Vector Algebra — Class 12 Maths Notes

Topics covered in Vector Algebra

1. 1.Introduction of Vector Algebra

   Introduction to Vector Algebra

   Vector algebra is a fascinating area of mathematics that allows us to solve problems involving direction and magnitude. Imagine you're playing a video game where you need to move an object from one point to another. Vectors would tell you not only how far to move the object but also in which direction. This concept is fundamental in physics, engineering, and even in everyday tasks like navigation.

   What is a Vector?

   A vector is a quantity that has both magnitude (size) and direction. Unlike regular numbers (scalars) that have only magnitude, vectors help us understand movements and forces in space. For example, when you push a shopping cart, you apply a force in a specific direction; this force can be represented by a vector.

   Components of a Vector

   Vectors are usually represented in two ways:

   1. Graphically, as arrows pointing in a specific direction, where the length of the arrow shows the magnitude.
   2. Algebraically, using coordinates. For instance, in a 2D space, a vector can be described by 𝑣=(𝑥,𝑦)v=(x,y), where 𝑥x and 𝑦y represent how far you move in each direction.

   Real-World Example

   Consider navigating a boat in a river. The direction and speed (velocity) of the boat can be described using a vector. This helps in setting the correct course to effectively counteract currents and winds.

   Application in Careers

   Vector algebra is crucial in various fields:

   • Engineering: Used in designing structures and understanding forces.
   • Computer Graphics: Helps in simulating real-world scenarios for games and animations.
   • Physics: Essential for describing the laws of motion and forces.
   • Robotics: Used in programming robots to move and operate in their environments.

   Vectors provide a powerful way to handle and manipulate these physical quantities in mathematical form, making them indispensable in both academics and numerous professional fields.

2. 2.Some Basic Concepts of Vector Algebra

   Basic Concepts of Vector Algebra

   Vector algebra is integral to understanding and manipulating quantities that have both magnitude and direction. Here are some foundational concepts and their definitions that will help you get a firm grasp on vector algebra:

   1. Vector

   A vector is a mathematical object that has both a magnitude (length) and a direction. It is often represented by an arrow, where the direction of the arrow indicates the direction of the vector and the length of the arrow represents its magnitude.

   2. Scalar

   A scalar is a quantity that has only magnitude but no direction. Common examples of scalars are numbers like 5, -3, or 0.5. In vector algebra, scalars are used to scale vectors, affecting their magnitude but not their direction.

   3. Magnitude (or Modulus) of a Vector

   The magnitude of a vector is a measure of its length. For a vector 𝑣=(𝑥,𝑦)v=(x,y) in two-dimensional space, the magnitude is calculated using the formula ∣𝑣∣=𝑥2+𝑦2∣v∣=x2+y2​. This is derived from the Pythagorean theorem.

   4. Direction of a Vector

   The direction of a vector describes the angle it makes with a reference axis, typically the horizontal axis. It is important for determining where the vector points in a plane or space.

   5. Unit Vector

   A unit vector is a vector that has a magnitude of 1. It is often used to specify a direction without specifying a magnitude. A vector can be converted into a unit vector by dividing it by its magnitude.

   6. Zero Vector

   The zero vector, denoted as 00, has zero magnitude and no specific direction. It is the only vector that does not have a well-defined direction.

   7. Vector Addition

   Vectors can be added together to produce a resultant vector. This operation is performed by adding the corresponding components of the vectors. If 𝑢=(𝑎,𝑏)u=(a,b) and 𝑣=(𝑥,𝑦)v=(x,y), then 𝑢+𝑣=(𝑎+𝑥,𝑏+𝑦)u+v=(a+x,b+y).

   8. Scalar Multiplication

   Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative, which also reverses the direction). For a vector 𝑣=(𝑥,𝑦)v=(x,y) and a scalar 𝑘k, the product is 𝑘𝑣=(𝑘𝑥,𝑘𝑦)kv=(kx,ky).

   9. Dot Product (Scalar Product)

   The dot product of two vectors is a scalar and is calculated by multiplying their corresponding components and then summing the products. For vectors 𝑢=(𝑎,𝑏)u=(a,b) and 𝑣=(𝑥,𝑦)v=(x,y), the dot product is 𝑢⋅𝑣=𝑎𝑥+𝑏𝑦u⋅v=ax+by.

