Math 1201 (MHU) 1
Lecture 1-3:
Vector Algebra & Linear dependence and independence of vectors
Course Plan / Outline
A. Course Information
1. Course Title: Engineering Mathematics II
2. Course Code: Math 1201
3. Prerequisite: Math 1101: Engineering Mathematics I
4. Semester: 1st year Even Semester, Sections A, B and C
5. Credit: 3.00
6. Class Hours: Three Hours per Week
B. Instructors’ Details
1. Name of the Instructors: Dr. Md. Helal Uddin Molla (MHU) and Md. Dalim Haque (MDH)
2. Designation: Professor and Assistant Professor
3. Office Room: 221 and 228, Mathematics Department, RUET.
4. Contact Number: +8801730875698 and +8801774015578
5. Email: helal.mathru@yahoo.com and dalimru10@gmail.com
6. Weblink: https://www.math.ruet.ac.bd/helalmathru, https://www.math.ruet.ac.bd/dalimru10
7. Consultation Hours:
C. Course Rationale:
The study of the course Math-1201:“Vector analysis, Fourier analysis and Laplace transforms” plays an
important role in electrical and electronic engineering. Under the study of vector analysis, the students will gain
knowledge about the vector field, position of a moving particle, its velocity and acceleration, Vector calculus
(Gradient, Divergence and Curl and their related theorems) are important to study Electro-Magnetic field.
Fourier Series is important to study vibration, frequency analysis and signal processing. Laplace transforms is
widely used to solve linear differential systems in circuit analysis.
D. Course Contents:
Vector Analysis: Review of vector algebra: Addition and subtraction of vectors, Scalar and vector product of
two vectors and their geometrical interpretation, Triple products and multiple products, Linear dependents and
independents of vectors. Vector Calculus: Differentiation and Integration of Vectors together with elementary
applications, Definition of line, surface and volume integrals, Gradient of scalar function, Divergence and Curl
of vector functions, various formulae: Green’s theorem, Gauss’s theorem, Stoke ’s theorem.
Fourier Analysis: Real and complex form of Fourier series, Finite transform, Fourier Integral, Fourier
transforms and their uses in solving boundary value problems of wave equations.
Laplace Transforms: Definition Laplace transforms of some elementary functions, Sufficient conditions for
existence of Laplace Transforms, Inverse Laplace Transforms, Laplace Transforms of derivatives. The unit step
function, Periodic function, Some special theorems on Laplace Transforms, Partial fractions, Solutions of
differential equations by Laplace Transforms, Evaluation of improper integrals.
Learning Materials:
(a) Text Book:
1. Vector Analysis Author: M.R. Spiegel, S. Lipschutz
2. Fourier Analysis with Applications to Boundary value Problems by M. R. Spiegel
3. Laplace Transformation Author: M.R. Spiegel
Math 1201 (MHU) 2
(b) Reference Books:
1. Advanced Engineering Mathematics, Author: H. K. Dass.
(c) Lecture Notes/ Hand notes:
(i) Hand note on the Vector Analysis, Fourier and Laplace Transform.
(d) Online Resources:
(i) http://www.uop.edu.pk/ocontents/Vector%20Analysis%20Schaum.pdf
(ii) https://physics.csusb.edu/~prenteln/notes/transform_notes.pdf
(e) Others:
(i) Instruction to know the differential and integral calculus.
Assessment and Marks Distribution
Students will be assessed on the basis of their overall performance in all the exams (class tests, final exam,
assignments, projects, and presentations. Final numeric reward will be the compilation of:
1. Continuous Assessment (Class Tests, Assignments, Projects, Presentations, etc.): 40%
a. Class Tests: 20%
b. Attendance 10%
c. Others i.e., Assignments/Projects/Presentations: 10%
2. Final Exam: 60%
Chapter-1:
Vector Algebra & Linear dependence and independence of vectors
Vector: A Vector is a quantity having both magnitude and direction. Such as velocity, acceleration force etc.
Scalar: A scalar is a quantity having only magnitude but no direction such as mass, length, time etc.
The dot or scalar product: The dot or scalar product of two vectors 𝐴⃗ and 𝐵⃗ denoted by 𝐴⃗ • 𝐵⃗ is defined as
the product of the magnitudes of 𝐴⃗ and 𝐵⃗ and the cosine of the angle 𝜃 between them. In symbols
𝐴⃗ • 𝐵⃗ = 𝐴𝐵 𝑐𝑜𝑠 𝜃. If 𝐴⃗ • 𝐵⃗ = 0 then the vectors 𝐴⃗ and 𝐵⃗ are perpendicular to each other.
