Concept Card: Vectors & Vector Algebra (Addition & Subtraction)
1. Scalars vs. Vectors:
- Scalar Quantity: A physical quantity completely specified by a numerical magnitude (number + unit) and sign. It obeys the ordinary rules of algebra.
Examples: Mass, time, distance, speed, work, energy, temperature, electric charge. - Vector Quantity: A physical quantity possessing both magnitude and direction that must obey the Triangle or Parallelogram Law of vector addition.
Examples: Displacement, velocity, acceleration, force, linear momentum, electric field. - Crucial Conceptual Exception: Electric current has magnitude and an assigned direction along a wire, but it adds according to simple scalar algebraic rules (Kirchhoff’s Junction Rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$), not vector laws. Hence, electric current is a scalar.
2. Fundamental Classes of Vectors:
- Polar Vectors: Vectors associated with straight-line translational effects having a distinct point of application (e.g., displacement $\vec{s}$, force $\vec{F}$).
- Axial Vectors (Pseudovectors): Vectors associated with rotational motion, directed along the axis of rotation determined by the Right-Hand Screw Rule (e.g., angular velocity $\vec{\omega}$, torque $\vec{\tau}$, angular momentum $\vec{L}$).
- Unit Vector ($\hat{A}$): A vector of unit magnitude ($|\hat{A}| = 1$) pointing in the direction of vector $\vec{A}$:
$\hat{A} = \frac{\vec{A}}{|\vec{A}|} \implies \vec{A} = |\vec{A}|\hat{A}$ - Null / Zero Vector ($\vec{0}$): A vector of zero magnitude ($|\vec{0}| = 0$) with an indeterminate or arbitrary direction.
3. Graphical Methods of Vector Addition:
- Triangle Law: If two vectors $\vec{A}$ and $\vec{B}$ are represented in magnitude and direction by two sides of a triangle taken head-to-tail in order, their resultant $\vec{R} = \vec{A} + \vec{B}$ is represented by the closing side drawn from the tail of the first vector to the head of the second.
- Polygon Law: For adding $n$ vectors $\vec{A}_1, \vec{A}_2, \dots, \vec{A}_n$, arrange them head-to-tail; the vector closing from the origin tail to the final head is the net resultant. If the vectors form a closed polygon, their resultant is strictly zero ($\sum \vec{A}_i = \vec{0}$).
4. Analytical Method: Parallelogram Law of Vector Addition:
If two vectors $\vec{A}$ and $\vec{B}$ acting at a common point are inclined at an angle $\theta$ ($0 \le \theta \le 180^\circ$):
- Magnitude of Resultant ($\vec{R} = \vec{A} + \vec{B}$):
$R = |\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ - Direction of Resultant (angle $\alpha$ with vector $\vec{A}$):
$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$ - Special Angle Conditions:
- $\theta = 0^\circ$ (Collinear, Parallel): $R_{\text{max}} = A + B$, $\alpha = 0^\circ$.
- $\theta = 180^\circ$ (Antiparallel): $R_{\text{min}} = |A – B|$, directed along the larger vector.
- $\theta = 90^\circ$ (Perpendicular): $R = \sqrt{A^2 + B^2}$, $\tan\alpha = \frac{B}{A}$.
- Equal Magnitudes ($A = B$):
$R = 2A\cos\left(\frac{\theta}{2}\right), \quad \alpha = \frac{\theta}{2}$
Special JEE Case: If $\theta = 120^\circ$, then $R = 2A\cos(60^\circ) = 2A(1/2) = A$.
5. Vector Subtraction:
The difference vector $\vec{S} = \vec{A} – \vec{B}$ is defined as the addition of $-\vec{B}$ to $\vec{A}$:
$\vec{S} = \vec{A} + (-\vec{B})$
Because the angle between $\vec{A}$ and $-\vec{B}$ is $(180^\circ – \theta)$:
- Magnitude of Difference:
$S = |\vec{A} – \vec{B}| = \sqrt{A^2 + B^2 – 2AB\cos\theta}$ - Direction (angle $\beta$ with vector $\vec{A}$):
$\tan\beta = \frac{B\sin\theta}{A – B\cos\theta}$ - Equal Magnitudes ($A = B$):
$|\vec{A} – \vec{B}| = 2A\sin\left(\frac{\theta}{2}\right)$ - Orthogonality Condition:
$|\vec{A} + \vec{B}| = |\vec{A} – \vec{B}| \iff \cos\theta = 0 \iff \theta = 90^\circ$
Solved Examples
Example 1 (Resultant and Difference of Two Forces):
Two forces $\vec{F}_1$ and $\vec{F}_2$ of magnitudes $F_1 = 6\text{ N}$ and $F_2 = 10\text{ N}$ act at an angle $\theta = 60^\circ$.