   10. Cross Product (Vector Product)

   The cross product of two vectors results in a vector that is perpendicular to both of the original vectors (only defined in three dimensions). If 𝑢=(𝑎,𝑏,𝑐)u=(a,b,c) and 𝑣=(𝑥,𝑦,𝑧)v=(x,y,z), then 𝑢×𝑣=(𝑏𝑧−𝑐𝑦,𝑐𝑥−𝑎𝑧,𝑎𝑦−𝑏𝑥)u×v=(bz−cy,cx−az,ay−bx).

   These concepts form the backbone of vector algebra and are essential for further studies in fields such as physics, engineering, and computer graphics.

3. 3.Types of Vectors

   Types of Vectors

   Vectors come in various forms, each having specific characteristics and applications. Understanding these different types is crucial in fields ranging from physics to engineering and computer graphics. Here are some of the most common types of vectors:

   1. Zero Vector

   • Description: A zero vector has zero magnitude and no particular direction. It is denoted as 00.
   • Application: It is used in physics to represent a state of no movement or force.

   2. Unit Vector

   • Description: A unit vector has a magnitude of exactly one. It often represents the direction of a vector without regard to its magnitude.
   • Application: Unit vectors are crucial in defining directions in space, especially in 3D modeling and physics.

   3. Position Vector

   • Description: A position vector points from the origin of a coordinate system to the position of a point in space.
   • Application: In physics and engineering, position vectors are used to describe the position of points in space, aiding in the analysis of movements and forces.

   4. Free Vector

   • Description: A free vector can be moved anywhere in space without its effect being altered. It is not fixed in position and only defined by its direction and magnitude.
   • Application: Used widely in physics, free vectors are essential for understanding forces that are not dependent on the point of application, like gravitational force.

   5. Sliding Vector

   • Description: A sliding vector can slide along its line of action without its impact on the system changing. Its exact point of application is not fixed.
   • Application: Useful in mechanical engineering for analyzing forces in structures, like beams and frames.

   6. Fixed Vector

   • Description: A fixed vector has a specific point of application that cannot be changed. Its position is fixed in space.
   • Application: In robotics and mechanical systems, fixed vectors are used to describe forces and movements that occur at specific points.

   7. Collinear Vectors

   • Description: Collinear vectors are vectors that lie along the same line, whether they point in the same or opposite directions.
   • Application: These vectors are used in mechanics to solve problems involving parallel forces.

   8. Coplanar Vectors

   • Description: Coplanar vectors lie in the same plane. They can be parallel, anti-parallel, or at any angle to each other within the plane.
   • Application: In engineering, understanding coplanar vectors helps analyze forces and motions within a single plane, crucial for structural analysis.

   9. Orthogonal Vectors

   • Description: Orthogonal vectors are perpendicular to each other. The dot product of orthogonal vectors is zero.
   • Application: Used in computer graphics and physics, orthogonal vectors are fundamental in defining coordinate systems and calculating forces.

   10. Parallel and Anti-parallel Vectors

   • Description: Parallel vectors have the same direction, while anti-parallel vectors have opposite directions but lie along the same line.
   • Application: These vectors are important in fields such as fluid dynamics and electromagnetism, where alignment of forces and fields plays a critical role.

   Understanding these types of vectors enriches your ability to solve complex problems in various scientific and engineering domains.

4. 4.Addition of Vectors

   Addition of Vectors:

   Vector addition is a critical operation in physics, engineering, and various other fields where quantities have both magnitude and direction. Vectors can be added graphically or algebraically.

   Graphical Addition of Vectors

   Tip-to-Tail Method: This method involves placing the tail of the second vector at the tip of the first vector. The resultant vector, which is the sum of the vectors, is drawn from the tail of the first vector to the tip of the last vector.

   Algebraic Addition of Vectors

   In algebraic terms, vectors are added component-wise:

   • If 𝐴=(𝑎1,𝑎2)A=(a1​,a2​) and 𝐵=(𝑏1,𝑏2)B=(b1​,b2​), then 𝐴+𝐵=(𝑎1+𝑏1,𝑎2+𝑏2)A+B=(a1​+b1​,a2​+b2​).