The cross or vector product: The cross or vector product of two vectors 𝐴⃗ and 𝐵⃗ denoted by 𝐴⃗ × 𝐵⃗ is defined
as the product of the magnitudes of 𝐴⃗ and 𝐵⃗ and the sine of the angle 𝜃 between them. The direction of the
vector𝐴⃗ × 𝐵⃗is perpendicular to the plane of 𝐴⃗ and 𝐵⃗. In symbols 𝐴⃗ × 𝐵⃗ = 𝐴𝐵 𝑠𝑖𝑛 𝜃 𝜂̂ . Where 𝜂̂ is a unit vector
indicating the direction of 𝐴⃗ × 𝐵⃗. If 𝐴⃗ × 𝐵⃗ = 0 then the vectors 𝐴⃗ and 𝐵⃗ are parallel to each other.
Note: i) If three vector 𝐴⃗, 𝐵⃗ and 𝐶⃗ are coplanar then 𝐴⃗ • 𝐵⃗ × 𝐶⃗ = 0.
ii) If three vector 𝐴⃗, 𝐵⃗ and 𝐶⃗ are formed a parallelepiped then volume 𝐴⃗ • 𝐵⃗ × 𝐶⃗ = 𝐴⃑𝐵⃗𝐶⃗ .
Question-1: Find the angle between the vectors 𝐴⃗ = 𝚤̂ + 2𝚥̂ − 𝑘 and 𝐵⃗ = −𝚤̂ + 𝚥̂ − 2𝑘 .
Solution: Given that 𝐴⃗ = 𝚤̂ + 2𝚥̂ − 𝑘 and 𝐵⃗ = −𝚤̂ + 𝚥̂ − 2𝑘
We know, 𝐴⃗ • 𝐵⃗ = 𝐴𝐵 𝑐𝑜𝑠 𝜃
⃗• ⃗
or, 𝑐𝑜𝑠 𝜃 =
̂ ̂ • ̂ ̂
or, 𝑐𝑜𝑠 𝜃 =
√ √
or, 𝑐𝑜𝑠 𝜃 =
√ √
Math 1201 (MHU) 3
or, 𝑐𝑜𝑠 𝜃 =
or, 𝜃 = Ans.
Question-2: Find the unit vector perpendicular to both 𝐴⃗ = 2𝚤̂ + 3𝚥̂ − 𝑘 and 𝐵⃗ = 3𝚤̂ − 2𝚥̂ + 𝑘 using dot
product.
Solution: Given that 𝐴⃗ = 2𝚤̂ + 3𝚥̂ − 𝑘 and 𝐵⃗ = 3𝚤̂ − 2𝚥̂ + 𝑘
Let, 𝐶 = 𝑥𝚤̂ + 𝑦𝚥̂ + 𝑧𝑘 be the required unit vector.
then 𝐶⃗ • 𝐶⃗ = 1
or, 𝑥 + 𝑦 + 𝑧 = 1 ........................ (i)
Since 𝐴⃗ is perpendicular to 𝐶⃗
So, 𝐴⃗ • 𝐶⃗ = 0
or, 2𝚤̂ + 3𝚥̂ − 𝑘 • 𝑥𝚤̂ + 𝑦𝚥̂ + 𝑧𝑘 = 0
or, 2𝑥 + 3𝑦 − 𝑧 = 0 ........................... (ii)
and Since 𝐵⃗ is perpendicular to 𝐶⃗
So, 𝐵⃗ • 𝐶⃗ = 0
or, 3𝚤̂ − 2𝚥̂ + 𝑘 • 𝑥𝚤̂ + 𝑦𝚥̂ + 𝑧𝑘 = 0
or, 3𝑥 − 2𝑦 + 𝑧 = 0 ........................... (iii)
Solving (ii) and (iii) we get, = =
∴ 𝑦 = −5𝑥, 𝑧 = −13𝑥
Putting these values of y and z in (i) we get,
𝑥 = ∴𝑥=±
√
So, 𝑦 = ± , 𝑧=±
√ √
̂ ̂
Therefore the required unit vector is 𝐶⃗ = ± Ans.
