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System of Particles and Rotational Motion - Vector Product of Two Vectors

Grade 11CBSEPhysics

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

🔑Concepts

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The vector product (or cross product) of two vectors A⃗\vec{A} and B⃗\vec{B} is a vector C⃗\vec{C} such that its magnitude is C=ABsin⁡θC = AB \sin \theta, where θ\theta is the angle between the two vectors.

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The direction of the resulting vector C⃗\vec{C} is perpendicular to the plane containing A⃗\vec{A} and B⃗\vec{B} and is determined by the Right-Hand Thumb Rule.

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The vector product is non-commutative, meaning A⃗×B⃗=−(B⃗×A⃗)\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}).

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The vector product follows the distributive law: A⃗×(B⃗+C⃗)=A⃗×B⃗+A⃗×C⃗\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}.

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For parallel or anti-parallel vectors, the vector product is a null vector because sin⁡(0∘)=sin⁡(180∘)=0\sin(0^\circ) = \sin(180^\circ) = 0.

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The magnitude of the vector product ∣A⃗×B⃗∣|\vec{A} \times \vec{B}| represents the area of the parallelogram formed with A⃗\vec{A} and B⃗\vec{B} as adjacent sides.

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In terms of unit vectors, the cross product follows a cyclic order: i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, and k^×i^=j^\hat{k} \times \hat{i} = \hat{j}.

📐Formulae

A⃗×B⃗=(ABsin⁡θ)n^\vec{A} \times \vec{B} = (AB \sin \theta) \hat{n}

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}

i^×i^=j^×j^=k^×k^=0\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0

i^×j^=k^,j^×k^=i^,k^×i^=j^\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}

Area of a triangle=12∣A⃗×B⃗∣\text{Area of a triangle} = \frac{1}{2} |\vec{A} \times \vec{B}|

sin⁡θ=∣A⃗×B⃗∣∣A⃗∣∣B⃗∣\sin \theta = \frac{|\vec{A} \times \vec{B}|}{|\vec{A}| |\vec{B}|}

💡Examples

Problem 1:

Find the vector product of A⃗=2i^+3j^+k^\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k} and B⃗=i^−j^+2k^\vec{B} = \hat{i} - \hat{j} + 2\hat{k}.

Solution:

A⃗×B⃗=∣i^j^k^2311−12∣\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{vmatrix} =i^(3×2−1×(−1))−j^(2×2−1×1)+k^(2×(−1)−3×1)= \hat{i}(3 \times 2 - 1 \times (-1)) - \hat{j}(2 \times 2 - 1 \times 1) + \hat{k}(2 \times (-1) - 3 \times 1) =i^(6+1)−j^(4−1)+k^(−2−3)= \hat{i}(6 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 - 3) =7i^−3j^−5k^= 7\hat{i} - 3\hat{j} - 5\hat{k}

Explanation:

To find the cross product, we use the determinant method with unit vectors in the first row, components of A⃗\vec{A} in the second, and components of B⃗\vec{B} in the third. We expand the determinant to find the resulting vector.

Problem 2:

Calculate the area of a parallelogram whose adjacent sides are given by the vectors P⃗=3i^+4j^\vec{P} = 3\hat{i} + 4\hat{j} and Q⃗=−3i^+7j^\vec{Q} = -3\hat{i} + 7\hat{j}.

Solution:

First, find P⃗×Q⃗\vec{P} \times \vec{Q}: P⃗×Q⃗=∣i^j^k^340−370∣\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 0 \\ -3 & 7 & 0 \end{vmatrix} =i^(0)−j^(0)+k^(21−(−12))=33k^= \hat{i}(0) - \hat{j}(0) + \hat{k}(21 - (-12)) = 33\hat{k} The magnitude is: ∣P⃗×Q⃗∣=02+02+332=33 sq. units|\vec{P} \times \vec{Q}| = \sqrt{0^2 + 0^2 + 33^2} = 33 \text{ sq. units}

Explanation:

The area of a parallelogram is equal to the magnitude of the cross product of the two vectors representing its adjacent sides. Since both vectors are in the xyxy-plane (no zz component), the cross product only has a kk component.

Vector Product of Two Vectors Class 11 Notes & Examples