   Properties of Vector Addition

   1. Commutative Property: 𝐴+𝐵=𝐵+𝐴A+B=B+A

      • This property states that the order in which vectors are added does not change the resultant vector.

   2. Associative Property: (𝐴+𝐵)+𝐶=𝐴+(𝐵+𝐶)(A+B)+C=A+(B+C)

      • This implies that when adding three or more vectors, the way in which pairs of vectors are grouped does not affect the resultant vector.

   Proof of Commutative Property

   • Let 𝐴=(𝑎1,𝑎2)A=(a1​,a2​) and 𝐵=(𝑏1,𝑏2)B=(b1​,b2​).
   • Then 𝐴+𝐵=(𝑎1+𝑏1,𝑎2+𝑏2)A+B=(a1​+b1​,a2​+b2​).
   • Similarly, 𝐵+𝐴=(𝑏1+𝑎1,𝑏2+𝑎2)B+A=(b1​+a1​,b2​+a2​).
   • Since addition of real numbers is commutative, 𝑎1+𝑏1=𝑏1+𝑎1a1​+b1​=b1​+a1​ and 𝑎2+𝑏2=𝑏2+𝑎2a2​+b2​=b2​+a2​.
   • Therefore, 𝐴+𝐵=𝐵+𝐴A+B=B+A.

   Examples of Vector Addition

   Example 1:

   • Suppose two vectors 𝐴A and 𝐵B are given by 𝐴=(4,3)A=(4,3) and 𝐵=(1,2)B=(1,2).
   • Find 𝐴+𝐵A+B.
     • Solution: 𝐴+𝐵=(4+1,3+2)=(5,5)A+B=(4+1,3+2)=(5,5).

   Graphical Representation:

   • Draw 𝐴A as a vector 4 units right and 3 units up.
   • Draw 𝐵B from the tip of 𝐴A, 1 unit right and 2 units up.
   • The resultant vector 𝑅R from the origin to the tip of 𝐵B measures 5 units right and 5 units up.

   Example 2:

   • Consider vectors 𝐶=(−3,1)C=(−3,1) and 𝐷=(4,−2)D=(4,−2).
   • Find 𝐶+𝐷C+D.
     • Solution: 𝐶+𝐷=(−3+4,1−2)=(1,−1)C+D=(−3+4,1−2)=(1,−1).

   Graphical Representation:

   • Draw 𝐶C as a vector 3 units left and 1 unit up.
   • From the tip of 𝐶C, draw 𝐷D 4 units right and 2 units down.
   • The resultant vector 𝑅R starts at the origin and ends at the tip of 𝐷D, measuring 1 unit right and 1 unit down.

   These examples illustrate how vectors can be added both algebraically and graphically, utilizing their properties to solve complex problems in a straightforward manner.

5. 5.Multiplication of a Vector by a Scalar

   Multiplication of a Vector by a Scalar

   Multiplying a vector by a scalar is a fundamental operation in vector algebra that involves scaling the magnitude of the vector without changing its direction (unless the scalar is negative, in which case the direction is reversed). This operation is used widely across various fields, such as physics, engineering, and computer graphics.

   How It Works

   When a vector 𝑣=(𝑥,𝑦)v=(x,y) is multiplied by a scalar 𝑘k, the resulting vector 𝑘𝑣kv is calculated by multiplying each component of the vector by the scalar:

   𝑘𝑣=(𝑘𝑥,𝑘𝑦)kv=(kx,ky)

   This results in a new vector whose length is ∣𝑘∣∣k∣ times the length of the original vector 𝑣v, and whose direction is the same as 𝑣v if 𝑘k is positive, and opposite if 𝑘k is negative.

   Properties of Scalar Multiplication

   1. Distributive Property Over Vector Addition: 𝑘(𝑢+𝑣)=𝑘𝑢+𝑘𝑣k(u+v)=ku+kv

      • Scalar multiplication distributes over vector addition.

   2. Distributive Property Over Scalar Addition: (𝑘+𝑚)𝑣=𝑘𝑣+𝑚𝑣(k+m)v=kv+mv

      • The sum of two scalars multiplied by a vector is the same as each scalar multiplied by the vector and then added.

   3. Associative Property: 𝑘(𝑚𝑣)=(𝑘𝑚)𝑣k(mv)=(km)v

      • The product of two scalars multiplied by a vector is the same as one scalar times the product of the other scalar and the vector.