√
Question-3: Show that the vectors 𝚤̂ + 𝚥̂ + 𝑘 , 𝚤̂ − 𝑘 and 𝚤̂ − 2𝚥̂ + 𝑘 are mutually orthogonal. Find x, y and z
if 𝚤̂ + 𝚥̂ + 2𝑘, −𝚤̂ + 𝑧𝑘 and 2𝚤̂ + 𝑥𝚥̂ + 𝑦𝑘 are mutually orthogonal.
Solution: First Part: Let, 𝐴⃗ = 𝚤̂ + 𝚥̂ + 𝑘 ,
𝐵⃗ = 3𝚤̂ − 2𝚥̂ + 𝑘
𝐶⃗ = 𝚤̂ − 2𝚥̂ + 𝑘
Now, 𝐴⃗ • 𝐵⃗ = 𝚤̂ + 𝚥̂ + 𝑘 • 𝚤̂ − 𝑘 = 1 − 1 = 0
𝐵⃗ • 𝐶⃗ = 𝚤̂ − 𝑘 • 𝚤̂ − 2𝚥̂ + 𝑘 = 1 − 1 = 0
𝐶⃗ • 𝐴⃗ = 𝚤̂ − 2𝚥̂ + 𝑘 • 𝚤̂ + 𝑗 + 𝑘 = 1 − 2 + 1 = 0
Since 𝐴⃗ • 𝐵⃗ = 𝐵⃗ • 𝐶⃗ = 𝐶⃗ • 𝐴⃗ = 0
Hence the vectors 𝚤̂ + 𝚥̂ + 𝑘, 𝚤̂ − 𝑘 and 𝚤̂ − 2𝚥̂ + 𝑘 are mutually orthogonal. (Proved)
2nd Part: Let, 𝑃⃗ = 𝚤̂ + 𝚥̂ + 2𝑘
𝑄⃗ = −𝚤̂ + 𝑧𝑘
𝑅⃗ = 2𝚤̂ + 𝑥𝚥̂ + 𝑦𝑘
Since 𝑃⃗, 𝑄⃗ and 𝑅⃗ are mutually orthogonal.
So, 𝑃⃗ • 𝑄⃗ = 𝑄⃗ • 𝑅⃗ = 𝑅⃗ • 𝑃⃗ = 0
Now, 𝑃⃗ • 𝑄⃗ = 0
or, 𝚤̂ + 𝚥̂ + 2𝑘 • −𝚤̂ + 𝑧𝑘 = 0
or, −1 + 2𝑧 = 0
Math 1201 (MHU) 4
or, 𝑧 =
And, 𝑄⃗ • 𝑅⃗ = 0
or, −𝚤̂ + 𝑧𝑘 • 2𝚤̂ + 𝑥𝚥̂ + 𝑦𝑘 = 0
or, −2 + 𝑦𝑧 = 0
or, 𝑦 = 4
Again 𝑅⃗ • 𝑃⃗ = 0
or, 2𝚤̂ + 𝑥𝚥̂ + 𝑦𝑘 • 𝚤̂ + 𝚥̂ + 2𝑘 = 0
or, 2 + 𝑥 + 2𝑦 = 0
or, 2 + 𝑥 + 8 = 0
or, 𝑥 = −10
Hence the required values are 𝑥 = −10, 𝑦 = 4 and 𝑧 = Ans.
Question-4: Find the vector Perpendicular to ̂ + 𝟐𝒌and ̂ + ̂ − 𝒌 and hence find the area of the
triangle with these two vectors as adjacent sides.
Solution: Let 𝐴⃗ = 𝚤̂ + 2𝑘 and 𝐵⃗ = 𝚤̂ + 𝚥̂ − 𝑘
Now any vector perpendicular to 𝐴⃗ and 𝐵⃗ is 𝐴⃗ × 𝐵⃗
∴ 𝐴⃗ × 𝐵⃗ = 𝚤̂ + 2𝑘 × 𝚤̂ + 𝚥̂ − 𝑘
𝚤̂ 𝚥̂ 𝑘
= 1 0 2 =−2𝚤̂ + 3𝚥̂ + 𝑘 Ans.
1 1 −1
The area of the triangle with the two vectors as adjacent sides = 𝐴⃗ × 𝐵⃗ = −2𝚤̂ + 3𝚥̂ + 𝑘 = √14 Ans.
