Vector Analysis

Scalars and Vectors

  • Scalars have magnitude only
  • Vectors have both magnitude and direction
  • For a physical quantity to be vector:
    • Necessary condition: should have direction
    • Sufficient condition: Should follow sum of vector addition
Q: Which is scalar quantity?

a) Temperature gradient

b) intensity of magnetization

c) intensity of radiation

d) current density

Correct: c

Intensity of radiation has only magnitude.

Q: Angular momentum is

a) a scalar

b) a polar vector

c) an axial vector

d) same as torque

Correct: c

Some vectors

  • All gradients of scalars
    • potential gradient
    • pressure gradient
  • All Field strength (intensity) are vectors
    • Electric Field Strength
    • Magnetic Field Strength
    • Gravitational Field Strength
  • Flux of scalar is vector
    • Electric Flux
    • Magnetic Flux
  • All Potentials are scalars
  • Small length and area are vector

Unit Vector

A vector with unit magnitude

\hat{a} = \frac{\overline{a}}{|\overline{a}|}
a^=a‾∣a‾∣\hat{a} = \frac{\overline{a}}{|\overline{a}|}

dimensionless and unitless. Possess only direction.

Triangle Law of Vector

vectors a and b represented in magnitude and direction by two sides of a triangle taken in same order

 

then their resultant c is given in magnitude and direction by 3rd side taken in reverse order.

\overline{a}
a‾\overline{a}
\overline{b}
b‾\overline{b}
\overline{c}
c‾\overline{c}

Lami's Theorm

co-planar vectors a, b, c are in equilibrium then

\overline{a}
a‾\overline{a}
\overline{b}
b‾\overline{b}
\overline{c}
c‾\overline{c}
\frac{a}{cos\alpha} = \frac{b}{cos\beta} = \frac{c}{cos\gamma}
acosα=bcosβ=ccosγ\frac{a}{cos\alpha} = \frac{b}{cos\beta} = \frac{c}{cos\gamma}
\alpha
α\alpha
\beta
β\beta
\gamma
γ\gamma
cos\beta = \frac{b^{2}-a^{2}-c^{2}}{2ac}
cosβ=b2−a2−c22accos\beta = \frac{b^{2}-a^{2}-c^{2}}{2ac}

Parallelogram law

\overline{a}
a‾\overline{a}
\overline{b}
b‾\overline{b}
\overline{c} = \overline{a} + \overline{b}
c‾=a‾+b‾\overline{c} = \overline{a} + \overline{b}
\theta
θ\theta
\alpha
α\alpha
\beta
β\beta

vectors a and b acting at a point represented in mag. and direction by 2 sides of a parallelogram then

\overline{c} = \overline{a} + \overline{b}
c‾=a‾+b‾\overline{c} = \overline{a} + \overline{b}

O

A

B

C

|\overline{a}+\overline{b}|=OC=\sqrt{a^2+2abcos\theta + b^2}
∣a‾+b‾∣=OC=a2+2abcosθ+b2|\overline{a}+\overline{b}|=OC=\sqrt{a^2+2abcos\theta + b^2}
\alpha = tan^{-1}(\frac{bsin\theta}{a+bcos\theta}) with \space \overline{a}
α=tan−1(bsinθa+bcosθ)with a‾\alpha = tan^{-1}(\frac{bsin\theta}{a+bcos\theta}) with \space \overline{a}
\beta = tan^{-1}(\frac{asin\theta}{b+acos\theta}) with \space \overline{b}
β=tan−1(asinθb+acosθ)with b‾\beta = tan^{-1}(\frac{asin\theta}{b+acos\theta}) with \space \overline{b}

If a > b, 

If a < b, 

\alpha > \beta
α>β\alpha > \beta
\alpha > \beta
α>β\alpha > \beta

If a = b, 

\alpha = \beta = \frac{\theta}{2}
α=β=θ2\alpha = \beta = \frac{\theta}{2}

Parallelogram law

\overline{a}
a‾\overline{a}
\overline{b}
b‾\overline{b}
\overline{c} = \overline{a} + \overline{b}
c‾=a‾+b‾\overline{c} = \overline{a} + \overline{b}
\theta
θ\theta
\alpha
α\alpha
\beta
β\beta