   4. Identity Element of Scalar Multiplication: 1𝑣=𝑣1v=v

      • Multiplying a vector by 1 leaves it unchanged.

   Examples

   Example 1:

   • Given 𝑣=(3,4)v=(3,4) and 𝑘=2k=2,
   • Calculate 2𝑣2v.
     • Solution: 2𝑣=(2×3,2×4)=(6,8)2v=(2×3,2×4)=(6,8).

   Example 2:

   • Given 𝑣=(5,−3)v=(5,−3) and 𝑘=−3k=−3,
   • Calculate −3𝑣−3v.
     • Solution: −3𝑣=(−3×5,−3×−3)=(−15,9)−3v=(−3×5,−3×−3)=(−15,9).
     • Here, the vector not only scales up in magnitude but also reverses direction due to the negative scalar.

   These examples demonstrate how multiplying a vector by a scalar changes its magnitude, and potentially its direction, providing a powerful tool for scaling vectors in applications ranging from physics simulations to graphical transformations.

6. 6.Components of a Vector & Section Formula

   Components of a Vector

   A vector in mathematics, especially in two or three dimensions, can be broken down into components, which represent its influence in each dimension of a coordinate system.

   Definition

   In a two-dimensional Cartesian coordinate system, any vector 𝑣v can be represented as 𝑣=(𝑥,𝑦)v=(x,y), where 𝑥x and 𝑦y are the horizontal and vertical components of the vector, respectively. In three dimensions, the vector is represented as 𝑣=(𝑥,𝑦,𝑧)v=(x,y,z), with 𝑧z being the depth component along with 𝑥x and 𝑦y.

   Explanation

   The components of a vector describe how far the vector extends along each axis of the coordinate system. For instance, the vector 𝑣=(3,4)v=(3,4) extends 3 units along the x-axis and 4 units along the y-axis. The magnitude of this vector can be found using the Pythagorean theorem: ∣𝑣∣=𝑥2+𝑦2=32+42=5∣v∣=x2+y2​=32+42​=5.

   Examples

   1. Vector 𝑎=(5,0)a=(5,0):
      • This vector lies entirely along the x-axis, indicating movement or force applied 5 units to the right, with no vertical component.
   2. Vector 𝑏=(0,−7)b=(0,−7):
      • This vector lies entirely along the y-axis, indicating a movement or force of 7 units downward.
   3. Vector 𝑐=(−3,4)c=(−3,4):
      • This vector extends 3 units left and 4 units up, forming a right triangle with sides of length 3 and 4, and a hypotenuse of 5 units.

   Section Formula

   The section formula is used in coordinate geometry to find the coordinates of a point that divides a line segment joining two points in a given ratio.

   Definition and Theorem

   Given two points 𝐴(𝑥1,𝑦1)A(x1​,y1​) and 𝐵(𝑥2,𝑦2)B(x2​,y2​) and a point 𝑃P that divides the line segment 𝐴𝐵AB in the ratio 𝑚:𝑛m:n, the coordinates of 𝑃P can be calculated as:

   𝑃=(𝑚𝑥2+𝑛𝑥1𝑚+𝑛,𝑚𝑦2+𝑛𝑦1𝑚+𝑛)P=(m+nmx2​+nx1​​,m+nmy2​+ny1​​)

   This is known as the section formula. The point 𝑃P is called the internal division point if 𝑚m and 𝑛n are positive.

   Examples

   1. Dividing in the Ratio 2:3:
      • Let 𝐴=(1,2)A=(1,2) and 𝐵=(5,8)B=(5,8). Find the point 𝑃P dividing 𝐴𝐵AB in the ratio 2:3.
      • Solution: 𝑃=(2×5+3×12+3,2×8+3×22+3)=(135,225)P=(2+32×5+3×1​,2+32×8+3×2​)=(513​,522​).
   2. Midpoint (1:1 ratio):
      • For 𝐴=(−2,3)A=(−2,3) and 𝐵=(4,−1)B=(4,−1), find the midpoint.
      • Solution: 𝑃=(−2+42,3−12)=(1,1)P=(2−2+4​,23−1​)=(1,1).
   3. External Division:
      • To find the external division point in the ratio 1:2 for 𝐴=(3,4)A=(3,4) and 𝐵=(5,8)B=(5,8).
      • Solution: 𝑃=(1×5−2×31−2,1×8−2×41−2)=(−1,0)P=(1−21×5−2×3​,1−21×8−2×4​)=(−1,0).