Question-5: Find the volume of a tetrahedron whose one vertex is at the origin and the other three
vertices are (3,2,1), (2,3, −1) and (−1,2,3).
Solution: Let 𝐴⃗ = 3𝚤̂ + 2𝚥̂ + 𝑘 , 𝐵⃗ = 2𝚤̂ + 3𝚥̂ − 𝑘 and 𝐶⃗ = −𝚤̂ + 2𝚥̂ + 3𝑘
We know, the volume of a tetrahedron, V = 𝐴⃑𝐵⃗ 𝐶⃗ = 𝐴⃗ • 𝐵⃗ × 𝐶⃗
𝚤̂ 𝚥̂ 𝑘
= 𝐴⃗ • 2 3 −1
−1 2 3
= (3𝚤̂ + 2𝚥̂ + 𝑘 ) • (11𝚤̂ − 5𝚥̂ + 7𝑘 )
= 5 Ans.
Question-6: Find the volume of a tetrahedron whose vertices are 𝑃(3,4,5), 𝐴(2,1,1), 𝐵(2,1,5) and
𝐶(1,4,2).
Solution: Given 𝑃(3,4,5), 𝐴(2,1,1), 𝐵(2,1,5) and 𝐶(1,4,2).
⎯
∴ 𝑃𝐴 = −𝚤̂ − 3𝚥̂ − 4𝑘 , 𝑃𝐵 = −𝚤̂ − 3𝚥̂ and 𝑃𝐶 = −2𝚤̂ − 3𝑘
⎯
We know, the volume of a tetrahedron, V = 𝑃𝐴 • 𝑃𝐵 × 𝑃𝐶
Math 1201 (MHU) 5
𝚤̂ 𝚥̂ 𝑘
= 𝑃𝐴 • −1 −3 0
−2 0 −3
= (−𝚤̂ − 3𝚥̂ − 4𝑘 ) • (9𝚤̂ − 3𝚥̂ − 6𝑘 )
= 4 Ans.
Question-7: An automobile travels 3km due north, then 5km due northeast. Represent these
displacements graphically and determine the resultant displacement analytically.
Solution:
N Q
𝐵⃗
P 𝐶⃗ R
𝐴⃗
W O E
S
⎯
Let, 𝑂𝑃 or, 𝐴⃗ represents displacement of 3km due north
⎯
and 𝑃𝑄 or, 𝐵⃗ represents displacement of 5km due northeast.
⎯
Also the vector 𝑂𝑄 or, 𝐶⃗ represents the resultant displacement.
From the triangle 𝛥𝑂𝑃𝑄 we have by the law of cosine, 𝐶 = 𝐴 + 𝐵 − 2𝐴𝐵 𝑐𝑜𝑠 ∠ 𝑂𝑃𝑄
or, 𝐶 = √3 + 5 − 2 × 3 × 5 𝑐𝑜𝑠 1 35
or, 𝐶 = 7.43
Also by the law of sine, =
∠ ∠
∠
or, 𝑠𝑖𝑛 ∠ 𝑂𝑄𝑃 = = = 0.2855
.
∴ ∠𝑂𝑄𝑃 = 16 35′
So, ∠𝑂𝑄𝑃 = ∠𝑄𝑂𝑅 = 16 35′
⎯
Thus the resultant displacement or vector 𝑂𝑄 has magnitude 7.43
and direction = (45 + 16 35′ ) = 61 35′ northeast.
Question-8: A Particle is subjected to forces 3 kg-wt, 4 kg-wt, 5 kg-wt respectively acting in directions
⎯ ⎯ ⎯
parallel to the edges 𝑨𝑩, 𝑩𝑪, 𝑪𝑨 of an equilateral 𝜟𝑨𝑩𝑪. Find the resultant force acting on the particle.
⎯ ⎯
Solution: Let, 𝐵𝐶 direction be parallel to 𝚤̂ and 𝚥̂ ⊥ to 𝐵𝐶 at B. 𝚥̂
⎯
3 kg-wt force acting along 𝐴𝐵 = 3(𝚤̂ 𝑐𝑜𝑠 1 20 + 𝚥̂ 𝑐𝑜𝑠 2 10 ) A
√
= − 𝚤̂ − 𝚥̂
⎯
4 kg-wt force acting along 𝐵𝐶 = 4𝚤̂
𝚤̂
5 kg-wt force acting along 𝐶𝐴 = 5(𝚤̂ 𝑐𝑜𝑠 2 40 + 𝚥̂ 𝑐𝑜𝑠 3 30 )
Math 1201 (MHU) 6
√
= − 𝚤̂ + 𝚥̂ B C
√ √
So, The total force = − 𝚤̂ − 𝚥̂ + 4𝚤̂ + − 𝚤̂ + 𝚥̂ = √3𝚥̂
⎯
= √3kg-wt to 𝐵𝐶 Ans.
