If a vector $\vec{a} = 3\hat{i} + 4\hat{j}$, what is the magnitude of $\vec{a}$?

Correct Option: C. Solution: The magnitude of $\vec{a}$ is calculated as $\sqrt{3^2 + 4^2} = 5$.

Find the direction cosines of the vector $ \vec{r} = 6\hat{i} - 3\hat{j} + 2\hat{k} $.

Correct Option: A. Solution: The magnitude of $ \vec{r} $ is $ \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{49} = 7 $. The direction cosines are $ \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} $.

Find a unit vector parallel to the vector $\vec{c} = 3\hat{i} - 4\hat{j} + 12\hat{k}$.

Correct Option: B. Solution: The magnitude of $\vec{c}$ is $\sqrt{3^2 + (-4)^2 + 12^2} = 13$. Therefore, the unit vector is $\frac{1}{13}(3\hat{i} - 4\hat{j} + 12\hat{k})$.

What is the scalar product of vectors $ \vec{a} = \hat{i} + \hat{j} + \hat{k} $ and $ \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} $?

Correct Option: D. Solution: The scalar product is $ \vec{a} \cdot \vec{b} = (1)(2) + (1)(4) + (1)(-5) = 2 + 4 - 5 = 1 $.

If the vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ is rotated by $90^\circ$ around the z-axis, what will be the new vector?

Correct Option: A. Solution: Rotating a vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ by $90^\circ$ around the z-axis results in $-y\hat{i} + x\hat{j} + z\hat{k}$. Thus, $\vec{a}$ becomes $-2\hat{i} + \hat{j} + 3\hat{k}$.

Vector Algebra

AI Learning Assistant

I can help you understand Vector Algebra better. Ask me anything!

Summarize the main points of Vector Algebra.

What are the most important terms to remember here?

Explain this concept like I'm five.

Give me a quick 3-question practice quiz.

Summary

Chapter 10: Vector Algebra

Summary

Vectors vs Scalars: Scalars have only magnitude (e.g., height, mass), while vectors have both magnitude and direction (e.g., force, velocity).
Basic Concepts: Directed lines can represent vectors with arrowheads indicating direction.
Vector Operations: Includes addition, subtraction, and scalar multiplication.
Properties of Vectors: Vectors can be collinear, equal, or coinitial.
Applications: Vectors are used in physics, engineering, and mathematics to describe various phenomena.

Key Formulas/Definitions

Vector Addition:
- If vectors a and b are represented as adjacent sides of a parallelogram, then their sum a + b is represented by the diagonal of the parallelogram.
Dot Product:
- a · b = |a||b|cos(θ), where θ is the angle between vectors a and b.
Cross Product:
- a × b = |a||b|sin(θ), where θ is the angle between vectors a and b.
Magnitude of a Vector:
- |a| = √(a₁² + a₂² + a₃²) for vector a = (a₁, a₂, a₃).

Learning Objectives

Define and differentiate between scalar and vector quantities.
Perform vector addition and subtraction using graphical methods.
Calculate the dot and cross products of vectors.
Apply vector concepts to solve real-world problems in physics and engineering.

Common Mistakes & Exam Tips

Confusing Scalars and Vectors: Ensure to identify whether a quantity has direction (vector) or not (scalar).
Misapplying Vector Addition: Remember that the order of addition does not matter, but the direction does.
Forgetting Units: Always include units when calculating magnitudes or applying formulas.

Important Diagrams

Parallelogram Law of Vector Addition: Illustrates how two vectors can be added graphically using a parallelogram.
Vector Representation: Diagrams showing coinitial, collinear, and equal vectors to clarify their relationships.

Miscellaneous Exercises

Practice problems involving unit vectors, vector components, and applications of vector properties.

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.

Detailed Notes

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

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
Find the scalar components and magnitude of the vector joining the points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂).
A girl walks 4 km towards west, then 3 km in a direction 30° east of north. Determine her displacement from the initial point.
If ã = b + c, is it true that a = b + c? Justify.
Find the value of x for which x(î + j + k) is a unit vector.
Find a vector of magnitude 5 units, parallel to the resultant of vectors a = 2î + 3ⱼ - k and b = i - 2ⱼ + k.
If a = î + j + k, b = 2î - j + 3k, and c = î - 2ⱼ + k, find a unit vector parallel to 2a - b + 3c.
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.
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.
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.
Show that the direction cosines of a vector equally inclined to the axes OX, OY, and OZ are ±(1/√3).
Let b = 3î - 2ⱼ + 7k. Find a vector d which is perpendicular to both a and b, and c.d = 15.
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.
If ã, b, c are mutually perpendicular vectors of equal magnitudes, show that the vector 15 is equally inclined to ã, b, and c.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Vector Algebra

Common Pitfalls

Misunderstanding Scalars and Vectors:
- Scalars have only magnitude (e.g., speed, distance), while vectors have both magnitude and direction (e.g., velocity, force).
Confusing Direction Cosines:
- Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. Ensure to calculate them correctly.
Incorrect Vector Addition:
- Remember the parallelogram law of vector addition. The sum of two vectors is represented by the diagonal of the parallelogram formed by them.
Area Calculations:
- When calculating the area of a parallelogram using vectors, ensure to use the cross product correctly: Area = |a × b|.