Special cases:

O

A

B

C

\theta = 0 ^{o}, then \space |\overline{a}+\overline{b}| = a + b
θ=0o,then ∣a‾+b‾∣=a+b\theta = 0 ^{o}, then \space |\overline{a}+\overline{b}| = a + b

when 

\theta = 90 ^{o}, then \space |\overline{a}+\overline{b}| = \sqrt{a^2+b^2}
θ=90o,then ∣a‾+b‾∣=a2+b2\theta = 90 ^{o}, then \space |\overline{a}+\overline{b}| = \sqrt{a^2+b^2}

when 

\theta = 180 ^{o}, then \space |\overline{a}+\overline{b}| = a - b
θ=180o,then ∣a‾+b‾∣=a−b\theta = 180 ^{o}, then \space |\overline{a}+\overline{b}| = a - b

when 

|\overline{a}| = |\overline{b}| = A
∣a‾∣=∣b‾∣=A|\overline{a}| = |\overline{b}| = A

when 

then 

|\overline{a}+\overline{b}| = 2Acos\frac{\theta}{2}
∣a‾+b‾∣=2Acosθ2|\overline{a}+\overline{b}| = 2Acos\frac{\theta}{2}
so, \space a-b \leq |\overline{a}+\overline{b}| \leq a+b
so, a−b≤∣a‾+b‾∣≤a+bso, \space a-b \leq |\overline{a}+\overline{b}| \leq a+b

Vector Difference

\overline{b}
b‾\overline{b}
\overline{a}
a‾\overline{a}
\overline{a} + \overline{b}
a‾+b‾\overline{a} + \overline{b}
-\overline{a}
−a‾-\overline{a}
-\overline{b}
−b‾-\overline{b}
\overline{a} - \overline{b}
a‾−b‾\overline{a} - \overline{b}
\overline{b} - \overline{a}
b‾−a‾\overline{b} - \overline{a}
\alpha
α\alpha
\theta
θ\theta
\alpha = tan^{-1}(\frac{bsin\theta}{a-bcos\theta}) \space with \space \overline{a}
α=tan−1(bsinθa−bcosθ) with a‾\alpha = tan^{-1}(\frac{bsin\theta}{a-bcos\theta}) \space with \space \overline{a}

Special cases:

\theta = 0 ^{o}, then \space |\overline{a}-\overline{b}| = a - b
θ=0o,then ∣a‾−b‾∣=a−b\theta = 0 ^{o}, then \space |\overline{a}-\overline{b}| = a - b

when 

\theta = 90 ^{o}, then \space |\overline{a}-\overline{b}| = \sqrt{a^2+b^2}
θ=90o,then ∣a‾−b‾∣=a2+b2\theta = 90 ^{o}, then \space |\overline{a}-\overline{b}| = \sqrt{a^2+b^2}

when 

\theta = 180 ^{o}, then \space |\overline{a}+\overline{b}| = a + b
θ=180o,then ∣a‾+b‾∣=a+b\theta = 180 ^{o}, then \space |\overline{a}+\overline{b}| = a + b

when 

|\overline{a}| = |\overline{b}| = A
∣a‾∣=∣b‾∣=A|\overline{a}| = |\overline{b}| = A

when 

then 

|\overline{a}-\overline{b}| = 2Asin\frac{\theta}{2}
∣a‾−b‾∣=2Asinθ2|\overline{a}-\overline{b}| = 2Asin\frac{\theta}{2}
so, \space a-b \leq |\overline{a}-\overline{b}| \leq a+b
so, a−b≤∣a‾−b‾∣≤a+bso, \space a-b \leq |\overline{a}-\overline{b}| \leq a+b
|\overline{a}-\overline{b}|=OC=\sqrt{a^2 - 2abcos\theta + b^2}
∣a‾−b‾∣=OC=a2−2abcosθ+b2|\overline{a}-\overline{b}|=OC=\sqrt{a^2 - 2abcos\theta + b^2}

