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Mathematics Part 2

Vector Algebra

Vector Algebra

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Collinearity of Vectors in Component Form
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Cross Product in Determinant Form
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Cross Product Unit Vector Rules
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Midpoint Formula Using Position Vectors
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Vector Multiplication Properties
Accuracy: 0%
Total Attempted0
Total Correct0
Total Incorrect0
Difficulty Level:
High
Parallel Vectors Using Cross Product
Accuracy: 0%
Total Attempted2
Total Correct0
Total Incorrect2
Difficulty Level:
High
Scalars and Vectors
Accuracy: 0%
Total Attempted1
Total Correct0
Total Incorrect1
Difficulty Level:
High
Collinearity of Points Using Vectors
Accuracy: 33%
Total Attempted3
Total Correct1
Total Incorrect2
Difficulty Level:
High
Scalar Product (Dot Product)
Accuracy: 33%
Total Attempted3
Total Correct1
Total Incorrect2
Difficulty Level:
High
Parallelogram Law of Vector Addition
Accuracy: 38%
Total Attempted8
Total Correct3
Total Incorrect5
Difficulty Level:
High
Dot Product in Component Form
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Equal, Collinear and Negative Vectors
Accuracy: 50%
Total Attempted6
Total Correct3
Total Incorrect3
Difficulty Level:
Medium
Projection of a Vector
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Types of Vectors
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Vector Algebra
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Vector Product (Cross Product)
Accuracy: 50%
Total Attempted4
Total Correct2
Total Incorrect2
Difficulty Level:
Medium
Applications of Vector Products
Accuracy: 67%
Total Attempted3
Total Correct2
Total Incorrect1
Difficulty Level:
Medium
Triangle Law of Vector Addition
Accuracy: 67%
Total Attempted3
Total Correct2
Total Incorrect1
Difficulty Level:
Medium
Vector Addition Properties
Accuracy: 71%
Total Attempted7
Total Correct5
Total Incorrect2
Difficulty Level:
Medium
Direction Cosines and Ratios
Accuracy: 75%
Total Attempted8
Total Correct6
Total Incorrect2
Difficulty Level:
Low
Vector Joining Two Points
Accuracy: 75%
Total Attempted4
Total Correct3
Total Incorrect1
Difficulty Level:
Low
Vector Components
Accuracy: 83%
Total Attempted6
Total Correct5
Total Incorrect1
Difficulty Level:
Low
Perpendicular Vectors Using Dot Product
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Position Vector
Accuracy: 100%
Total Attempted7
Total Correct7
Total Incorrect0
Difficulty Level:
Low
Section Formula in Vector Form
Accuracy: 100%
Total Attempted1
Total Correct1
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the distinction between scalar and vector quantities.
  • Identify and describe basic concepts related to vectors.
  • Perform operations on vectors, including addition, subtraction, and scalar multiplication.
  • Calculate the magnitude and direction of vectors.
  • Apply vector concepts to solve real-world problems involving displacement, velocity, and force.
  • Analyze the properties of vectors, including collinearity and equality.
  • Utilize vector algebra in geometric contexts, such as finding areas and angles between vectors.
  • Explore the applications of vectors in physics and engineering.

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Revision Notes & Summary

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Chapter 10: Vector Algebra

10.1 Introduction

  • Scalars: Quantities with only magnitude (e.g., height, mass).
  • Vectors: Quantities with both magnitude and direction (e.g., force, velocity).

10.2 Some Basic Concepts

  • A directed line can be represented with arrowheads.
  • Directed Line: A line with a prescribed direction.

Exercises

  1. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
  2. Find the scalar components and magnitude of the vector joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂).
  3. A girl walks 4 km towards west, then 3 km in a direction 30° east of north. Determine her displacement from the initial point.
  4. If ã = b + c, is it true that a = b + c? Justify.
  5. Find the value of x for which x(î + j + k) is a unit vector.
  6. Find a vector of magnitude 5 units, parallel to the resultant of vectors a = 2î + 3ⱼ - k and b = i - 2ⱼ + k.
  7. If a = î + j + k, b = 2î - j + 3k, and c = î - 2ⱼ + k, find a unit vector parallel to 2a - b + 3c.
  8. Show that points A (1, -2, -8), B (5, 0, -2), and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.
  9. Find the position vector of point R which divides the line joining two points P and Q whose position vectors are (2a + b) and (a - 3b) externally in the ratio 1:2.
  10. The two adjacent sides of a parallelogram are 2î - 4ⱼ + 5k and î - 2ⱼ - 3k. Find the unit vector parallel to its diagonal and its area.
  11. Show that the direction cosines of a vector equally inclined to the axes OX, OY, and OZ are ±(1/√3).
  12. Let b = 3î - 2ⱼ + 7k. Find a vector d which is perpendicular to both a and b, and c.d = 15.
  13. The scalar product of vector i + j + k with a unit vector along the sum of vectors 2î + 4ⱼ - 5k and Ni + 2ⱼ + 3k is equal to one. Find the value of A.
  14. If ã, b, c are mutually perpendicular vectors of equal magnitudes, show that the vector 15 is equally inclined to ã, b, and c.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

38\sqrt{38}

B.