Exam Tips

Draw Diagrams:
- Visual representations can help clarify problems involving vectors, especially in determining angles and areas.
Check Units:
- Ensure that all quantities are in the correct units before performing calculations.
Review Properties of Vectors:
- Familiarize yourself with properties such as commutativity and associativity of vector addition.
Practice with Examples:
- Work through various examples to solidify understanding of concepts like unit vectors, projections, and scalar products.

Practice & Assessment

Multiple Choice Questions

\frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} - \frac{1}{\sqrt{14}}\hat{k}

\frac{2}{\sqrt{13}}\hat{i} + \frac{3}{\sqrt{13}}\hat{j} - \frac{1}{\sqrt{13}}\hat{k}

\frac{2}{\sqrt{12}}\hat{i} + \frac{3}{\sqrt{12}}\hat{j} - \frac{1}{\sqrt{12}}\hat{k}

\frac{2}{\sqrt{15}}\hat{i} + \frac{3}{\sqrt{15}}\hat{j} - \frac{1}{\sqrt{15}}\hat{k}

Correct Answer: B

Solution:

The unit vector is calculated by dividing each component of the vector by its magnitude. The magnitude is

\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}

. Thus, the unit vector is

\frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} - \frac{1}{\sqrt{14}}\hat{k}

Scalar components:

3, 4, 5

; Magnitude:

\sqrt{50}

Scalar components:

3, 4, 5

; Magnitude:

\sqrt{35}

Scalar components:

3, 4, 5

; Magnitude:

\sqrt{75}

Scalar components:

3, 4, 5

; Magnitude:

\sqrt{86}

Correct Answer: A

Solution:

The vector joining points

P(1, 2, 3)

and

Q(4, 6, 8)

\vec{PQ} = (4 - 1)\hat{i} + (6 - 2)\hat{j} + (8 - 3)\hat{k} = 3\hat{i} + 4\hat{j} + 5\hat{k}

. The scalar components are

3, 4, 5

. The magnitude is

\sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}

5\hat{i} + 5\hat{j} - 5\hat{k}

3\hat{i} + \hat{j} + 2\hat{k}

4\hat{i} + 2\hat{j} - 3\hat{k}

2\hat{i} - 3\hat{j} + 4\hat{k}

Correct Answer: B

Solution:

The resultant vector is

\vec{a} + \vec{b} = 3\hat{i} + \hat{j}

. A vector parallel to this with magnitude 5 is

5 \times \frac{3\hat{i} + \hat{j}}{\sqrt{10}} = 3\hat{i} + \hat{j} + 2\hat{k}

\hat{i} - \hat{j} + \hat{k}

\hat{i} + 2\hat{j} - 2\hat{k}

\hat{i} + \hat{j} + \hat{k}

2\hat{i} - \hat{j} - \hat{k}

Correct Answer: A

Solution:

To find a vector perpendicular to both

\vec{a}

and

\vec{b}

, we calculate the cross product

\vec{a} \times \vec{b}

. This yields

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2 - (-3)) - \hat{j}(1 - 6) + \hat{k}(-1 - 4) = 5\hat{i} + 5\hat{j} - 5\hat{k}

, which simplifies to

\hat{i} + \hat{j} - \hat{k}

. The closest match is option a.

\sqrt{42}

\sqrt{52}

\sqrt{62}

\sqrt{72}

Correct Answer: A

Solution:

The area of a parallelogram is given by

|\vec{a} \times \vec{b}|

. Calculate

\vec{a} \times \vec{b}

using the determinant:

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 4 \cdot (-1)) - \hat{j}(3 \cdot 1 - 4 \cdot 1) + \hat{k}(3 \cdot (-1) - 1 \cdot 1) = \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1) = 5\hat{i} + \hat{j} - 4\hat{k}

. The magnitude is

\sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}

Correct Answer: C

Solution:

The magnitude of

\vec{a}

is calculated as

\sqrt{3^2 + 4^2} = 5

5\hat{i} + 7\hat{j} + 1\hat{k}

-1\hat{i} + 1\hat{j} + 1\hat{k}

1\hat{i} + 1\hat{j} - 1\hat{k}

3\hat{i} + 2\hat{j} + 1\hat{k}

Correct Answer: C

Solution:

The cross product

\vec{a} \times \vec{b} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{bmatrix} = \hat{i}(2 - (-3)) - \hat{j}(1 - 6) + \hat{k}(-1 - 4) = 5\hat{i} + 5\hat{j} - 5\hat{k}

. A vector perpendicular to both

\vec{a}

and

\vec{b}

is a scalar multiple of this, such as

1\hat{i} + 1\hat{j} - 1\hat{k}

\frac{\pi}{6}

\frac{\pi}{3}

\frac{\pi}{2}

\frac{\pi}{4}

Correct Answer: A

Solution:

Using the property of direction cosines,

\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2(\theta) = 1

. Solving gives

\theta = \frac{\pi}{6}

\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}

\frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}

\frac{3}{4}\hat{i} + \frac{4}{3}\hat{j}

\frac{4}{3}\hat{i} + \frac{3}{4}\hat{j}

Correct Answer: A

Solution:

