Vectors

A quantity is a vector only if it has a direction and combines by the triangle (or parallelogram) law. Direction alone is not enough.

• Vectors: displacement, velocity, acceleration, force, momentum, impulse, torque, angular momentum, area (its direction is the outward normal), electric field.
• Scalars: distance, speed, mass, time, work, energy, power, pressure, temperature, charge and electric current.

Current flows “in a direction” along a wire, yet two currents meeting at a junction simply add as numbers. They do not obey the triangle law, so current is a scalar. Pressure acts on every surface normally but has no single direction of its own, so it is also a scalar.

A component is the shadow of the vector on an axis. Always ask: “which axis is the angle measured from?”

1. Angle θ measured from the x-axis: \(A_x = A\cos\theta\), \(A_y = A\sin\theta\).
2. Angle α measured from the y-axis: \(A_x = A\sin\alpha\), \(A_y = A\cos\alpha\). The cosine always goes with the axis the angle touches.
3. Component along any line making angle φ with the vector: \(A\cos\varphi\). It is largest (equal to A) for φ = 0, zero for φ = 90° and negative for φ > 90°.

So a component can never be larger in size than the vector itself. Rotating the axes changes the components but never the magnitude \(\sqrt{A_x^2 + A_y^2}\).

From the components, \(\tan\theta = A_y/A_x\). A calculator returns an angle between −90° and 90° only, so fix the quadrant from the signs:

• (+, +): quadrant I, θ as given.
• (−, +): quadrant II, θ = 180° − reference angle.
• (−, −): quadrant III, θ = 180° + reference angle.
• (+, −): quadrant IV, θ = 360° − reference angle.

Example: \((-3, 4)\) has reference angle 53°, so the vector points at 127° from +x.

If \(\vec A\) makes angles α, β, γ with the x, y and z axes, then \(\cos\alpha = A_x/A\), \(\cos\beta = A_y/A\), \(\cos\gamma = A_z/A\). These three numbers are the components of the unit vector \(\hat A\), so

This identity is the quick test for “can a vector make these angles with the axes?” For example, 45°, 45°, 45° gives \(3 \times \tfrac12 = \tfrac32 \ne 1\), so no vector does that.

To write a vector of magnitude B along \(\vec A\), multiply the unit vector by B: \(\vec B = B\hat A = B\vec A/A\). A 10 N force along \(3\hat i + 4\hat j\) is \(10(3\hat i + 4\hat j)/5 = (6\hat i + 8\hat j)\) N.

In 3D the magnitude is \(A = \sqrt{A_x^2 + A_y^2 + A_z^2}\). Every later chapter (forces, momentum, fields) uses this to switch between “magnitude and direction” and “î, ĵ, k̂ form”.

• Position depends on where you put the origin. Displacement does not, because the origin cancels in \(\vec r_2 - \vec r_1\).
• Displacement depends only on the end points, not on the path taken. Distance travelled depends on the path.
• For a closed path (back to the start), \(\Delta\vec r = 0\) though the distance travelled is not zero.

Put \(\vec A\) along the x-axis and let \(\vec B\) make angle θ with it (both drawn from one tail).

1. Components of the sum: \(R_x = A + B\cos\theta\) and \(R_y = B\sin\theta\).
2. Square and add: \(R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta = A^2 + B^2 + 2AB\cos\theta\).
3. The angle α of \(\vec R\) with \(\vec A\) follows from \(\tan\alpha = R_y/R_x = \dfrac{B\sin\theta}{A + B\cos\theta}\).

The resultant always leans toward the larger vector. If A = B, it bisects the angle.

• θ = 0°: R = A + B (maximum).
• θ = 180°: R = |A − B| (minimum), along the larger vector.
• θ = 90°: \(R = \sqrt{A^2+B^2}\).
• Equal magnitudes A: \(R = 2A\cos(\theta/2)\). So R = A at 120°, \(R=\sqrt2A\) at 90°, \(R=\sqrt3A\) at 60°.
• Always \(|A-B| \le R \le A+B\). Three vectors can add to zero only if each magnitude is no larger than the sum of the other two.

As θ grows from 0° to 180°, R falls steadily from A + B to |A − B|.

\(\vec A - \vec B = \vec A + (-\vec B)\). Reversing \(\vec B\) changes the angle to 180° − θ, so \(|\vec A-\vec B|^2 = A^2 + B^2 - 2AB\cos\theta\). For equal magnitudes, \(|\vec A-\vec B| = 2A\sin(\theta/2)\).

This is the change in velocity when only the direction turns by θ. It is used again in circular motion and in bouncing-ball impulse questions. A useful consequence: \(|\vec A+\vec B| = |\vec A-\vec B|\) only when \(\cos\theta = 0\), that is, when the vectors are perpendicular.

• Commutative: \(\vec A\cdot\vec B = \vec B\cdot\vec A\). Distributive: \(\vec A\cdot(\vec B+\vec C) = \vec A\cdot\vec B + \vec A\cdot\vec C\).
• \(\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1\) and \(\hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0\).
• \(\vec A\cdot\vec A = A^2\), so a magnitude can always be found as \(\sqrt{\vec A\cdot\vec A}\).

Expanding \((A_x\hat i + A_y\hat j + A_z\hat k)\cdot(B_x\hat i + B_y\hat j + B_z\hat k)\) gives nine terms. Six of them contain a product of different unit vectors and vanish, leaving \(A_xB_x + A_yB_y + A_zB_z\).

