Physics vectors

Scalars vs. Vectors

• Physical quantities in daily life
  • Scalars: magnitude + unit only (e.g.a0mass, time, length, work).
  • Vectors: magnitude + unit + direction (e.g.a0displacement, velocity, force, acceleration, torque, electric flux).
• Real-world illustration
  • Road signboard lists distances (scalar). To actually reach a destination you need both distance and direction e28094 together they form the vector "displacement".

Graphical Representation of a Vector

• Depicted by a line segment with an arrowhead.
  • Length (drawn to scale) $d7$ magnitude.
  • Arrowhead $d7$ direction.
• Vector addition previously learned; current focus: components & products.

Rectangular Components of a Vector (2-D)

• Definitions
  • Components = effective parts of a vector along chosen axes.
  • Along $x$-axis: horizontal component, $A_x$.
  • Along $y$-axis: vertical component, $A_y$.
  • Since $x$ and $y$ axes are perpendicular, these are called rectangular components.
• Vector in component form
  • $\vec A = \vec A<em>x + \vec A</em>y$
  • Using unit vectors $\hat i$ (along $+x$) & $\hat j$ (along $+y$):
    $\vec A = A<em>x \hat i + A</em>y \hat j$

2.1.1 Finding Rectangular Components of a Given Vector

• Given $|\vec A| = A$ and the angle $\theta$ between $\vec A$ and the horizontal:
  • Using right–triangle trigonometry (Fig.a02.1):
  • $\sin\theta = \frac{A<em>y}{A} \;\;\Rightarrow\;\; A</em>y = A\sin\theta$
  • $\cos\theta = \frac{A<em>x}{A} \;\;\Rightarrow\;\; A</em>x = A\cos\theta$
• Both formulas allow conversion from polar (magnitude, direction) to rectangular form.

2.1.2 Constructing a Vector from Its Components

• If $Ax$ and $Ay$ are known:
  • Magnitude (via Pythagoras):
    $A = \sqrt{A<em>x^{2} + A</em>y^{2}}$
  • Direction angle:
    $\theta = \tan^{-1}\left(\frac{A<em>y}{A</em>x}\right)$
  • Together these invert the previous process (rectangular $\rightarrow$ polar).

Worked Example 2.1

• Problem: A peddler pushes a trolley with force $F = 50\,\text{N}$ at $\theta = 30^{\circ}$ to the horizontal. Find $Fx$ and $Fy$.
• Solution
  • $F_x = F\cos\theta = 50\cos30^{\circ}=43.3\,\text{N}$
  • $F_y = F\sin\theta = 50\sin30^{\circ}=25\,\text{N}$

Assignment 2.1 (Self-Check)

• Fatima pulls a bag on a ramp; given $Fx = 12\,\text{N}$, $Fy = 5\,\text{N}$. Compute $|\vec F|$ and $\theta$ using the same formulas.

Product of Two Vectors

• Multiplying two vectors can yield either a scalar (dot product) or another vector (cross product).

Scalar (Dot) Product $(\vec A\cdot\vec B)$

• Definition
  • $\vec A \cdot \vec B = AB\cos\theta$
  • $A,B$ = magnitudes; $\theta$ = smaller angle between them.
• Alternative viewpoints (Fig.a02.3):
  • $\vec A\cdot\vec B$ = magnitude of one vector $\times$ component of the other along the first:
    $\vec A\cdot\vec B = (A\cos\theta)\,B = (\text{component of }\vec A\text{ along }\vec B)\,B$
    $= A\,(\text{component of }\vec B\text{ along }\vec A)$

Everyday Examples (all scalars produced via dot product)

• Work: $W = \vec F\cdot\vec d$ (force $\times$ displacement)
• Power: $P = \vec F\cdot\vec v$ (force $\times$ velocity)
• Electric flux: $\Phi_E = \vec E\cdot\vec A$ (field $\times$ vector area)

Properties of the Dot Product

• Commutative: $\vec A\cdot\vec B = \vec B\cdot\vec A$
• Zero for perpendicular vectors: $\vec A\perp\vec B \;\Rightarrow\; \vec A\cdot\vec B = 0$
  • Includes $\hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0$.
• Maximum for parallel vectors: $\theta = 0^{\circ}\;\Rightarrow\; \vec A\cdot\vec B = AB$
  • For a vector with itself: $\vec A\cdot\vec A = A^{2}$ and $\hat i\cdot\hat i = 1$, etc.
• Negative for antiparallel vectors: $\theta = 180^{\circ}\;\Rightarrow\; \vec A\cdot\vec B = -AB$
• Component (algebraic) formula: For
  $\vec A = A<em>x\hat i + A</em>y\hat j + A<em>z\hat k$$\vec B = B</em>x\hat i + B<em>y\hat j + B</em>z\hat k$
  $\boxed{\vec A\cdot\vec B = A<em>xB</em>x + A<em>yB</em>y + A<em>zB</em>z}$