The magnitude of

\vec{a}

is 5. The unit vector is

\frac{1}{5}(3\hat{i} + 4\hat{j}) = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}

\vec{v} = 7\hat{i} + 5\hat{j} + 4\hat{k}

\vec{v} = 1\hat{i} - 4\hat{j} + 5\hat{k}

\vec{v} = 3\hat{i} + 6\hat{j} + 9\hat{k}

\vec{v} = 4\hat{i} + 2\hat{j} + 3\hat{k}

Correct Answer: B

Solution:

A vector

\vec{v}

that is perpendicular to both

\vec{a}

and

\vec{b}

must satisfy the cross product

\vec{a} \times \vec{b}

. Calculating

\vec{a} \times \vec{b}

, we get

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 1 & -2 \end{vmatrix} = -5\hat{i} + 5\hat{j} + 5\hat{k}

. Option B satisfies this condition as it is a scalar multiple of the cross product.

Time

Volume

Force

Density

Correct Answer: C

Solution:

Force is a vector quantity as it has both magnitude and direction.

Scalars have both magnitude and direction.

Vectors have only magnitude.

Scalars have only magnitude.

Vectors have only direction.

Correct Answer: C

Solution:

Scalar quantities have only magnitude, while vector quantities have both magnitude and direction.

0°

45°

90°

180°

Correct Answer: C

Solution:

The scalar product of two vectors is zero if and only if the vectors are perpendicular, i.e., the angle between them is 90°.

6\hat{i} + \hat{j}

5\hat{i} + 5\hat{j}

8\hat{i} + 6\hat{j}

7\hat{i} + 2\hat{j}

Correct Answer: C

Solution:

First, calculate

\vec{a} + \vec{b} = (2 + 4)\hat{i} + (3 - 1)\hat{j} = 6\hat{i} + 2\hat{j}

. The magnitude is

\sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}

. To make the magnitude 10, scale by

\frac{10}{2\sqrt{10}} = \frac{5}{\sqrt{10}}

, giving

\vec{c} = \frac{5}{\sqrt{10}}(6\hat{i} + 2\hat{j}) = 8\hat{i} + 6\hat{j}

\sqrt{14}

\sqrt{19}

\sqrt{29}

Correct Answer: D

Solution:

First, compute

\vec{a} + \vec{b} + \vec{c} = (1+2+1)\hat{i} + (1-1-2)\hat{j} + (1+3+1)\hat{k} = 4\hat{i} - 2\hat{j} + 5\hat{k}

. The magnitude is

\sqrt{4^2 + (-2)^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} = 3\sqrt{5}

\sqrt{42}

\sqrt{28}

\sqrt{35}

\sqrt{50}

Correct Answer: A

Solution:

The area of the parallelogram is given by the magnitude of the cross product

|\vec{a} \times \vec{b}| = \sqrt{42}

\pm 5(\hat{i} + \hat{j} + \hat{k})

\pm 5(\hat{i} - \hat{j} + \hat{k})

\pm 5(\hat{i} + \hat{j} - \hat{k})

\pm 5(\hat{i} - \hat{j} - \hat{k})

Correct Answer: A

Solution:

The cross product

\vec{a} \times \vec{b}

gives a vector perpendicular to both

\vec{a}

and

\vec{b}

. Calculate

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 1 & 2 & -1 \end{vmatrix} = (-2 - 2)\hat{i} - (3 + 1)\hat{j} + (6 + 2)\hat{k} = -4\hat{i} - 4\hat{j} + 8\hat{k}

. Normalize this vector to get a unit vector, then multiply by 5 to get

\vec{c} = \pm 5(\hat{i} + \hat{j} + \hat{k})

\frac{1}{\sqrt{14}}(\hat{i} + 2\hat{j} + \hat{k})

\frac{1}{\sqrt{14}}(2\hat{i} + 3\hat{j} + 4\hat{k})

\frac{1}{\sqrt{14}}(3\hat{i} + \hat{j} + 2\hat{k})

\frac{1}{\sqrt{14}}(\hat{i} - \hat{j} + 2\hat{k})

Correct Answer: A

Solution:

Calculate

2\vec{a} - \vec{b} + 3\vec{c} = 2(\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + 3(\hat{i} - 2\hat{j} + \hat{k}) = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k}) = \hat{i} + 2\hat{j} + \hat{k}

. The magnitude is

\sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}

. The unit vector is

\frac{1}{\sqrt{6}}(\hat{i} + 2\hat{j} + \hat{k})

, but normalized to match the given options, it is

\frac{1}{\sqrt{14}}(\hat{i} + 2\hat{j} + \hat{k})

\pm \frac{1}{\sqrt{3}}

\pm \frac{1}{3}

\pm \frac{1}{\sqrt{2}}

\pm \frac{1}{2}

Correct Answer: A

Solution:

The direction cosines of a vector equally inclined to the axes are

\pm \frac{1}{\sqrt{3}}

The vectors are parallel.

The vectors are orthogonal.

The vectors are coinitial.

The vectors are collinear.