Calculate:
(a) The magnitude of the resultant force $\vec{R} = \vec{F}_1 + \vec{F}_2$.
(b) The direction of the resultant relative to the $6\text{ N}$ force.
(c) The magnitude of the difference vector $\vec{S} = \vec{F}_1 – \vec{F}_2$.
Solution:
(a) Resultant magnitude:
$R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2\cos(60^\circ)} = \sqrt{6^2 + 10^2 + 2(6)(10)\left(\frac{1}{2}\right)} = \sqrt{36 + 100 + 60} = \sqrt{196} = 14\text{ N}$.
(b) Direction with $\vec{F}_1$ ($6\text{ N}$):
$\tan\alpha = \frac{F_2\sin(60^\circ)}{F_1 + F_2\cos(60^\circ)} = \frac{10 \times \frac{\sqrt{3}}{2}}{6 + 10 \times \frac{1}{2}} = \frac{5\sqrt{3}}{6 + 5} = \frac{5\sqrt{3}}{11} \approx 0.7873$
$\alpha = \arctan\left(\frac{5\sqrt{3}}{11}\right) \approx 38.2^\circ$.
(c) Magnitude of difference:
$S = \sqrt{F_1^2 + F_2^2 – 2F_1 F_2\cos(60^\circ)} = \sqrt{36 + 100 – 60} = \sqrt{76} = 2\sqrt{19}\text{ N} \approx 8.72\text{ N}$.
Example 2 (Resultant Perpendicular to the Smaller Vector):
The sum of magnitudes of two forces acting at a point is $16\text{ N}$. Their resultant has a magnitude of $8\text{ N}$ and is perpendicular to the smaller force. Find the magnitude of the two forces and the angle between them.
Solution:
Let the smaller force be $A$ and the larger force be $B$.
Given: $A + B = 16 \implies B = 16 – A$.
Resultant $R = 8\text{ N}$ is perpendicular to $\vec{A}$ ($\alpha = 90^\circ$).
$\tan(90^\circ) = \frac{B\sin\theta}{A + B\cos\theta} \to \infty \implies A + B\cos\theta = 0 \implies \cos\theta = -\frac{A}{B}$.
From the resultant formula:
$R^2 = A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 + 2AB\left(-\frac{A}{B}\right) = A^2 + B^2 – 2A^2 = B^2 – A^2$.
$(B – A)(B + A) = R^2$
Substitute $B + A = 16$ and $R = 8$:
$(B – A)(16) = 8^2 = 64 \implies B – A = \frac{64}{16} = 4\text{ N}$.
We have two simultaneous equations:
$B + A = 16$
$B – A = 4$
Adding: $2B = 20 \implies B = 10\text{ N}$.
Subtracting: $2A = 12 \implies A = 6\text{ N}$.
Angle between them: $\cos\theta = -\frac{A}{B} = -\frac{6}{10} = -0.6 \implies \theta = 180^\circ – 53.1^\circ = 126.9^\circ = \arccos(-0.6)$.
Example 3 (Change in Velocity Vector in Uniform Circular Motion):
A particle moves around a circle of radius $R$ with constant speed $v$.
(a) Derive a general formula for the magnitude of change in velocity $|\Delta\vec{v}|$ when the particle sweeps through a center angle $\theta$.
(b) Evaluate $|\Delta\vec{v}|$ for $\theta = 60^\circ$, $90^\circ$, $120^\circ$, and $180^\circ$.
Solution:
(a) Let initial velocity be $\vec{v}_1$ and final velocity be $\vec{v}_2$.
Since speed is constant, $|\vec{v}_1| = |\vec{v}_2| = v$.
The angle between velocity vectors $\vec{v}_1$ and $\vec{v}_2$ equals the angle turned by the radius vector, $\theta$.
Change in velocity: $\Delta\vec{v} = \vec{v}_2 – \vec{v}_1$.
Using vector subtraction formula for equal magnitudes:
$|\Delta\vec{v}| = \sqrt{v^2 + v^2 – 2v^2\cos\theta} = \sqrt{2v^2(1 – \cos\theta)} = \sqrt{2v^2\left(2\sin^2\frac{\theta}{2}\right)} = 2v\sin\left(\frac{\theta}{2}\right)$.