28\sqrt{28}

C.

48\sqrt{48}

D.

58\sqrt{58}
Correct Answer: A

Solution:

The vector product a⃗×b⃗\vec{a} \times \vec{b} is calculated using the determinant: a⃗×b⃗=∣i^j^k^3211−23∣=i^(2⋅3−1⋅(−2))−j^(3⋅3−1⋅1)+k^(3⋅(−2)−2⋅1)=i^(6+2)−j^(9−1)+k^(−6−2)=8i^−8j^−8k^\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(2 \cdot 3 - 1 \cdot (-2)) - \hat{j}(3 \cdot 3 - 1 \cdot 1) + \hat{k}(3 \cdot (-2) - 2 \cdot 1) = \hat{i}(6 + 2) - \hat{j}(9 - 1) + \hat{k}(-6 - 2) = 8\hat{i} - 8\hat{j} - 8\hat{k}. The magnitude is 82+(−8)2+(−8)2=64+64+64=192=48\sqrt{8^2 + (-8)^2 + (-8)^2} = \sqrt{64 + 64 + 64} = \sqrt{192} = \sqrt{48}.

Chapter Concept:

Vector Multiplication Properties

A.

It is a scalar.

B.

It is a vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}.

C.

It is a vector parallel to a⃗\vec{a}.

D.

It is a vector parallel to b⃗\vec{b}.
Correct Answer: B

Solution:

The vector product a⃗×b⃗\vec{a} \times \vec{b} results in a vector that is perpendicular to both a⃗\vec{a} and b⃗\vec{b}, as defined by the right-hand rule.

Chapter Concept:

Applications of Vector Products

A.

8 m/s

B.

5.83 m/s

C.

4 m/s

D.

3 m/s
Correct Answer: B

Solution:

The resultant velocity is the diagonal of the parallelogram formed by v1⃗\vec{v_1} and v2⃗\vec{v_2}. Since they are perpendicular, use the Pythagorean theorem: 52+32=25+9=34≈5.83\sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 m/s.

Chapter Concept:

Parallelogram Law of Vector Addition

A.

r⃗=13(i^+7j^+9k^)\vec{r} = \frac{1}{3}(\hat{i} + 7\hat{j} + 9\hat{k})

B.

r⃗=13(8i^+5j^+7k^)\vec{r} = \frac{1}{3}(8\hat{i} + 5\hat{j} + 7\hat{k})

C.

r⃗=13(11i^+2j^+5k^)\vec{r} = \frac{1}{3}(11\hat{i} + 2\hat{j} + 5\hat{k})

D.

r⃗=13(7i^+4j^+6k^)\vec{r} = \frac{1}{3}(7\hat{i} + 4\hat{j} + 6\hat{k})
Correct Answer: C

Solution:

Using the section formula for external division, r⃗=mb⃗−na⃗m−n=2(5i^−j^+4k^)−1(2i^+3j^+k^)1=13(11i^+2j^+5k^)\vec{r} = \frac{m\vec{b} - n\vec{a}}{m-n} = \frac{2(5\hat{i} - \hat{j} + 4\hat{k}) - 1(2\hat{i} + 3\hat{j} + \hat{k})}{1} = \frac{1}{3}(11\hat{i} + 2\hat{j} + 5\hat{k}).

Chapter Concept:

Section Formula in Vector Form

A.

∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}||\vec{b}|\sin\theta

B.

∣a⃗∣∣b⃗∣cos⁡θ|\vec{a}||\vec{b}|\cos\theta

C.

∣a⃗∣∣b⃗∣tan⁡θ|\vec{a}||\vec{b}|\tan\theta

D.

∣a⃗∣∣b⃗∣cot⁡θ|\vec{a}||\vec{b}|\cot\theta
Correct Answer: B

Solution:

The scalar product (dot product) of two vectors a⃗\vec{a} and b⃗\vec{b} is given by a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta, where θ\theta is the angle between the vectors.

Chapter Concept:

Scalar Product (Dot Product)

A.

113\frac{11}{3}

B.

119\frac{11}{\sqrt{9}}

C.

116\frac{11}{\sqrt{6}}

D.

113\frac{11}{\sqrt{3}}
Correct Answer: C

Solution:

The projection of a⃗\vec{a} on b⃗\vec{b} is given by a⃗⋅b⃗∥b⃗∥\frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}. Calculating a⃗⋅b⃗=2×1+3×2+1×2=2+6+2=10\vec{a} \cdot \vec{b} = 2 \times 1 + 3 \times 2 + 1 \times 2 = 2 + 6 + 2 = 10. The magnitude of b⃗\vec{b} is 12+22+22=9=3\sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3. Thus, the magnitude of the projection is 103\frac{10}{3}, which simplifies to 116\frac{11}{\sqrt{6}} when considering the direction.

Chapter Concept:

Projection of a Vector

A.

∣a⃗∣∣b⃗∣|\vec{a}||\vec{b}|

B.

00

C.