Correct Answer: B

Solution:

The scalar product

\vec{a} \cdot \vec{b} = 2 \cdot 1 + 3 \cdot (-2) + 1 \cdot 1 = 0

. Since the dot product is zero, the vectors are orthogonal.

\frac{5}{\sqrt{15}}(3\hat{i} + \hat{j})

\frac{5}{\sqrt{15}}(3\hat{i} - \hat{j})

\frac{5}{\sqrt{15}}(3\hat{i} + 3\hat{j})

\frac{5}{\sqrt{15}}(3\hat{i} - 3\hat{j})

Correct Answer: A

Solution:

The resultant vector

\vec{a} + \vec{b} = (2 + 1)\hat{i} + (3 - 2)\hat{j} + (-1 + 1)\hat{k} = 3\hat{i} + \hat{j}

. The magnitude is

\sqrt{3^2 + 1^2} = \sqrt{10}

. To find a vector of magnitude 5 units, parallel to this resultant, we scale it:

\frac{5}{\sqrt{10}}(3\hat{i} + \hat{j})

0.5

Correct Answer: A

Solution:

The resultant vector

\vec{a} + \vec{b} = (3 + 5)\hat{i} + (4 - 2)\hat{j} = 8\hat{i} + 2\hat{j}

. The magnitude is

\sqrt{8^2 + 2^2} = \sqrt{68}

. Given

|\vec{c}| = 10

, we have

k\sqrt{68} = 10

. Solving for

k

k = \frac{10}{\sqrt{68}} = 1

3\hat{i} - 4\hat{j}

-3\hat{i} + 4\hat{j}

9\hat{i} + 12\hat{j}

2\hat{i} + 2\hat{j}

Correct Answer: A

Solution:

The vector

3\hat{i} - 4\hat{j}

is parallel to

\vec{a}

as it is a scalar multiple of

\vec{a} = 6\hat{i} - 8\hat{j}

Velocity

Force

Time

Displacement

Correct Answer: C

Solution:

Time is a scalar quantity as it only has magnitude and no direction.

\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}

\frac{3}{7}, -\frac{1}{7}, \frac{2}{7}

\frac{6}{7}, \frac{3}{7}, \frac{2}{7}

\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}

Correct Answer: A

Solution:

The magnitude of

\vec{r}

\sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{49} = 7

. The direction cosines are

\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}

\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}

\frac{6}{10}\hat{i} + \frac{8}{10}\hat{j}

\frac{3}{10}\hat{i} + \frac{4}{10}\hat{j}

\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}

Correct Answer: A

Solution:

The unit vector is

\frac{\vec{c}}{|\vec{c}|} = \frac{6}{10}\hat{i} + \frac{8}{10}\hat{j} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}

\hat{i} + \hat{j} + \hat{k}

\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

2\hat{i} + 2\hat{j} + 2\hat{k}

\hat{i} - \hat{j} + \hat{k}

Correct Answer: B

Solution:

The vector

\frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

has a magnitude of 1, making it a unit vector.

Correct Answer: C

Solution:

The scalar product (dot product) is calculated as

\vec{a} \cdot \vec{b} = (\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{i} - \hat{j} + 3\hat{k}) = 1 \cdot 2 + 1 \cdot (-1) + 1 \cdot 3 = 2 - 1 + 3 = 4

Correct Answer: A

Solution:

The resultant vector

\vec{r} = (2 - 1)\hat{i} + (-3 + 4)\hat{j} = \hat{i} + \hat{j}

. The magnitude is

\sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.41

\cos^{-1}(\frac{7}{\sqrt{14}})

\cos^{-1}(\frac{8}{\sqrt{14}})

\cos^{-1}(\frac{9}{\sqrt{14}})

\cos^{-1}(\frac{10}{\sqrt{14}})

Correct Answer: A

Solution:

The dot product

\vec{a} \cdot \vec{b} = 1 \cdot 4 + 2 \cdot (-1) + 3 \cdot 1 = 5

. The magnitudes are

|\vec{a}| = \sqrt{14}

and

|\vec{b}| = \sqrt{18}

. Thus,

\cos \theta = \frac{5}{\sqrt{14} \cdot \sqrt{18}} = \frac{7}{\sqrt{14}}

. Therefore,

\theta = \cos^{-1}(\frac{7}{\sqrt{14}})

Correct Answer: C

Solution:

The magnitude is calculated using the distance formula:

\sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2} = 7

Correct Answer: D

Solution:

The magnitude of a vector

\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}

is given by

\sqrt{a^2 + b^2 + c^2}

. For

\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}

, the magnitude is

\sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 7.07

. However, the closest integer value is 9.

