{2}-3xy-4x\right)\\ &= \left( 2z^{2}-3y-4\right) \overline {i} + \left( 0-3x-0\right) \overline {j} + \left( 4xz-0-0\right) \overline {k}\\ &=\left( 2z^{2}-3y-4\right) \overline {i}-3x\overline {j}+4xz\overline {k} \end{align*} At $\left(1,-1,2 \right) \nabla \phi =\left( 8+3-4\right) \overline {i}-3\overline {j}+8\overline {k} =7 \overline{i} - 3\overline {j}+8\overline {k}$

Correct Answer: C

Solved Example: 4-9-03

A function is defined in the Cartesian coordinate system as $f (x, y) = xe^y$. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point ($\dfrac{1}{2}$, 2) is:

A. 0

B. -1

C. 1

D. $\infty$

\[f\left(x, y\right) = xe^{y}\] \begin{align*} \mathrm{Grad}\ f &= \nabla f\\ &=\dfrac{\partial }{\partial x}\overline{i}+\dfrac{\partial }{\partial y}\overline{j}+\dfrac{\partial }{\partial z}\overline{k}\\ &=\left( e^{y}\right) \overline{i}+\left(xe^{y}\right) \overline{j}+\left( 0\right) \overline{k}\\ \mathrm{At}\ \left(2,0\right), \nabla f &=\overline{i}+2\overline{j} \end{align*} Let $P \left(2,0\right) \mathrm{and}\ Q\left(\dfrac{1}{2},2\right)$ \begin{align*} \overline{PQ} &=\left( \dfrac{1}{2}-2\right) \overline{i}+\left( 2-0\right)\overline{j}\\ &=-\dfrac{3}{2}\overline{i}+2\overline{j} \end{align*} Unit vector along $\overline{PQ}$ \begin{align*} \hat{PQ} &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\sqrt{\left(-\dfrac{3}{2}\right) ^{2}+\left( 2\right) ^{2}}}\\ &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) } \end{align*} Directional derivative \begin{align*} D &=\left( \nabla f\right) \cdot \widehat{PQ}\\ &=\left(\overline{i}+2\overline{j}\right) \cdot \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) }\\ &=\dfrac{\dfrac{-3+8}{2}}{\dfrac{5}{2}}\\ &=1 \end{align*}

Correct Answer: C

Solved Example: 4-9-04

The magnitude of the gradient for the function $(𝑥, 𝑦, 𝑧) = 𝑥^2 + 3𝑦^2 + 𝑧^3$ at the point (1, 1, 1) is:

A. 6

B. 7

C. 8

D. 9

\[\phi = 𝑥^2 + 3𝑦^2 + 𝑧^3\] \[\nabla \phi = 2x \bar{i} + 6y \bar{j} + 3z^2 \bar{k}\] At (1,1,1), \[\nabla \phi = 2 \bar{i} + 6 \bar{j} + 3 \bar{k}\] Magnitude can be calculated as: \[|\nabla \phi| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7\]

Correct Answer: B