Question-9: A Particle is acted on by constant forces 𝟑 ̂ + 𝟐 ̂ + 𝟓𝒌 and 𝟐 ̂ + ̂ − 𝟑𝒌 and is displaced from
a point whose position vector is 𝟐 ̂ − ̂ − 𝟑𝒌to a point whose position vector 𝟒 ̂ − 𝟑 ̂ + 𝟕𝒌. Calculate the
work done.
Solution: Total force = 3𝚤̂ + 2𝚥̂ + 5𝑘 + 2𝚤̂ + 𝚥̂ − 3𝑘 = 5𝚤̂ + 3𝚥̂ + 2𝑘
Displacement = 4𝚤̂ − 3𝚥̂ + 7𝑘 − 2𝚤̂ − 𝚥̂ − 3𝑘 = 2𝚤̂ − 2𝚥̂ + 10𝑘
Therefore, Work done = Force• Displacement
= 5𝚤̂ + 3𝚥̂ + 2𝑘 • 2𝚤̂ − 2𝚥̂ + 10𝑘 = 10 − 6 + 20 = 24 Ans.
Question-10: Force of magnitude 5 and 3 units acting in the directions 𝟔 ̂ + 𝟐 ̂ + 𝟑𝒌 and 𝟑 ̂ − 𝟐 ̂ + 𝟔𝒌
respectively act on a particle which is displaced from a point (2, 2, -1) to (4, 3, 1). find the work done.
Solution:
̂ ̂
1st force of magnitude 5 units, acting in the direction 6𝚤̂ + 2𝚥̂ + 3𝑘= 5 = 6𝚤̂ + 2𝚥̂ + 3𝑘
√
̂ ̂
2nd force of magnitude 3 units, acting in the direction 3𝚤̂ − 2𝚥̂ + 6𝑘 = 3 = 3𝚤̂ − 2𝚥̂ + 6𝑘
√
So, the resultant force = 6𝚤̂ + 2𝚥̂ + 3𝑘 + 3𝚤̂ − 2𝚥̂ + 6𝑘 = 39𝚤̂ + 4𝚥̂ + 33𝑘
Displacement = 4𝚤̂ + 3𝚥̂ + 𝑘 − 2𝚤̂ + 2𝚥̂ − 𝑘 = 2𝚤̂ + 𝚥̂ + 2𝑘
Therefore, Work done = Force• Displacement
= 39𝚤̂ + 4𝚥̂ + 33𝑘 • 2𝚤̂ + 𝚥̂ + 2𝑘 = (78 + 4 + 66) = Ans.
Question-11: Find a vector of magnitude 5 parallel to yz plane and perpendicular to the vector
𝟐 ̂ + 𝟑 ̂ + 𝒌.
⃗
Solution: Let 𝐴 = 𝑏𝚥̂ + 𝑐𝑘be any vector parallel to yz plane and 𝐵⃗ = 2𝚤̂ + 3𝚥̂ + 𝑘.
Since 𝐴⃗ ⊥ 𝐵⃗
So, 𝐴⃗ • 𝐵⃗ = 0
or, 𝑏𝚥̂ + 𝑐𝑘 • 2𝚤̂ + 3𝚥̂ + 𝑘 = 0
or, 3𝑏 + 𝑐 = 0
or, 𝑐 = −3𝑏.......................... (i)
Again 𝐴⃗ = 5
or, √𝑏 + 𝑐 = 5
or, 𝑏 + (−3𝑏) = 5
or, 𝑏√10 = 5
or, 𝑏 = =
√
So, from (i) 𝑐 = −3
Hence the required vector is 𝐴⃗ = 𝚥̂ − 3𝑘 Ans.
Question-12: Find the projection of the vector 𝟒 ̂ − 𝟑 ̂ + 𝒌 on the line passing through the points
(2, 3, -1) and (-2, -4, 3).