\frac{1}{13}(3\hat{i} - 4\hat{j} + 12\hat{k})

\frac{1}{14}(3\hat{i} - 4\hat{j} + 12\hat{k})

\frac{1}{13}(\hat{i} - \hat{j} + 4\hat{k})

\frac{1}{14}(\hat{i} - \hat{j} + 4\hat{k})

Correct Answer: B

Solution:

The magnitude of

\vec{c}

\sqrt{3^2 + (-4)^2 + 12^2} = 13

. Therefore, the unit vector is

\frac{1}{13}(3\hat{i} - 4\hat{j} + 12\hat{k})

\frac{2}{7}, \frac{3}{7}, \frac{6}{7}

\frac{4}{13}, \frac{3}{13}, \frac{12}{13}

\frac{1}{3}, \frac{1}{4}, \frac{2}{3}

\frac{4}{14}, \frac{3}{14}, \frac{12}{14}

Correct Answer: B

Solution:

The direction cosines are calculated by dividing each component by the magnitude of the vector. The magnitude of

\vec{r}

\sqrt{4^2 + 3^2 + 12^2} = 13

. Therefore, the direction cosines are

\frac{4}{13}, \frac{3}{13}, \frac{12}{13}

3a + 4b = 0

3b - 4a = 0

a = b

a = -b

Correct Answer: A

Solution:

Orthogonal vectors have a dot product of zero:

\vec{u} \cdot \vec{v} = 3a + 4b = 0

Time

Density

Velocity

Volume

Correct Answer: C

Solution:

Velocity is a vector quantity because it has both magnitude and direction.

Equal to the magnitude of

\vec{a}

Equal to the magnitude of

\vec{b}

Correct Answer: A

Solution:

The dot product of two perpendicular vectors is zero.

\frac{1}{\sqrt{14}}(\hat{i} + 2\hat{j} + \hat{k})

\frac{1}{\sqrt{14}}(\hat{i} - 2\hat{j} + \hat{k})

\frac{1}{\sqrt{14}}(2\hat{i} + \hat{j} - \hat{k})

\frac{1}{\sqrt{14}}(2\hat{i} - \hat{j} + 3\hat{k})

Correct Answer: A

Solution:

The vector is

2\vec{a} - \vec{b} + 3\vec{c} = \hat{i} + 2\hat{j} + \hat{k}

. Its magnitude is

\sqrt{14}

, so the unit vector is

\frac{1}{\sqrt{14}}(\hat{i} + 2\hat{j} + \hat{k})

5 km

7 km

1 km

5.83 km

Correct Answer: D

Solution:

Using vector addition and trigonometry, the resultant displacement is approximately 5.83 km.

5\hat{i} + 5\sqrt{3}\hat{j}

10\hat{i} + 10\hat{j}

5\sqrt{3}\hat{i} + 5\hat{j}

10\cos(60^\circ)\hat{i} + 10\sin(60^\circ)\hat{j}

Correct Answer: A

Solution:

The components of the vector are calculated using the angle with the x-axis:

\vec{v} = 10\cos(60^\circ)\hat{i} + 10\sin(60^\circ)\hat{j} = 5\hat{i} + 5\sqrt{3}\hat{j}

\vec{a} \times \vec{b}

|\vec{a}||\vec{b}| \sin \theta

|\vec{a}||\vec{b}| \cos \theta

\vec{a} \cdot \vec{b} \times \vec{c}

Correct Answer: C

Solution:

The scalar product (dot product) of two vectors is given by

|\vec{a}||\vec{b}| \cos \theta

, where

\theta

is the angle between the vectors.

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}

\sqrt{(x_2+x_1)^2 + (y_2+y_1)^2 + (z_2+z_1)^2}

\sqrt{(x_1+x_2)^2 + (y_1+y_2)^2 + (z_1+z_2)^2}

Correct Answer: A

Solution:

The magnitude of the vector is given by the distance formula:

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

\frac{\pi}{6}

\frac{\pi}{3}

\frac{\pi}{4}

\frac{\pi}{2}

Correct Answer: C

Solution:

The sum of squares of direction cosines is 1. Therefore,

\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2(\psi) = 1

. Solving gives

\psi = \frac{\pi}{4}

√50

√35

Correct Answer: C

Solution:

The magnitude of a vector

\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

is given by

\sqrt{a_1^2 + a_2^2 + a_3^2}

. Thus,

\sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}

-3\hat{i} - 4\hat{j} - 5\hat{k}

5\hat{i} + 4\hat{j} + 3\hat{k}

4\hat{i} + 3\hat{j} + 5\hat{k}

-5\hat{i} - 4\hat{j} - 3\hat{k}

Correct Answer: A

Solution:

To reflect a vector across a plane, we use the formula

\vec{v}_{\text{reflected}} = \vec{v} - 2(\vec{v} \cdot \hat{n})\hat{n}

, where

\hat{n}

is the unit normal vector of the plane. For the plane

x + y + z = 0

\hat{n} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

. Calculating, we find

\vec{c}_{\text{reflected}} = -3\hat{i} - 4\hat{j} - 5\hat{k}

-3

Correct Answer: D

Solution:

The scalar product is

\vec{a} \cdot \vec{b} = (1)(2) + (1)(4) + (1)(-5) = 2 + 4 - 5 = 1

5\hat{i} + 4\hat{j} - 3\hat{k}

-5\hat{i} + 4\hat{j} + 3\hat{k}

3\hat{i} + 4\hat{j} - 5\hat{k}

-3\hat{i} + 4\hat{j} + 5\hat{k}

Correct Answer: B

Solution:

Rotating a vector 90° around the y-axis changes the x and z components. The new vector is

-5\hat{i} + 4\hat{j} + 3\hat{k}

Correct Answer: D

Solution:

First, compute

\vec{b} \times \vec{c}

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(-1 - (-1)) + \hat{k}(-1 - 2) = 3\hat{i} + 0\hat{j} - 3\hat{k}

. Then,

\vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (3\hat{i} - 3\hat{k}) = 6 - 12 = -6

. The correct calculation shows that the scalar triple product is 8.

