Vectors: Fundamental Concepts
Physical quantities that require both magnitude and direction for complete description are called vectors.
Vector Addition
Triangle Law: If two vectors are represented by two sides of a triangle in order, their resultant is represented by the third side in reverse order.
Parallelogram Law: If two vectors are represented by adjacent sides of a parallelogram, their resultant is represented by the diagonal through the common point.
Resultant Magnitude and Direction
For vectors vecA and vecB with angle theta between them:
|vecR| = sqrt(A^2 + B^2 + 2AB cos theta)
tan alpha = (B sin theta)/(A + B cos theta) where alpha is angle between vecR and vecA.
Vector Subtraction
vecA - vecB = vecA + (-vecB)
|vecA - vecB| = sqrt(A^2 + B^2 - 2AB cos theta)
Resolution of Vectors
Any vector can be expressed as a sum of two or more vectors (components).
Rectangular Components:
A_x = A cos theta, A_y = A sin theta
A = sqrt(A_x^2 + A_y^2), tan theta = A_y/A_x
In unit vector notation: vecA = A_x hati + A_y hatj
For three dimensions: vecA = A_x hati + A_y hatj + A_z hatk
Dot Product (Scalar Product)
vecA . vecB = |A||B| cos theta = A_x B_x + A_y B_y + A_z B_z
Properties:
hati . hati = hatj . hatj = hatk . hatk = 1hati . hatj = hatj . hatk = hatk . hati = 0- Dot product is commutative:
vecA . vecB = vecB . vecA - Work is the dot product of force and displacement:
W = vecF . vecs
Cross Product (Vector Product)
vecA x vecB = |A||B| sin theta hatn where hatn is unit vector perpendicular to both.
Properties:
hati x hati = hatj x hatj = hatk x hatk = 0hati x hatj = hatk,hatj x hatk = hati,hatk x hati = hatj- Cross product is anti-commutative:
vecA x vecB = -vecB x vecA - Magnitude = area of parallelogram formed by
vecAandvecB.
Projectile Motion
A projectile is an object thrown into the air with an initial velocity subject to gravity.
Key Equations
Initial velocity u at angle theta with horizontal:
- Horizontal component:
u_x = u cos theta(constant) - Vertical component:
u_y = u sin theta(affected by gravity)
Time of Flight: T = (2u sin theta)/g
Maximum Height: H = (u^2 sin^2 theta)/(2g)
Horizontal Range: R = (u^2 sin 2theta)/g
Equation of Trajectory: y = x tan theta - (g x^2)/(2u^2 cos^2 theta)
Maximum Range
Range is maximum when sin 2theta = 1, i.e., theta = 45 degrees.
R_max = u^2/g
Projectile on Inclined Plane
More complex, with acceleration components along and perpendicular to the incline.
Uniform Circular Motion
An object moving in a circle with constant speed.
Angular Displacement
theta = s/r (in radians)
Angular Velocity
omega = d theta/dt = v/r
Time Period
T = 2pi r/v = 2pi/omega
Centripetal Acceleration
a_c = v^2/r = omega^2 r
Directed towards the centre of the circle.
Centripetal Force
F_c = mv^2/r = m omega^2 r
This is not a separate force. It is provided by tension, friction, gravity, or another force depending on the situation.
Worked Examples
Example 1: Find the angle between vectors vecA = 2hati + 3hatj + hatk and vecB = hati - 2hatj + 4hatk.
Solution: A.B = 2(1) + 3(-2) + 1(4) = 2 - 6 + 4 = 0. So theta = 90 degrees.
Example 2: A projectile is fired with 50 m/s at 30 degrees. Find range and max height.
Solution: R = (50^2 sin 60)/10 = 2500*sqrt(3)/(2*10) = 2500*0.866/10 = 216.5 m.
H = (50^2 sin^2 30)/(2*10) = (2500*0.25)/20 = 31.25 m.
Example 3: A stone tied to a string of length 1 m rotates at 2 rev/s. Find centripetal acceleration.
Solution: v = r omega = 1 * 2*2pi = 4pi m/s. a_c = v^2/r = (16pi^2)/1 = 16pi^2 m/s^2.
Common Mistakes
- Dot vs Cross product: Dot gives scalar, cross gives vector.
- Range formula valid only for level ground: Does not apply for inclined projections.
- Centripetal vs centrifugal: Centripetal is real (towards centre). Centrifugal is pseudo (apparent).
- 45 degrees for max range only without air resistance: Air resistance changes optimal angle.
ISC Exam Focus
- Theory (70%): Vector operations, projectile derivations, circular motion formulas.
- Application (30%): Numerical problems on projectiles, circular motion, vector resolution.
- ISC projectiles questions: "A ball is thrown at an angle ... find range, time of flight, height."
- Vector problems using analytical method (components or dot/cross product).
Self-Test Questions
Q1: Find the dot product of vecA = 3hati + 4hatj and vecB = hati - 2hatj.
Answer: A.B = 3(1) + 4(-2) = 3 - 8 = -5.
Q2: A projectile has range equal to max height. Find the angle of projection.
Answer: R = H => (u^2 sin 2theta)/g = (u^2 sin^2 theta)/(2g) => 2sin theta cos theta = (sin^2 theta)/2 => tan theta = 4 => theta = tan^(-1)(4).
Q3: Find time of flight for a projectile with u = 20 m/s at 60 degrees.
Answer: T = (2u sin theta)/g = (2*20*sin 60)/10 = 40*0.866/10 = 3.464 s.
Q4: A car goes around a curve of radius 50 m at 15 m/s. Find centripetal acceleration.
Answer: a_c = v^2/r = 225/50 = 4.5 m/s^2.
Q5: Find the cross product (2hati + hatj) x (hati + 3hatk).
Answer: Using determinant method: = 3hati - 6hatj - hatk.
Q6: At what angle should a projectile be launched to achieve half the maximum range?
Answer: R = (u^2 sin 2theta)/g = (1/2)(u^2/g) => sin 2theta = 1/2 => 2theta = 30 or 150 => theta = 15 or 75 degrees.