Solution: Let 𝐴⃗ = 4𝚤̂ − 3𝚥̂ + 𝑘
any vector passing through the points 𝐵⃗ = (−2 − 2)𝚤̂ + (−4 − 3)𝚥̂ + (3 + 1)𝑘 = 4𝚤̂ − 7𝚥̂ + 4𝑘 .
⃗• ⃗ ̂ ̂ • ̂ ̂
So, the projection of 𝐴⃗ on 𝐵⃗ = =
⃗
= 1 Ans.
√
Math 1201 (MHU) 7
Question-13: Find the angle which the vector 𝑨⃗ = 𝟑 ̂ − 𝟔 ̂ + 𝟐𝒌 makes with the coordinate axes.
Solution: Let 𝛼, 𝛽, 𝛾 be the angle which the vector 𝑨⃗ = 𝟑 ̂ − 𝟔 ̂ + 𝟐𝒌 makes with the positive x, y, z axes
respectively.
Then, 𝐴⃗ • 𝚤̂ = 𝐴⃗ |𝚤̂| 𝑐𝑜𝑠 𝛼
̂ ̂ •̂
or, 𝑐𝑜𝑠 𝛼 = =
√ .√
or, 𝛼 = 64. 6
Similarly 𝛽 = 149 , 𝛾 = 73. 4 Ans.
H.W. Question-14: An Aeroplan travels 200km due west, then 150km due 600 northwest. Represent
these displacements graphically and determine the resultant displacement analytically.
Ans. 304.1Km, 25017 © northwest.
H.W. Question-15: Force of magnitude 5, 3 and 1 kg acting in the directions 𝟔 ̂ + 𝟐 ̂ + 𝟑𝒌, 𝟑 ̂ − 𝟐 ̂ + 𝟔𝒌
and 𝟐 ̂ − 𝟑 ̂ − 𝟔𝒌respectively act on a particle which is displaced from a point (2, -1, -3) to (5, -1, 1). find
the work done by the force, the unit of length being meter. Ans. 33 Joules.
H.W. Question-16: Find a vector of magnitude 9 which is perpendicular to both vectors𝟒 ̂ − ̂ + 𝟑𝒌 and
−𝟐 ̂ + ̂ − 𝟐𝒌. Ans. 3 −𝚤̂ + 2𝚥̂ + 2𝑘 .
H.W. Question-17: Find the projection of the vector 𝑨⃗ = ̂ − 𝟐 ̂ + 𝒌 on the vector 𝐵⃗ = 4𝚤̂ − 4𝚥̂ + 7𝑘
Ans: 19/9
H.W. Question-18: Find a unit vector parallel to xy plane and perpendicular to the vector 𝟒 ̂ − 𝟑 ̂ + 𝒌.
𝟏
Ans. (𝟑 ̂ + 𝟒 ̂)
𝟓
H.W. Question-19: Find a unit vector perpendicular to both vectors 𝑨⃗ = 𝟒 ̂ − ̂ + 𝟑𝒌 and
𝟏
𝑩⃗ = −𝟐 ̂ + ̂ − 𝟐𝒌. Ans. − ̂ + 𝟐 ̂ + 𝟐𝒌
𝟑
H.W. Question-20: Find the area of the triangle formed by the points whose position vectors are 𝟑 ̂ + ̂,
𝟓 ̂ + 𝟐 ̂ + 𝒌and ̂ − 𝟐 ̂ + 𝟑𝒌. Ans. √𝟐𝟗
H.W. Question-21: Find the volume of a parallelepiped whose edges are represents by 𝑨⃗ = 𝟐 ̂ − 𝟑 ̂ + 𝟒𝒌,
𝑩⃗ = ̂ + 𝟐 ̂ − 𝒌and 𝑪⃗ = 𝟑 ̂ − ̂ + 𝟐𝒌. Ans. 𝟕
Linearly dependent and independent Vectors
The Vectors 𝐴⃗, 𝐵⃗, 𝐶⃗, ⋯ ⋯ are called linearly dependent if we can find a set of scalars a, b, c, ------- not all zero,
so that 𝑎𝐴⃗ + 𝑏𝐵⃗ + 𝑐𝐶⃗ + ⋯ ⋯ = 0 otherwise they are linearly independent.