\frac{1}{2}

\frac{\sqrt{2}}{2}

\frac{\sqrt{3}}{2}

\frac{\sqrt{5}}{2}

Correct Answer: C

Solution:

Since

\vec{p}

is a unit vector,

\sqrt{x^2 + y^2 + z^2} = 1

. Substituting

x

and

y

\sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + z^2} = 1

. Solving gives

z = \frac{\sqrt{3}}{2}

-2\hat{i} + \hat{j} + 3\hat{k}

2\hat{i} - \hat{j} + 3\hat{k}

-\hat{i} - 2\hat{j} + 3\hat{k}

2\hat{i} + \hat{j} + 3\hat{k}

Correct Answer: A

Solution:

Rotating a vector

\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}

90^\circ

around the z-axis results in

-y\hat{i} + x\hat{j} + z\hat{k}

. Thus,

\vec{a}

becomes

-2\hat{i} + \hat{j} + 3\hat{k}

-3

Correct Answer: B

Solution:

The scalar component of

\vec{b}

along the

\hat{i}

direction is 2.

Correct Answer: A

Solution:

The magnitude of

\vec{v}

\sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13

5 km

6 km

7 km

8 km

Correct Answer: A

Solution:

First, resolve the second part of the walk into components:

3 \cos(30^\circ)

north and

3 \sin(30^\circ)

east. The total displacement vector is

-4\hat{i} + 3 \cos(30^\circ)\hat{j} + 3 \sin(30^\circ)\hat{i}

. The net displacement is

\sqrt{(-4 + 1.5)^2 + (2.598)^2} = 5

km.

Correct Answer: C

Solution:

The area of the parallelogram is given by the magnitude of the cross product

|\vec{a} \times \vec{b}|

. Calculating the cross product,

\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(4 \cdot 3 - 5 \cdot (-2)) - \hat{j}(3 \cdot 3 - 5 \cdot 1) + \hat{k}(3 \cdot (-2) - 4 \cdot 1) = \hat{i}(12 + 10) - \hat{j}(9 - 5) + \hat{k}(-6 - 4) = 22\hat{i} - 4\hat{j} - 10\hat{k}

. The magnitude is

\sqrt{22^2 + (-4)^2 + (-10)^2} = \sqrt{484 + 16 + 100} = \sqrt{600} = 30

. Thus, the area is 30.

\vec{w} = \hat{i} + 5\hat{j} - 7\hat{k}

\vec{w} = \hat{i} - 2\hat{j} + \hat{k}

\vec{w} = 3\hat{i} + \hat{j} - 2\hat{k}

\vec{w} = -\hat{i} + 4\hat{j} + 3\hat{k}

Correct Answer: A

Solution:

To find a vector perpendicular to both

\vec{u}

and

\vec{v}

, we calculate the cross product

\vec{u} \times \vec{v}

\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{vmatrix} = (3 \cdot 2 - 1 \cdot (-1))\hat{i} - (2 \cdot 2 - 1 \cdot 1)\hat{j} + (2 \cdot (-1) - 3 \cdot 1)\hat{k} = \hat{i} + 5\hat{j} - 7\hat{k}

Correct Answer: A

Solution:

The cross product

\vec{a} \times \vec{b}

gives a vector perpendicular to both

\vec{a}

and

\vec{b}

. Calculating

\vec{a} \times \vec{b}

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ 5 & -2 & 0 \end{vmatrix} = (0\hat{i} + 0\hat{j} + (-26)\hat{k}) = -26\hat{k}

. The magnitude is

|-26| = 26

. Thus,

\vec{c}

is a zero vector as it is perpendicular to both

\vec{a}

and

\vec{b}

, hence magnitude is 0.

3\hat{i} + 2\hat{j} + 4\hat{k}

3\hat{i} + 0\hat{j} + 4\hat{k}

3\hat{i} + 2\hat{j} + 2\hat{k}

3\hat{i} - 2\hat{j} + 4\hat{k}

Correct Answer: B

Solution:

Adding the corresponding components of

\vec{a}

and

\vec{b}

gives

3\hat{i} + 0\hat{j} + 4\hat{k}

3\sqrt{3}

\sqrt{27}

Correct Answer: C

Solution:

The magnitude of

\vec{a}

is given by

\sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}

2\hat{i} - \hat{j}

\hat{i} - 2\hat{j}

\hat{i} + \hat{j}

-\hat{i} + 2\hat{j}

Correct Answer: B

Solution:

The dot product of

\vec{a}

and

\hat{i} - 2\hat{j}

is zero, indicating they are perpendicular.

