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Kinematics - Vectors

Grade 11ICSEPhysics

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

🔑Concepts

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A Scalar quantity has only magnitude (e.g., mass, speed), while a Vector quantity has both magnitude and direction and follows vector laws of addition (e.g., velocity, force).

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A Unit Vector is a vector with a magnitude of 11 and is used to specify direction. In Cartesian coordinates, these are represented as i^\hat{i}, j^\hat{j}, and k^\hat{k} for the xx, yy, and zz axes respectively.

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The Triangle Law of Vector Addition states that if two vectors are represented by two sides of a triangle in order, their resultant is represented by the third side taken in the opposite order.

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

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Resolution of a Vector: A vector A⃗\vec{A} can be resolved into rectangular components: Ax=Acos⁡θA_x = A\cos\theta and Ay=Asin⁡θA_y = A\sin\theta, where θ\theta is the angle made with the xx-axis.

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Dot Product (Scalar Product): The product A⃗⋅B⃗=ABcos⁡θ\vec{A} \cdot \vec{B} = AB\cos\theta results in a scalar. It is commutative: A⃗⋅B⃗=B⃗⋅A⃗\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}.

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Cross Product (Vector Product): The product A⃗×B⃗=(ABsin⁡θ)n^\vec{A} \times \vec{B} = (AB\sin\theta) \hat{n} results in a vector perpendicular to the plane containing A⃗\vec{A} and B⃗\vec{B}. It is non-commutative: A⃗×B⃗=−(B⃗×A⃗)\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}).

📐Formulae

∣R⃗∣=A2+B2+2ABcos⁡θ|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}

tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}

a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}

A⃗⋅B⃗=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z

cos⁡θ=A⃗⋅B⃗∣A⃗∣∣B⃗∣\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}

A⃗×B⃗=∣i^j^k^AxAyAzBxByBz∣\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}

v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

💡Examples

Problem 1:

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

Solution:

Given A=3 NA = 3\,N, B=4 NB = 4\,N, and θ=90∘\theta = 90^\circ. The magnitude is R=32+42+2(3)(4)cos⁡90∘=9+16+0=5 NR = \sqrt{3^2 + 4^2 + 2(3)(4)\cos 90^\circ} = \sqrt{9 + 16 + 0} = 5\,N. The direction is tan⁡α=4sin⁡90∘3+4cos⁡90∘=43\tan\alpha = \frac{4\sin 90^\circ}{3 + 4\cos 90^\circ} = \frac{4}{3}. Therefore, α=tan⁡−1(1.33)≈53.13∘\alpha = \tan^{-1}(1.33) \approx 53.13^\circ.

Explanation:

Since the vectors are perpendicular, the parallelogram law simplifies to the Pythagorean theorem for magnitude, and the tangent of the angle is the ratio of the components.

Problem 2:

Find the unit vector parallel to the vector A⃗=3i^−4j^\vec{A} = 3\hat{i} - 4\hat{j}.

Solution:

First, find the magnitude: ∣A⃗∣=32+(−4)2=9+16=5|\vec{A}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5. The unit vector is A^=A⃗∣A⃗∣=3i^−4j^5=0.6i^−0.8j^\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{3\hat{i} - 4\hat{j}}{5} = 0.6\hat{i} - 0.8\hat{j}.

Explanation:

A unit vector is obtained by dividing the original vector by its own magnitude, ensuring the new vector has a magnitude of 11 while preserving direction.

Problem 3:

Calculate the work done if a force F⃗=(2i^+3j^) N\vec{F} = (2\hat{i} + 3\hat{j})\,N causes a displacement s⃗=(5i^+4j^) m\vec{s} = (5\hat{i} + 4\hat{j})\,m.

Solution:

Work done is the dot product of force and displacement: W=F⃗⋅s⃗=(2)(5)+(3)(4)=10+12=22 JW = \vec{F} \cdot \vec{s} = (2)(5) + (3)(4) = 10 + 12 = 22\,J.

Explanation:

Work is a scalar quantity calculated using the dot product formula AxBx+AyByA_x B_x + A_y B_y.

Vectors - Revision Notes & Key Formulas | ICSE Class 11 Physics