Question-22: If 𝒂, 𝒃, 𝒄are non coplanar vectors determine whether the vectors 𝑷⃗ = 𝟐𝒂 − 𝟑𝒃 + 𝒄,
𝑸⃗ = 𝟑𝒂 − 𝟓𝒃 + 𝟐𝒄 and 𝑹⃗ = 𝟒𝒂 − 𝟓𝒃 + 𝒄 are linearly dependent or independent.
Solution: Given that, 𝑃⃗ = 2𝑎 − 3𝑏 + 𝑐̂ , 𝑄⃗ = 3𝑎 − 5𝑏 + 2𝑐̂ and 𝑅⃗ = 4𝑎 − 5𝑏 + 𝑐̂
Let, 𝑥𝑃⃗ + 𝑦𝑄⃗ + 𝑧𝑅⃗ = 0 where x, y, z are scalar quantities.
or, 𝑥 2𝑎 − 3𝑏 + 𝑐̂ + 𝑦 3𝑎 − 5𝑏 + 2𝑐̂ + 𝑧 4𝑎 − 5𝑏 + 𝑐̂ = 0
or, (2𝑥 + 3𝑦 + 4𝑧)𝑎 − (3𝑥 + 5𝑦 + 5𝑧)𝑏 + (𝑥 + 2𝑦 + 𝑧)𝑐̂ = 0
Since, 𝑎, 𝑏, 𝑐̂ are non coplanar vectors.
So, 2𝑥 + 3𝑦 + 4𝑧 = 0.......................(i)
3𝑥 + 5𝑦 + 5𝑧 = 0 ..........................(ii)
𝑥 + 2𝑦 + 𝑧 = 0 ...............................(iii)
Solving Eq. (ii) and (iii) we get, = =
∴ 𝑥 = −5𝑧and 𝑥 = −5𝑧
Putting these values of x and y in Eq.(i) we get, −10𝑧 + 6𝑧 + 4𝑧 = 0
or, 0 = 0
⃗ ⃗ ⃗
Hence, the vectors 𝑃, 𝑄 , 𝑅 are linearly dependent.
Math 1201 (MHU) 8
Question-23: Determine whether the following vectors are linearly dependent or independent:
𝑨⃗ = 𝟐 ̂ + ̂ − 𝟑𝒌, 𝑩⃗ = ̂ − 𝟒𝒌, 𝑪⃗ = 𝟒 ̂ + 𝟑 ̂ − 𝒌
Solution: Given that, 𝐴⃗ = 2𝚤̂ + 𝚥̂ − 3𝑘 , 𝐵⃗ = 𝚤̂ − 4𝑘 and 𝐶⃗ = 4𝚤̂ + 3𝚥̂ − 𝑘
Let, 𝑥𝐴⃗ + 𝑦𝐵⃗ + 𝑧𝐶⃗ = 0 where x, y, z are scalar quantities.
or, 𝑥 2𝚤̂ + 𝚥̂ − 3𝑘 + 𝑦 𝚤̂ − 4𝑘 + 𝑧 4𝚤̂ + 3𝚥̂ − 𝑘 = 0
or, (2𝑥 + 𝑦 + 4𝑧)𝚤̂ + (𝑥 + 3𝑧)𝚥̂ − (3𝑥 + 4𝑦 + 𝑧)𝑘 = 0
Since, 𝚤̂, 𝚥̂, 𝑘are non coplanar vectors.
So, 2𝑥 + 𝑦 + 4𝑧 = 0....................... (i)
𝑥 + 3𝑧 = 0 .......................... (ii)
3𝑥 + 4𝑦 + 𝑧 = 0 ............................... (iii)
Solving Eq. (ii) and (iii) we get, = =
∴ 𝑥 = −3𝑧and 𝑥 = −5𝑧
Putting these values of x and y in Eq.(i) we get, −6𝑧 + 2𝑧 + 4𝑧 = 0
or, 0 = 0
Hence, the vectors 𝐴⃗, 𝐵⃗, 𝐶⃗ are linearly dependent.
H.W. Question-24: Determine whether the following vectors are linearly dependent or independent:
𝑨⃗ = ̂ − 𝟑 ̂ + 𝟐𝒌, 𝑩⃗ = 𝟐 ̂ − 𝟒 ̂ − 𝒌, 𝑪⃗ = 𝟑 ̂ + 𝟐 ̂ − 𝒌 Ans: Linearly independent.