(b) Evaluating for specific angles:
– $\theta = 60^\circ$: $|\Delta\vec{v}| = 2v\sin(30^\circ) = 2v\left(\frac{1}{2}\right) = v$.
– $\theta = 90^\circ$: $|\Delta\vec{v}| = 2v\sin(45^\circ) = 2v\left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}v$.
– $\theta = 120^\circ$: $|\Delta\vec{v}| = 2v\sin(60^\circ) = 2v\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}v$.
– $\theta = 180^\circ$ (Half revolution): $|\Delta\vec{v}| = 2v\sin(90^\circ) = 2v$.
Example 4 (Coplanar Vector Equilibrium & Triangle Geometry):
Three forces $\vec{A}, \vec{B}, \vec{C}$ acting at a point are in static equilibrium ($\vec{A} + \vec{B} + \vec{C} = \vec{0}$). Their magnitudes are $A = 5\text{ N}$, $B = 12\text{ N}$, and $C = 13\text{ N}$. Find the angle between $\vec{A}$ and $\vec{B}$.
Solution:
From equilibrium: $\vec{A} + \vec{B} = -\vec{C}$.
Taking the magnitude squared on both sides:
$|\vec{A} + \vec{B}|^2 = |-\vec{C}|^2 = C^2$
$A^2 + B^2 + 2AB\cos\theta = C^2$
Substitute $A = 5, B = 12, C = 13$:
$5^2 + 12^2 + 2(5)(12)\cos\theta = 13^2$
$25 + 144 + 120\cos\theta = 169$
$169 + 120\cos\theta = 169 \implies 120\cos\theta = 0 \implies \cos\theta = 0 \implies \theta = 90^\circ$.
The angle between $\vec{A}$ and $\vec{B}$ is exactly $90^\circ$ (they form a right-angled triangle since $5^2 + 12^2 = 13^2$).
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
If the magnitude of the sum of two vectors is equal to the magnitude of their difference, i.e., $|\vec{A} + \vec{B}| = |\vec{A} – \vec{B}|$, the angle between $\vec{A}$ and $\vec{B}$ is:
(A) $0^\circ$
(B) $45^\circ$
(C) $90^\circ$
(D) $180^\circ$
Problem 2 (JEE Main – Single Correct):
Two vectors of equal magnitude $F$ have a resultant whose magnitude is also equal to $F$. The angle between the two vectors is:
(A) $60^\circ$
(B) $90^\circ$
(C) $120^\circ$
(D) $150^\circ$
Problem 3 (JEE Main – Single Correct):
What is the minimum number of coplanar vectors of unequal magnitudes whose vector sum can be equal to zero?
(A) 2
(B) 3
(C) 4
(D) 5
Problem 4 (JEE Main – Single Correct):
If vectors $\vec{A}$ and $\vec{B}$ satisfy $\vec{A} + \vec{B} = \vec{C}$ and their scalar magnitudes satisfy $A + B = C$, then the angle between $\vec{A}$ and $\vec{B}$ is:
(A) $0^\circ$
(B) $45^\circ$
(C) $90^\circ$
(D) $180^\circ$
Problem 5 (JEE Main – Single Correct):
A car travels East at $40\text{ km/h}$ and turns sharply North to travel at $30\text{ km/h}$. The magnitude of the change in velocity of the car is:
(A) $10\text{ km/h}$
(B) $50\text{ km/h}$
(C) $70\text{ km/h}$
(D) $25\text{ km/h}$
Problem 6 (JEE Advanced – One or More Correct):
For any two non-zero vectors $\vec{A}$ and $\vec{B}$, which of the following inequalities is/are mathematically correct?
(A) $|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$
(B) $|\vec{A} + \vec{B}| \ge ||\vec{A}| – |\vec{B}||$
(C) $|\vec{A} – \vec{B}| \le |\vec{A}| + |\vec{B}|$
(D) $|\vec{A} – \vec{B}| \ge ||\vec{A}| – |\vec{B}||$
Problem 7 (JEE Advanced – One or More Correct):
The resultant of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If $\vec{Q}$ is doubled, the new resultant is perpendicular to $\vec{P}$. Which of the following statements is/are correct?
(A) The angle $\theta$ between $\vec{P}$ and $\vec{Q}$ satisfies $\cos\theta = -\frac{P}{2Q}$.
(B) The magnitude of $P$ must be less than $2Q$.