1. Angle: \(\cos\theta = \dfrac{\vec A\cdot\vec B}{AB}\). The sign of the dot product tells you at once whether θ is acute (+), right (0) or obtuse (−).
2. Scalar projection of \(\vec A\) on \(\vec B\): \(A\cos\theta = \dfrac{\vec A\cdot\vec B}{B}\). Divide by the magnitude of the vector you project onto.
3. Perpendicular test: set \(\vec A\cdot\vec B = 0\) and solve for the unknown component.
4. Resultant link: \(|\vec A+\vec B|^2 = (\vec A+\vec B)\cdot(\vec A+\vec B) = A^2 + B^2 + 2\vec A\cdot\vec B\).

Work \(W = \vec F\cdot\vec s\), power \(P = \vec F\cdot\vec v\), flux \(\Phi = \vec E\cdot\vec A\). In each case only the parallel part of one vector matters. A force perpendicular to the motion, such as the tension in a uniformly whirled stone or the normal force on a level road, does zero work.

Write \(\hat i \to \hat j \to \hat k \to \hat i\) in a circle. Going forward gives +: \(\hat i\times\hat j = \hat k\), \(\hat j\times\hat k = \hat i\), \(\hat k\times\hat i = \hat j\). Going backward gives −. Any unit vector crossed with itself is zero.

The full product is the determinant

Do not forget the minus sign on the middle (\(\hat j\)) term.

• \(|\vec A\times\vec B| = AB\sin\theta\) is the area of the parallelogram with sides \(\vec A\) and \(\vec B\). Half of it is the area of the triangle they form.
• \(\vec A\times\vec B\) is perpendicular to both, so \(\vec A\cdot(\vec A\times\vec B) = 0\) and \(\vec B\cdot(\vec A\times\vec B) = 0\) always.
• Unit vector normal to both: \(\hat n = \dfrac{\vec A\times\vec B}{|\vec A\times\vec B|}\). The opposite vector \(-\hat n\) is also normal.

Since \(\cos^2\theta + \sin^2\theta = 1\): \((\vec A\cdot\vec B)^2 + |\vec A\times\vec B|^2 = A^2B^2\). Also \(\tan\theta = |\vec A\times\vec B| / (\vec A\cdot\vec B)\).

• \(|\vec A\times\vec B| = \vec A\cdot\vec B\) means θ = 45°.
• \(|\vec A\times\vec B| = \sqrt3\,\vec A\cdot\vec B\) means θ = 60°.

Physics that uses the cross product: torque \(\vec\tau = \vec r\times\vec F\), angular momentum \(\vec L = \vec r\times\vec p\), velocity in rotation \(\vec v = \vec\omega\times\vec r\), and the magnetic force \(q\vec v\times\vec B\) in Class 12.

When three forces keep a particle in equilibrium, \(\vec F_1 + \vec F_2 + \vec F_3 = 0\). This has three useful consequences.

1. Drawn head to tail, the three arrows form a closed triangle.
2. Any one force is equal and opposite to the resultant of the other two: \(\vec F_3 = -(\vec F_1 + \vec F_2)\).
3. So \(|F_1 - F_2| \le F_3 \le F_1 + F_2\). A set like 2 N, 3 N, 6 N can never be in equilibrium.

Three non-parallel forces in equilibrium on a body must also be coplanar and their lines of action must pass through one point. Otherwise they would produce a net torque.

Resolve along directions that remove unknowns. For a block on an incline, take axes along and perpendicular to the slope. For a knot held by two strings, take one axis along a string if you want the other string’s tension directly. The equations are equally valid for any perpendicular pair; a good choice only reduces algebra.

If a particle is in equilibrium under many forces and one force \(\vec F\) is suddenly removed, the rest add up to \(-\vec F\). The particle then accelerates at \(\vec F/m\) in the direction opposite to the removed force. Likewise n equal forces drawn symmetrically from a point (at 360°/n to each other) form a closed regular polygon and sum to zero.

1. Draw \(\vec F_1\), \(\vec F_2\), \(\vec F_3\) head to tail. Equilibrium means the triangle closes.
2. If α is the angle between \(\vec F_2\) and \(\vec F_3\) at the knot, the interior angle of the triangle opposite \(F_1\) is 180° − α.
3. Sine rule: \(\dfrac{F_1}{\sin(180^\circ-\alpha)} = \dfrac{F_2}{\sin(180^\circ-\beta)} = \dfrac{F_3}{\sin(180^\circ-\gamma)}\).
4. Because \(\sin(180^\circ - x) = \sin x\), this is Lami’s theorem.

The largest force is the one opposite the angle whose sine is largest, which is the angle closest to 90°.

If a weight W hangs from the middle of a string and each half makes angle θ with the vertical, the vertical parts give \(2T\cos\theta = W\), so

At θ = 0 each half carries W/2. At θ = 60°, each tension equals W. As the string approaches horizontal (θ → 90°), T grows without limit. That is why a washing line or a power cable always sags: no finite tension can make it perfectly straight with a load on it.

Use Lami when there are exactly three forces and the angles between them are easy to read. Use components when there are four or more forces, or when the angles are given with different reference lines. Both methods must give the same answer, so one can check the other.