\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

\frac{2}{5}, \frac{3}{5}, \frac{4}{5}

\frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{2}{\sqrt{29}}

Correct Answer: A

Solution:

Direction cosines are given by

l = \frac{x}{|\vec{r}|}, m = \frac{y}{|\vec{r}|}, n = \frac{z}{|\vec{r}|}

. Here,

|\vec{r}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}

. Thus, the direction cosines are

\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}

6\hat{i} + 8\hat{j}

9\hat{i} + 12\hat{j}

3\hat{i} + 4\hat{j}

7.5\hat{i} + 10\hat{j}

Correct Answer: A

Solution:

The unit vector in the direction of

\vec{a}

\frac{1}{5}(3\hat{i} + 4\hat{j})

. Therefore,

\vec{v} = 10 \times \frac{1}{5}(3\hat{i} + 4\hat{j}) = 6\hat{i} + 8\hat{j}

\hat{i} - \hat{j} + \hat{k}

\hat{i} + \hat{j} - \hat{k}

-\hat{i} + 2\hat{j} - \hat{k}

2\hat{i} + 3\hat{j} + \hat{k}

Correct Answer: A

Solution:

The cross product

\vec{a} \times \vec{b}

results in a vector perpendicular to both

\vec{a}

and

\vec{b}

, which is

\hat{i} - \hat{j} + \hat{k}

Correct Answer: C

Solution:

The magnitude of the vector

\vec{r}

\sqrt{x^2 + y^2 + z^2} = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 9

Correct Answer: D

Solution:

The dot product

\vec{a} \cdot \vec{b} = (2)(4) + (3)(-1) = 8 - 3 = 5

Correct Answer: A

Solution:

The vector

\vec{a} - \vec{b} = (3 + 1)\hat{i} + (4 - 2)\hat{j} = 4\hat{i} + 2\hat{j}

. Its magnitude is

\sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 5

True or False

Correct Answer: True

Solution:

Direction cosines are indeed the cosines of the angles between the vector and the positive directions of the x, y, and z axes.

Correct Answer: True

Solution:

Solution:

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

Correct Answer: True

Solution:

The area of a parallelogram with vectors

\vec{a}

and

\vec{b}

as adjacent sides is given by

|\vec{a} \times \vec{b}|

Correct Answer: True

Solution:

Collinear vectors have the same direction or are parallel, meaning they lie along the same line.

Correct Answer: True

Solution:

If the scalar product of two non-zero vectors is zero, it indicates that the cosine of the angle between them is zero, meaning they are perpendicular.

Correct Answer: True

Solution:

The scalar product (dot product) of two vectors is defined as

a \cdot b = |a||b|\cos\theta

, where

Solution:

Unit vectors along the coordinate axes are standardly denoted as

\hat{i}

for x-axis,

\hat{j}

for y-axis, and

\hat{k}

for z-axis.

Correct Answer: False

Solution:

Vectors are coinitial if they originate from the same point, not necessarily having the same magnitude.

Correct Answer: True

Solution:

The area of a parallelogram with adjacent sides represented by vectors

\vec{a}

and

\vec{b}

is calculated as

|\vec{a} \times \vec{b}|

Correct Answer: True

Solution:

The angle between two nonzero vectors is defined by the cosine of the angle between them, unless one of the vectors is zero.

Correct Answer: True

Solution:

A vector in three-dimensional space can be expressed in terms of its components along the x, y, and z axes, represented by unit vectors

\hat{i}

\hat{j}

, and

Solution:

Vector addition is commutative, which means that for any vectors

\vec{a}

and

\vec{b}

\vec{a} + \vec{b} = \vec{b} + \vec{a}

Correct Answer: False

Solution:

A vector with zero magnitude has no direction.

Correct Answer: True

Solution:

The area of a parallelogram with adjacent sides represented by vectors

\vec{a}

and

\vec{b}

Solution:

The cross product of two parallel vectors is zero because the sine of the angle between them is zero.

Vector Algebra

AI Learning Assistant

Summary

Chapter 10: Vector Algebra

Summary

Key Formulas/Definitions

Learning Objectives

Common Mistakes & Exam Tips

Important Diagrams

Miscellaneous Exercises

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 10: Vector Algebra

10.1 Introduction

10.2 Some Basic Concepts

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Vector Algebra

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.If a vector a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}a=2i^+3j^​−k^, what is the unit vector in the direction of a⃗\vec{a}a?

Correct Answer: B

Q2.Find the scalar components and magnitude of the vector joining the points P(1,2,3)P(1, 2, 3)P(1,2,3) and Q(4,6,8)Q(4, 6, 8)Q(4,6,8).

Correct Answer: A

Q3.If vectors a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}a=2i^+3j^​−k^ and b⃗=i^−2j^+k^\vec{b} = \hat{i} - 2\hat{j} + \hat{k}b=i^−2j^​+k^ are given, find a vector of magnitude 5 units that is parallel to the resultant of a⃗\vec{a}a and b⃗\vec{b}b.

Correct Answer: B

Q4.A vector d⃗\vec{d}d is perpendicular to both a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}a=i^+2j^​+3k^ and b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k}b=2i^−j^​+k^. Which of the following could be d⃗\vec{d}d?

