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Motion in a Plane - Scalars and Vectors

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Scalars are physical quantities that have only magnitude and no direction. Examples include mass (mm), time (tt), and temperature (TT).

Vectors are physical quantities that have both magnitude and direction and obey the laws of vector addition. Examples include displacement (s\vec{s}), velocity (v\vec{v}), and force (F\vec{F}).

A Unit Vector is a vector of unit magnitude in a specific direction, defined as A^=AA\hat{A} = \frac{\vec{A}}{|\vec{A}|}. The unit vectors along the x,y,zx, y, z axes are i^,j^,k^\hat{i}, \hat{j}, \hat{k} respectively.

The Triangle Law of Vector Addition states that if two vectors are represented by two sides of a triangle taken in the same order, their resultant is represented by the third side taken in the opposite order: R=A+B\vec{R} = \vec{A} + \vec{B}.

The Parallelogram Law of Vector Addition states that if two vectors acting at a point are represented by the adjacent sides of a parallelogram, the diagonal passing through their point of intersection represents the resultant.

Resolution of a vector involves splitting a vector into its components. For a vector A\vec{A} making an angle θ\theta with the xx-axis, the components are Ax=AcosθA_x = A \cos \theta and Ay=AsinθA_y = A \sin \theta.

The Scalar (Dot) Product of two vectors A\vec{A} and B\vec{B} is a scalar quantity defined as AB=ABcosθ\vec{A} \cdot \vec{B} = AB \cos \theta. It is used to find the work done W=FdW = \vec{F} \cdot \vec{d}.

The Vector (Cross) Product of two vectors A\vec{A} and B\vec{B} is a vector C=A×B\vec{C} = \vec{A} \times \vec{B} with magnitude ABsinθAB \sin \theta and direction perpendicular to the plane containing both vectors, determined by the Right-Hand Thumb Rule.

📐Formulae

Magnitude of Resultant: R=A2+B2+2ABcosθ\text{Magnitude of Resultant: } R = \sqrt{A^2 + B^2 + 2AB \cos \theta}

Direction of Resultant: tanα=BsinθA+Bcosθ\text{Direction of Resultant: } \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}

Unit Vector: A^=AA=Axi^+Ayj^+Azk^Ax2+Ay2+Az2\text{Unit Vector: } \hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{A_x\hat{i} + A_y\hat{j} + A_z\hat{k}}{\sqrt{A_x^2 + A_y^2 + A_z^2}}

Scalar Product: AB=AxBx+AyBy+AzBz=ABcosθ\text{Scalar Product: } \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z = |\vec{A}||\vec{B}| \cos \theta

Vector Product: A×B=i^j^k^AxAyAzBxByBz\text{Vector Product: } \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}

Relative Velocity: vAB=vAvB\text{Relative Velocity: } \vec{v}_{AB} = \vec{v}_A - \vec{v}_B

💡Examples

Problem 1:

Two forces of magnitude 3N3\,N and 4N4\,N act on a body at an angle of 9090^\circ to each other. Find the magnitude of the resultant force.

Solution:

R=F12+F22+2F1F2cosθR = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} R=32+42+2(3)(4)cos90R = \sqrt{3^2 + 4^2 + 2(3)(4) \cos 90^\circ} R=9+16+0=25=5NR = \sqrt{9 + 16 + 0} = \sqrt{25} = 5\,N

Explanation:

Since the forces are perpendicular, cos90=0\cos 90^\circ = 0, reducing the formula to the Pythagorean theorem.

Problem 2:

Find the angle between two vectors A=3i^+4j^\vec{A} = 3\hat{i} + 4\hat{j} and B=4i^3j^\vec{B} = 4\hat{i} - 3\hat{j}.

Solution:

AB=(3)(4)+(4)(3)=1212=0\vec{A} \cdot \vec{B} = (3)(4) + (4)(-3) = 12 - 12 = 0 0=ABcosθ0 = |\vec{A}||\vec{B}| \cos \theta cosθ=0    θ=90\cos \theta = 0 \implies \theta = 90^\circ

Explanation:

The dot product of two non-zero vectors is zero only when the vectors are orthogonal (perpendicular) to each other.

Scalars and Vectors - Revision Notes & Key Formulas | CBSE Class 11 Physics