General Component

\alpha
α\alpha
\beta
β\beta
\overline{F_1}
F1‾\overline{F_1}
\overline{F_2}
F2‾\overline{F_2}
\overline{F}
F‾\overline{F}
\overline{F_1}
F1‾\overline{F_1}
\overline{F}
F‾\overline{F}
\overline{F_2}
F2‾\overline{F_2}
\theta
θ\theta
\overline{F_1} = \frac{sin\beta}{sin(\alpha+\beta)}F
F1‾=sinβsin(α+β)F\overline{F_1} = \frac{sin\beta}{sin(\alpha+\beta)}F
\overline{F_2} = \frac{sin\alpha}{sin(\alpha+\beta)}F
F2‾=sinαsin(α+β)F\overline{F_2} = \frac{sin\alpha}{sin(\alpha+\beta)}F
F_{1}^{2} + 2F_{1}F_{2}cos(\alpha + \beta) + F_{2}^{2} = F^{2}
F12+2F1F2cos(α+β)+F22=F2F_{1}^{2} + 2F_{1}F_{2}cos(\alpha + \beta) + F_{2}^{2} = F^{2}
\frac{F_{1}}{F_{2}} = \frac{sin\beta}{sin\alpha}
F1F2=sinβsinα\frac{F_{1}}{F_{2}} = \frac{sin\beta}{sin\alpha}

x

Y

\overline{F_x} = FCos\theta
Fx‾=FCosθ\overline{F_x} = FCos\theta
\overline{F_y} = FSin\theta
Fy‾=FSinθ\overline{F_y} = FSin\theta
F_{x}^2 + F_{y}^2 = F^2
Fx2+Fy2=F2F_{x}^2 + F_{y}^2 = F^2
\frac{F_x}{F_y} = Cot\theta
FxFy=Cotθ\frac{F_x}{F_y} = Cot\theta

Rectangular Component

Scalar(dot) Product

Vector(Cross) Product

\overline{a}.\overline{b} = abcos\theta = \overline{b}.\overline{a}
a‾.b‾=abcosθ=b‾.a‾\overline{a}.\overline{b} = abcos\theta = \overline{b}.\overline{a}
|\overline{a} * \overline{b}| = absin\theta, \overline{a} * \overline{b} \neq \overline{b} * \overline{a}
∣a‾∗b‾∣=absinθ,a‾∗b‾≠b‾∗a‾|\overline{a} * \overline{b}| = absin\theta, \overline{a} * \overline{b} \neq \overline{b} * \overline{a}
\overline{a}.\overline{b} = ab(max) if(\theta=0)
a‾.b‾=ab(max)if(θ=0)\overline{a}.\overline{b} = ab(max) if(\theta=0)
\overline{a}.\overline{b} = -ab(min) if(\theta=180)
a‾.b‾=−ab(min)if(θ=180)\overline{a}.\overline{b} = -ab(min) if(\theta=180)
\overline{a}.\overline{b} = 0 \space if(\theta=90)
a‾.b‾=0 if(θ=90)\overline{a}.\overline{b} = 0 \space if(\theta=90)
\overline{a}.\overline{a} = a^{2}
a‾.a‾=a2\overline{a}.\overline{a} = a^{2}
\hat{a}.\hat{a} = 1
a^.a^=1\hat{a}.\hat{a} = 1
|\overline{a} * \overline{b}| = ab(max) \space if \theta = 90^{o}
∣a‾∗b‾∣=ab(max) ifθ=90o|\overline{a} * \overline{b}| = ab(max) \space if \theta = 90^{o}
|\overline{a} * \overline{b}| = 0(min) \space if \theta = 90^{o}
∣a‾∗b‾∣=0(min) ifθ=90o|\overline{a} * \overline{b}| = 0(min) \space if \theta = 90^{o}
|\overline{a} * \overline{a}| = 0
∣a‾∗a‾∣=0|\overline{a} * \overline{a}| = 0

Cross product is never -ve

Vector Analysis

By Roshan Bhandari

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Vector Analysis

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