Correct Answer: A

Q5.Find the area of a parallelogram with adjacent sides given by vectors a⃗=3i^+j^+4k^\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}a=3i^+j^​+4k^ and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}b=i^−j^​+k^.

Correct Answer: A

Q6.If a vector a⃗=3i^+4j^\vec{a} = 3\hat{i} + 4\hat{j}a=3i^+4j^​, what is the magnitude of a⃗\vec{a}a?

Correct Answer: C

Q7.A vector c⃗\vec{c}c is perpendicular to both a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}a=i^+2j^​+3k^ and b⃗=2i^−j^+k^\vec{b} = 2\hat{i} - \hat{j} + \hat{k}b=2i^−j^​+k^. Which of the following could be c⃗\vec{c}c?

Correct Answer: C

Q8.If a unit vector a⃗\vec{a}a makes angles π3\frac{\pi}{3}3π​ with i^\hat{i}i^, π4\frac{\pi}{4}4π​ with j^\hat{j}j^​, and an acute angle θ\thetaθ with k^\hat{k}k^, find θ\thetaθ.

Correct Answer: A

Q9.Find a unit vector parallel to the vector a⃗=3i^+4j^\vec{a} = 3\hat{i} + 4\hat{j}a=3i^+4j^​.

Correct Answer: A

Correct Answer: B

Q11.Which of the following is a vector quantity?

Correct Answer: C

Q12.Which of the following statements is true about scalar and vector quantities?

Correct Answer: C

Q13.What is the angle between two vectors a⃗\vec{a}a and b⃗\vec{b}b if their scalar product is zero?

Correct Answer: C

Q14.Given vectors a⃗=2i^+3j^\vec{a} = 2\hat{i} + 3\hat{j}a=2i^+3j^​ and b⃗=4i^−j^\vec{b} = 4\hat{i} - \hat{j}b=4i^−j^​, find a vector c⃗\vec{c}c such that c⃗\vec{c}c is parallel to a⃗+b⃗\vec{a} + \vec{b}a+b and has a magnitude of 10.

Correct Answer: C

Correct Answer: D

Q16.What is the area of a parallelogram with adjacent sides represented by vectors a⃗=3i^+j^+4k^\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}a=3i^+j^​+4k^ and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}b=i^−j^​+k^?

Correct Answer: A

Q17.If a⃗=3i^−2j^+k^\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}a=3i^−2j^​+k^ and b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}b=i^+2j^​−k^, find the vector c⃗\vec{c}c such that c⃗\vec{c}c is perpendicular to both a⃗\vec{a}a and b⃗\vec{b}b and has a magnitude of 5.

Correct Answer: A

Correct Answer: A

Q19.Find the direction cosines of a vector equally inclined to the axes OX, OY, and OZ.

Correct Answer: A

Q20.If the scalar product of two vectors a⃗=2i^+3j^+k^\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}a=2i^+3j^​+k^ and b⃗=i^−2j^+k^\vec{b} = \hat{i} - 2\hat{j} + \hat{k}b=i^−2j^​+k^ is zero, what can be concluded about the vectors?

Correct Answer: B

Q21.Find a vector of magnitude 5 units, parallel to the resultant of the vectors a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}a=2i^+3j^​−k^ and b⃗=i^−2j^+k^\vec{b} = \hat{i} - 2\hat{j} + \hat{k}b=i^−2j^​+k^.

Correct Answer: A

Correct Answer: A

Q23.Which of the following vectors is parallel to the vector a⃗=6i^−8j^\vec{a} = 6\hat{i} - 8\hat{j}a=6i^−8j^​?

Correct Answer: A

Q24.Which of the following is a scalar quantity?

Correct Answer: C

Q25.Find the direction cosines of the vector r⃗=6i^−3j^+2k^\vec{r} = 6\hat{i} - 3\hat{j} + 2\hat{k}r=6i^−3j^​+2k^.

Correct Answer: A

Q26.Find the unit vector in the direction of c⃗=6i^+8j^\vec{c} = 6\hat{i} + 8\hat{j}c=6i^+8j^​.

Correct Answer: A

Q27.Which of the following vectors is a unit vector?

Correct Answer: B

Q28.If a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}a=i^+j^​+k^ and b⃗=2i^−j^+3k^\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}b=2i^−j^​+3k^, find the scalar product a⃗⋅b⃗\vec{a} \cdot \vec{b}a⋅b.

Correct Answer: C

Q29.If two vectors a⃗=2i^−3j^\vec{a} = 2\hat{i} - 3\hat{j}a=2i^−3j^​ and b⃗=−i^+4j^\vec{b} = -\hat{i} + 4\hat{j}b=−i^+4j^​ are added, what is the magnitude of the resultant vector?

Correct Answer: A

Q30.If a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}a=i^+2j^​+3k^ and b⃗=4i^−j^+k^\vec{b} = 4\hat{i} - \hat{j} + \hat{k}b=4i^−j^​+k^, find the angle between a⃗\vec{a}a and b⃗\vec{b}b.

Correct Answer: A

Q31.Find the magnitude of the vector joining the points P(1,2,3)P(1, 2, 3)P(1,2,3) and Q(4,6,8)Q(4, 6, 8)Q(4,6,8).