   Each of these points, from components of vectors to the section formula, plays a fundamental role in vector algebra and coordinate geometry, helping to solve real-world problems in physics, engineering, and beyond. This systematic breakdown should provide a clear understanding of how vectors operate in space and how specific points can be determined along a segment.

7. 7.Exercise Questions

   1. Find a vector in the direction of vector 5𝑖−𝑗+2𝑘5i−j+2k which has magnitude 8 units.
   2. Show that the vectors 2𝑖−3𝑗+4𝑘2i−3j+4k and −4𝑖+6𝑗−8𝑘−4i+6j−8k are collinear.
   3. Find the direction cosines of the vector 𝑖+2𝑗+3𝑘i+2j+3k.
   4. Find the direction cosines of the vector joining the points A(1, 2, -3) and B(-1, -2, 1), directed from A to B.

   Now, let's solve each problem step by step.

   1. Find a Vector with Magnitude 8

   The given vector is 𝑣=5𝑖−𝑗+2𝑘v=5i−j+2k. First, we find its magnitude and then scale it to have a magnitude of 8 units.

   ∣𝑣∣=52+(−1)2+22=25+1+4=30∣v∣=52+(−1)2+22​=25+1+4​=30​

   To scale this vector to a magnitude of 8 units, we multiply by 83030​8​:

   𝑢=830𝑣=830(5𝑖−𝑗+2𝑘)u=30​8​v=30​8​(5i−j+2k) 𝑢=(4030𝑖−830𝑗+1630𝑘)u=(30​40​i−30​8​j+30​16​k)

   2. Check if Vectors are Collinear

   Two vectors are collinear if one is a scalar multiple of the other. Let's check if 𝑢=2𝑖−3𝑗+4𝑘u=2i−3j+4k and 𝑣=−4𝑖+6𝑗−8𝑘v=−4i+6j−8k satisfy this condition:

   Dividing components of 𝑣v by those of 𝑢u: −42=−2,6−3=−2,−84=−22−4​=−2,−36​=−2,4−8​=−2

   Since all ratios are equal, the vectors are collinear.

   3. Direction Cosines of 𝑖+2𝑗+3𝑘i+2j+3k

   Direction cosines are calculated by dividing each component of the vector by the vector's magnitude: 𝑣=𝑖+2𝑗+3𝑘v=i+2j+3k ∣𝑣∣=12+22+32=14∣v∣=12+22+32​=14​

   Direction cosines: (114,214,314)(14​1​,14​2​,14​3​)

   4. Direction Cosines from A to B

   Points A(1, 2, -3) and B(-1, -2, 1): 𝐴𝐵=[(−1−1)𝑖,(−2−2)𝑗,(1+3)𝑘]AB=[(−1−1)i,(−2−2)j,(1+3)k] 𝐴𝐵=−2𝑖−4𝑗+4𝑘AB=−2i−4j+4k ∣𝐴𝐵∣=(−2)2+(−4)2+42=36=6∣AB∣=(−2)2+(−4)2+42​=36​=6

   Direction cosines: (−26,−46,46)=(−13,−23,23)(6−2​,6−4​,64​)=(−31​,−32​,32​)

   These solutions address each problem as specified.

8. 8.Product of Two Vectors

   In mathematics, specifically in vector algebra, the product of two vectors can be defined in several ways depending on the application and the dimensionality of the space involved. The most commonly used products are the dot product (scalar product) and the cross product (vector product). Let’s explore both:
   

   1. Dot Product (Scalar Product)

   Definition:

   The dot product of two vectors results in a scalar (a single number) and is calculated by multiplying corresponding components of the vectors and summing the results.

