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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

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

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

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A Unit Vector is a vector of unit magnitude in a specific direction, defined as A^=A⃗∣A⃗∣\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.

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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: R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}.

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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.

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

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The Scalar (Dot) Product of two vectors A⃗\vec{A} and B⃗\vec{B} is a scalar quantity defined as A⃗⋅B⃗=ABcos⁡θ\vec{A} \cdot \vec{B} = AB \cos \theta. It is used to find the work done W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}.

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The Vector (Cross) Product of two vectors A⃗\vec{A} and B⃗\vec{B} is a vector C⃗=A⃗×B⃗\vec{C} = \vec{A} \times \vec{B} with magnitude ABsin⁡θAB \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+2ABcos⁡θ\text{Magnitude of Resultant: } R = \sqrt{A^2 + B^2 + 2AB \cos \theta}

Direction of Resultant: tan⁡α=Bsin⁡θA+Bcos⁡θ\text{Direction of Resultant: } \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}

Unit Vector: A^=A⃗∣A⃗∣=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: A⃗⋅B⃗=AxBx+AyBy+AzBz=∣A⃗∣∣B⃗∣cos⁡θ\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: A⃗×B⃗=∣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: v⃗AB=v⃗A−v⃗B\text{Relative Velocity: } \vec{v}_{AB} = \vec{v}_A - \vec{v}_B

💡Examples

Problem 1:

Two forces of magnitude 3 N3\,N and 4 N4\,N act on a body at an angle of 90∘90^\circ to each other. Find the magnitude of the resultant force.

Solution:

R=F12+F22+2F1F2cos⁡θR = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} R=32+42+2(3)(4)cos⁡90∘R = \sqrt{3^2 + 4^2 + 2(3)(4) \cos 90^\circ} R=9+16+0=25=5 NR = \sqrt{9 + 16 + 0} = \sqrt{25} = 5\,N

Explanation:

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

Problem 2:

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

Solution:

A⃗⋅B⃗=(3)(4)+(4)(−3)=12−12=0\vec{A} \cdot \vec{B} = (3)(4) + (4)(-3) = 12 - 12 = 0 0=∣A⃗∣∣B⃗∣cos⁡θ0 = |\vec{A}||\vec{B}| \cos \theta cos⁡θ=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.