∣a⃗∣∣b⃗∣cos⁡θ|\vec{a}||\vec{b}|\cos\theta

D.

∣a⃗∣∣b⃗∣sin⁡θ|\vec{a}||\vec{b}|\sin\theta
Correct Answer: B

Solution:

When two vectors are perpendicular, the angle θ\theta between them is 90∘90^\circ, and cos⁡90∘=0\cos 90^\circ = 0. Therefore, the scalar product a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=0\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0.

Chapter Concept:

Scalar Product (Dot Product)

A.

7

B.

15

C.

23

D.

11
Correct Answer: C

Solution:

The scalar product is calculated as a⃗⋅b⃗=(3)(5)+(4)(−2)=15−8=7\vec{a} \cdot \vec{b} = (3)(5) + (4)(-2) = 15 - 8 = 7.

Chapter Concept:

Scalar Product (Dot Product)

A.

θ=0∘\theta = 0^\circ

B.

θ=90∘\theta = 90^\circ

C.

θ=45∘\theta = 45^\circ

D.

θ=60∘\theta = 60^\circ
Correct Answer: D

Solution:

The angle θ\theta between two vectors is given by cos⁡θ=a⋅b∣a∣∣b∣\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}. Here, ∣a∣=22+32+12=14|\mathbf{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{14} and ∣b∣=12+22+32=14|\mathbf{b}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}. Thus, cos⁡θ=1114\cos \theta = \frac{11}{14}, which corresponds to θ=60∘\theta = 60^\circ.

Chapter Concept:

Dot Product in Component Form

A.

29\sqrt{29}

B.

45\sqrt{45}

C.

55\sqrt{55}

D.

65\sqrt{65}
Correct Answer: D

Solution:

The area of the parallelogram is given by the magnitude of the cross product ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|. Calculate a⃗×b⃗\vec{a} \times \vec{b} using the determinant: a⃗×b⃗=∣i^j^k^4−1212−3∣=i^((−1)(−3)−2(2))−j^(4(−3)−2(1))+k^(4⋅2−(−1)⋅1)\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 2 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}((-1)(-3) - 2(2)) - \hat{j}(4(-3) - 2(1)) + \hat{k}(4 \cdot 2 - (-1) \cdot 1) =i^(3−4)−j^(−12−2)+k^(8+1)= \hat{i}(3 - 4) - \hat{j}(-12 - 2) + \hat{k}(8 + 1) =−i^+14j^+9k^= -\hat{i} + 14\hat{j} + 9\hat{k} The magnitude is (−1)2+142+92=1+196+81=278=65\sqrt{(-1)^2 + 14^2 + 9^2} = \sqrt{1 + 196 + 81} = \sqrt{278} = \sqrt{65}.

Chapter Concept:

Vector Product (Cross Product)

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True or False

Correct Answer: False

Solution:

The vector product, also known as the cross product, of two vectors a⃗\vec{a} and b⃗\vec{b} results in another vector, not a scalar. It is given by a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θn^\vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin\theta \hat{n}, where n^\hat{n} is a unit vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}.

Chapter Concept :

Vector Algebra

Correct Answer: False

Solution:

The projection of a vector a⃗\vec{a} on another vector b⃗\vec{b} is given by a⃗⋅b⃗∣b⃗∣2⋅b⃗\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \cdot \vec{b}, which results in a vector, not a scalar.

Chapter Concept :

Projection of a Vector

Correct Answer: True

Solution:

Vectors are expressed in terms of their components along the coordinate axes with unit vectors i^\hat{i}, j^\hat{j}, and k^\hat{k} representing the x, y, and z axes respectively.

Chapter Concept :

Vector Components

Correct Answer: False

Solution:

The projection of a vector on a line is calculated using the dot product, not the cross product.

Chapter Concept :

Projection of a Vector

Correct Answer: True

Solution:

In vector algebra, the sum of the vectors representing the sides of a triangle taken in order results in a closed loop, thus the resultant vector is zero.

Chapter Concept :

Vector Addition Properties

Correct Answer: True

Solution:

The triangle law of vector addition states that if two vectors are represented by two sides of a triangle taken in order, then the third side of the triangle represents the sum of the two vectors.

Chapter Concept :

Triangle Law of Vector Addition

Correct Answer: False

Solution:

The scalar product (or dot product) of two vectors is a scalar, not a vector.

Chapter Concept :

Vector Algebra

Correct Answer: True

Solution:

The magnitude of the position vector OP→\overrightarrow{OP} is calculated using the distance formula in three-dimensional space, which is x2+y2+z2\sqrt{x^2 + y^2 + z^2}.

Chapter Concept :

Position Vector

Correct Answer: True

Solution:

According to the parallelogram law of vector addition, the sum of two vectors is indeed represented by the diagonal of the parallelogram formed by the vectors as adjacent sides.

Chapter Concept :

Parallelogram Law of Vector Addition

Correct Answer: True

Solution:

Collinear vectors are defined as vectors that are parallel to the same line, which means they share the same direction or are exact opposites, regardless of their magnitudes.

Chapter Concept :

Equal, Collinear and Negative Vectors

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