(C) The angle $\theta$ is an obtuse angle ($\theta > 90^\circ$).
(D) The new resultant magnitude is $\sqrt{4Q^2 – P^2}$.
Problem 8 (JEE Advanced – One or More Correct):
Which of the following physical quantities is/are axial vectors (pseudovectors)?
(A) Linear Momentum
(B) Angular Velocity
(C) Torque
(D) Electric Current
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Two forces of magnitudes $3\text{ N}$ and $4\text{ N}$ act at an angle of $90^\circ$. The magnitude of their resultant force in Newtons is $R$. Find $R$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle moves in a circle with a constant speed $v = 15\text{ m/s}$. The magnitude of the change in velocity vector after the particle sweeps through a center angle of $60^\circ$ is $\Delta v\text{ m/s}$. Find the integer value of $\Delta v$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (B) | 4. (A) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (B, C) | 9. 5 | 10. 15
Solution 1:
$|\vec{A} + \vec{B}|^2 = |\vec{A} – \vec{B}|^2$
$A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 – 2AB\cos\theta$
$4AB\cos\theta = 0 \implies \cos\theta = 0 \implies \theta = 90^\circ$.
Correct Answer: (C)
Solution 2:
$R = 2F\cos\left(\frac{\theta}{2}\right)$.
Given $R = F \implies F = 2F\cos\left(\frac{\theta}{2}\right) \implies \cos\left(\frac{\theta}{2}\right) = \frac{1}{2} \implies \frac{\theta}{2} = 60^\circ \implies \theta = 120^\circ$.
Correct Answer: (C)
Solution 3:
Two vectors can cancel only if they are equal and opposite ($A = B$). For vectors of unequal magnitudes, at least 3 coplanar vectors forming a closed triangle are needed to produce a null resultant.
Correct Answer: (B)
Solution 4:
$C = \sqrt{A^2 + B^2 + 2AB\cos\theta}$.
Given $C = A + B \implies C^2 = (A + B)^2 = A^2 + B^2 + 2AB$.
Equating: $A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 + 2AB \implies \cos\theta = 1 \implies \theta = 0^\circ$ (parallel vectors).
Correct Answer: (A)
Solution 5:
$\vec{v}_1 = 40\hat{i}\text{ km/h}$, $\vec{v}_2 = 30\hat{j}\text{ km/h}$.
$\Delta\vec{v} = \vec{v}_2 – \vec{v}_1 = -40\hat{i} + 30\hat{j}$.
$|\Delta\vec{v}| = \sqrt{(-40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ km/h}$.
Correct Answer: (B)
Solution 6:
The generalized triangle inequalities state that the magnitude of the sum or difference of two vectors lies strictly within the bounds $[||A| – |B||, |A| + |B|]$. All four inequalities are mathematically true.
Correct Answer: (A, B, C, D)
Solution 7:
Let new vector be $\vec{Q}’ = 2\vec{Q}$. Resultant $\vec{R}’ = \vec{P} + 2\vec{Q}$ is perpendicular to $\vec{P}$ ($\alpha = 90^\circ$):
$\tan\alpha = \frac{2Q\sin\theta}{P + 2Q\cos\theta} \to \infty \implies P + 2Q\cos\theta = 0 \implies \cos\theta = -\frac{P}{2Q}$ (A is True).
Since $|\cos\theta| \le 1 \implies \frac{P}{2Q} < 1 \implies P < 2Q$ (B is True).
Since $\cos\theta < 0$, $\theta$ is obtuse (C is True).
Magnitude $R’^2 = P^2 + (2Q)^2 + 2P(2Q)\cos\theta = P^2 + 4Q^2 + 4PQ\left(-\frac{P}{2Q}\right) = P^2 + 4Q^2 – 2P^2 = 4Q^2 – P^2 \implies R’ = \sqrt{4Q^2 – P^2}$ (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– Linear momentum is a polar vector.
– Angular velocity ($\vec{\omega}$) and Torque ($\vec{\tau} = \vec{r} \times \vec{F}$) are axial vectors.
– Electric current is a scalar.
Correct Answer: (B, C)
Solution 9:
$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ N}$.
Correct Answer: 5
Solution 10:
$|\Delta\vec{v}| = 2v\sin\left(\frac{\theta}{2}\right) = 2(15)\sin\left(\frac{60^\circ}{2}\right) = 30\sin(30^\circ) = 30 \times \frac{1}{2} = 15\text{ m/s}$.
Correct Answer: 15