   Formula:

   If 𝑎=𝑎1𝑖+𝑎2𝑗+𝑎3𝑘a=a1​i+a2​j+a3​k and 𝑏=𝑏1𝑖+𝑏2𝑗+𝑏3𝑘b=b1​i+b2​j+b3​k, then the dot product of 𝑎a and 𝑏b is given by: 𝑎⋅𝑏=𝑎1𝑏1+𝑎2𝑏2+𝑎3𝑏3a⋅b=a1​b1​+a2​b2​+a3​b3​

   Properties:

   • Commutative: 𝑎⋅𝑏=𝑏⋅𝑎a⋅b=b⋅a
   • Distributive over vector addition: 𝑎⋅(𝑏+𝑐)=𝑎⋅𝑏+𝑎⋅𝑐a⋅(b+c)=a⋅b+a⋅c
   • Scalar multiplication: 𝑐(𝑎⋅𝑏)=(𝑐𝑎)⋅𝑏=𝑎⋅(𝑐𝑏)c(a⋅b)=(ca)⋅b=a⋅(cb), where 𝑐c is a scalar.

   Physical Interpretation:

   The dot product measures the extent to which two vectors are aligned. It is maximally positive when the vectors are parallel, zero when the vectors are orthogonal, and maximally negative when the vectors are anti-parallel.

   2. Cross Product (Vector Product)

   Definition:

   The cross product of two vectors results in a third vector which is perpendicular to the plane containing the first two vectors. It is defined only in three-dimensional space.

   Formula:

   If 𝑎=𝑎1𝑖+𝑎2𝑗+𝑎3𝑘a=a1​i+a2​j+a3​k and 𝑏=𝑏1𝑖+𝑏2𝑗+𝑏3𝑘b=b1​i+b2​j+b3​k, then their cross product 𝑎×𝑏a×b is given by: 𝑎×𝑏=(𝑎2𝑏3−𝑎3𝑏2)𝑖−(𝑎1𝑏3−𝑎3𝑏1)𝑗+(𝑎1𝑏2−𝑎2𝑏1)𝑘a×b=(a2​b3​−a3​b2​)i−(a1​b3​−a3​b1​)j+(a1​b2​−a2​b1​)k

   Properties:

   • Anticommutative: 𝑎×𝑏=−𝑏×𝑎a×b=−b×a
   • Distributive over vector addition: 𝑎×(𝑏+𝑐)=𝑎×𝑏+𝑎×𝑐a×(b+c)=a×b+a×c
   • Not associative (but satisfies the Jacobi identity).

   Physical Interpretation:

   The magnitude of the cross product is equal to the area of the parallelogram formed by the two vectors. The direction is given by the right-hand rule, which determines the orientation of the resultant vector.

   Examples

   Dot Product Example: For vectors 𝑎=3𝑖+2𝑗+𝑘a=3i+2j+k and 𝑏=𝑖+𝑗+𝑘b=i+j+k, 𝑎⋅𝑏=3×1+2×1+1×1=6a⋅b=3×1+2×1+1×1=6

   Cross Product Example: For the same vectors, 𝑎×𝑏=∣𝑖𝑗𝑘321111∣=(2−1)𝑖−(3−1)𝑗+(3−2)𝑘=𝑖−2𝑗+𝑘a×b=∣∣​i31​j21​k11​∣∣​=(2−1)i−(3−1)j+(3−2)k=i−2j+k

   These vector products are crucial in fields ranging from physics and engineering to computer graphics and robotics.

9. 9.Exercise Questions

   1. Find ∣𝑎×𝑏∣∣a×b∣, if 𝑎=𝑖−7𝑗+7𝑘a=i−7j+7k and 𝑏=3𝑖−2𝑗+2𝑘b=3i−2j+2k.
   2. Find a unit vector perpendicular to each of the vector 𝑎+𝑏a+b and 𝑎−𝑏a−b, where 𝑎=3𝑖+2𝑗+2𝑘a=3i+2j+2k and 𝑏=𝑖+2𝑗−2𝑘b=i+2j−2k.
   3. If a unit vector 𝑎a makes angles 𝜋33π​ with 𝑖i, 𝜋44π​ with 𝑗j, and an acute angle 𝜃θ with 𝑘k, then find 𝜃θ and hence, the components of 𝑎a.
   4. Show that(𝑎−𝑏)×(𝑎+𝑏)=2(𝑎×𝑏)(a−b)×(a+b)=2(a×b)
   5. Find 𝜆λ and 𝜇μ if (2𝑖+6𝑗+27𝑘)×(𝜆𝑖+𝜇𝑗+𝜇𝑘)=0(2i+6j+27k)×(λi+μj+μk)=0.