Worked Example 2.2

• Given
  • Force $\vec F = (8\hat i - 2\hat j)\,\text{N}$
  • Velocity $\vec v = (3\hat i + 4\hat j)\,\text{m\,s}^{-1}$
• Power delivered
  $P = \vec F\cdot\vec v = 8\times3 + (-2)\times4 = 24 - 8 = 16\,\text{W}$

Assignment 2.2 (Self-Check)

• For $\vec A = 5\hat i + \hat j$ and $\vec B = 2\hat i + 4\hat j$ find:
  1. Projection of $\vec A$ on $\vec B$.
  2. Projection of $\vec B$ on $\vec A$.

Vector (Cross) Product $(\vec A\times\vec B)$

• Definition
  • $\vec A \times \vec B = AB\sin\theta\,\hat n$
  • $\hat n$ = unit vector perpendicular to the plane of $\vec A$ and $\vec B$.
• Right-Hand Rule (Fig.a02.5)
  • Curl fingers from first vector to second through the smaller angle; thumb points along $\hat n$ (direction of product).

Common Physical Quantities Based on Cross Product

• Torque: $\vec \tau = \vec r \times \vec F = rF\sin\theta\,\hat n$
• Angular momentum: $\vec L = \vec r \times \vec p = rp\sin\theta\,\hat n$

Properties of the Cross Product

• Anti-commutative: $\vec A\times\vec B = -\big(\vec B\times\vec A\big)$
  • Same magnitude, opposite direction.
• Zero for parallel or antiparallel vectors: $\theta = 0^{\circ}$ or $180^{\circ}$ yields $\vec A\times\vec B = 0$
  • Also $\vec A\times\vec A = 0$; similarly $\hat i\times\hat i = 0$, etc.
• Maximum for perpendicular vectors: $\theta = 90^{\circ}\;\Rightarrow\; |\vec A\times\vec B| = AB$
  • Basis-vector results: $\hat i\times\hat j = \hat k,\; \hat j\times\hat k = \hat i,\; \hat k\times\hat i = \hat j$

Component (Determinant) Formula

For
$\vec A = A<em>x\hat i + A</em>y\hat j + A<em>z\hat k$$\vec B = B</em>x\hat i + B<em>y\hat j + B</em>z\hat k$
$\boxed{\vec A\times\vec B = \begin{vmatrix} \hat i &amp; \hat j &amp; \hat k\ A<em>x & A</em>y &amp; A<em>z\ B</em>x &amp; B<em>y & B</em>z \end{vmatrix} = (A<em>yB</em>z - A<em>zB</em>y)\hat i + (A<em>zB</em>x - A<em>xB</em>z)\hat j + (A<em>xB</em>y - A<em>yB</em>x)\hat k}$

Geometric / Physical Significance

• Magnitude $|\vec A\times\vec B|$ equals the area of the parallelogram with sides $\vec A$ and $\vec B$ (Fig.a02.7).
• Direction $\hat n$ is a normal to that plane (found via right-hand rule).

Worked Example 2.3

• Side vectors of parallelogram
  • $\vec A = (\hat i + 6\hat j + 2\hat k)\,\text{m}$
  • $\vec B = (7\hat i + \hat j + 5\hat k)\,\text{m}$
• Area = |$\vec A\times\vec B$| (compute the determinant):
  • $\vec A\times\vec B = (6\cdot5 - 2\cdot1)\hat i + (2\cdot7 - 1\cdot5)\hat j + (1\cdot1 - 6\cdot7)\hat k$
  • $= (30 - 2)\hat i + (14 - 5)\hat j + (1 - 42)\hat k$
  • $= 28\hat i + 9\hat j - 41\hat k\;\text{m}^2$
  • Magnitude (if required): $A = \sqrt{28^{2} + 9^{2} + (-41)^{2}}\,\text{m}^2$

Assignment 2.3 (Self-Check)

• Given $|\vec A|=3.2$, $|\vec B|=5.2$, angle $\theta = 60^{\circ}$. Find $|\vec A\times\vec B|$ using $AB\sin\theta$.

Connections & Broader Context

• Math foundations: Dot & cross products rely on trigonometry, Pythagorean theorem, and vector component decomposition.
• Real-world relevance:
  • Navigation & GPS: displacement vectors; component breakdown along latitude/longitude.
  • Engineering: forces on trusses use components; torque uses cross product to size wrenches.
  • Physics: work–energy theorem (dot product) and rotational dynamics (cross product).
• Ethical / safety notes: Correct vector analysis prevents structural failure and ensures accurate navigation systems.