   Let's solve each problem step by step:

   Problem 1: Cross Product Magnitude

   Given vectors: 𝑎=𝑖−7𝑗+7𝑘a=i−7j+7k 𝑏=3𝑖−2𝑗+2𝑘b=3i−2j+2k

   Calculate 𝑎×𝑏a×b and then its magnitude: 𝑎×𝑏=∣𝑖𝑗𝑘1−773−22∣a×b=∣∣​i13​j−7−2​k72​∣∣​ =𝑖((−7)(2)−7(−2))−𝑗((1)(2)−7(3))+𝑘((1)(−2)−(−7)(3))=i((−7)(2)−7(−2))−j((1)(2)−7(3))+k((1)(−2)−(−7)(3)) =𝑖(0)−𝑗(−19)+𝑘(23)=i(0)−j(−19)+k(23) =19𝑗+23𝑘=19j+23k

   Magnitude: ∣𝑎×𝑏∣=192+232=361+529=890∣a×b∣=192+232​=361+529​=890​

   Problem 2: Perpendicular Unit Vector

   Given vectors: 𝑎=3𝑖+2𝑗+2𝑘a=3i+2j+2k 𝑏=𝑖+2𝑗−2𝑘b=i+2j−2k

   𝑎+𝑏=4𝑖+4𝑗a+b=4i+4j 𝑎−𝑏=2𝑖−4𝑘a−b=2i−4k

   The cross product of 𝑎+𝑏a+b and 𝑎−𝑏a−b gives a vector perpendicular to both: (𝑎+𝑏)×(𝑎−𝑏)=∣𝑖𝑗𝑘44020−4∣(a+b)×(a−b)=∣∣​i42​j40​k0−4​∣∣​ =𝑖(0+16)−𝑗(0+8)+𝑘(0−8)=i(0+16)−j(0+8)+k(0−8) =16𝑖−8𝑗−8𝑘=16i−8j−8k

   Normalize this vector to find the unit vector: Magnitude=162+82+82=384Magnitude=162+82+82​=384​ Unit vector=(16384,−8384,−8384)Unit vector=(384​16​,−384​8​,−384​8​)

   Problem 3: Direction Cosines

   Given angles 𝜋33π​ with 𝑖i, 𝜋44π​ with 𝑗j: Direction cosines 𝑙=cos⁡(𝜋3)=12l=cos(3π​)=21​, 𝑚=cos⁡(𝜋4)=22m=cos(4π​)=22​​.

   Using 𝑙2+𝑚2+𝑛2=1l2+m2+n2=1: (12)2+(22)2+𝑛2=1(21​)2+(22​​)2+n2=1 14+12+𝑛2=141​+21​+n2=1 𝑛2=14n2=41​ 𝑛=±12n=±21​ (selecting positive since 𝜃θ is acute)

   The components of 𝑎a are 12𝑖+22𝑗+12𝑘21​i+22​​j+21​k.

   Problem 4: Vector Identity

   Prove: (𝑎−𝑏)×(𝑎+𝑏)=2(𝑎×𝑏)(a−b)×(a+b)=2(a×b) Using distributive property: =𝑎×𝑎+𝑎×𝑏−𝑏×𝑎−𝑏×𝑏=a×a+a×b−b×a−b×b =0+𝑎×𝑏−𝑏×𝑎+0=0+a×b−b×a+0 =𝑎×𝑏+𝑎×𝑏=a×b+a×b =2(𝑎×𝑏)=2(a×b) (since 𝑏×𝑎=−𝑎×𝑏b×a=−a×b)

   Problem 5: Solving Vector Equation

   Solve: (2𝑖+6𝑗+27𝑘)×(𝜆𝑖+𝜇𝑗+𝜇𝑘)=0(2i+6j+27k)×(λi+μj+μk)=0 This equation implies the vectors are collinear, therefore: 𝜆2=𝜇6=𝜇272λ​=6μ​=27μ​

   Equating 𝜇66μ​ and 𝜇2727μ​: 𝜇=0μ=0 𝜆=2×0=0λ=2×0=0

   Thus, 𝜆=0λ=0 and 𝜇=0μ=0.