Define cross product of two vectors and interpret the product geometrically. If a , b , c are three non zero vectors, prove that: [6] a ( b + c) = a b + a c

Definition of Cross Product

Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors. The cross product (or vector product) of $\vec{a}$ and $\vec{b}$, denoted by $\vec{a}\times \vec{b}$, is defined as a vector $\vec{c}$ such that:

1. Magnitude: The magnitude of $\vec{c}$ is given by: $$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0\le \theta \le \pi$).

2. Direction: The direction of $\vec{c}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. It is determined by the Right-Hand Rule: if the fingers of the right hand curl from $\vec{a}$ to $\vec{b}$ through the angle $\theta$, the thumb points in the direction of $\vec{a}\times \vec{b}$.

If $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, then: $$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\hat{i}-(a_1{b}_3-a_3{b}_1)\hat{j}+(a_1{b}_2-a_2{b}_1)\hat{k}$$

Geometrical Interpretation

The cross product has two primary geometrical interpretations:

1. Area of a Parallelogram: The magnitude of the cross product, $|\vec{a}\times \vec{b}|$, represents the area of the parallelogram formed by the vectors $\vec{a}$ and $\vec{b}$ as adjacent sides. $$\text{Area}=|\vec{a}||\vec{b}|\sin \theta$$ Consequently, the area of the triangle formed by these vectors is half of this value: $\frac{1}{2}|\vec{a}\times \vec{b}|$.

2. Normal Vector: The vector $\vec{a}\times \vec{b}$ is normal (perpendicular) to the plane containing $\vec{a}$ and $\vec{b}$. This property is frequently used to find the equation of a plane passing through three non-collinear points.

Proof of Distributive Law

To Prove: $\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$

Proof:

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, and $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$.

Step 1: Evaluate the Left-Hand Side (LHS)

First, determine the vector $\vec{b}+\vec{c}$: $$\vec{b}+\vec{c}=(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$$

Now, compute the cross product $\vec{a}\times (\vec{b}+\vec{c})$: $$\text{LHS}=\vec{a}\times (\vec{b}+\vec{c})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1+c_1& b_2+c_2& b_3+c_3\end{array}\right|$$

Expanding the determinant along the first row: $$\text{LHS}=\hat{i}\left[a_2(b_3+c_3)-a_3(b_2+c_2)\right]-\hat{j}\left[a_1(b_3+c_3)-a_3(b_1+c_1)\right]+\hat{k}\left[a_1(b_2+c_2)-a_2(b_1+c_1)\right]$$

Distribute the terms inside the brackets: $$\text{LHS}=\hat{i}[(a_2{b}_3+a_2{c}_3)-(a_3{b}_2+a_3{c}_2)]-\hat{j}[(a_1{b}_3+a_1{c}_3)-(a_3{b}_1+a_3{c}_1)]+\hat{k}[(a_1{b}_2+a_1{c}_2)-(a_2{b}_1+a_2{c}_1)]$$

Group the terms containing $b$ and the terms containing $c$: $$\text{LHS}=\hat{i}[(a_2{b}_3-a_3{b}_2)+(a_2{c}_3-a_3{c}_2)]-\hat{j}[(a_1{b}_3-a_3{b}_1)+(a_1{c}_3-a_3{c}_1)]+\hat{k}[(a_1{b}_2-a_2{b}_1)+(a_1{c}_2-a_2{c}_1)]$$

Separate the expression into two parts, one involving $\vec{b}$ and one involving $\vec{c}$: $$\text{LHS}=\left[(a_2{b}_3-a_3{b}_2)\hat{i}-(a_1{b}_3-a_3{b}_1)\hat{j}+(a_1{b}_2-a_2{b}_1)\hat{k}\right]+\left[(a_2{c}_3-a_3{c}_2)\hat{i}-(a_1{c}_3-a_3{c}_1)\hat{j}+(a_1{c}_2-a_2{c}_1)\hat{k}\right]$$

Step 2: Evaluate the Right-Hand Side (RHS)

Compute $\vec{a}\times \vec{b}$: $$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|=(a_2{b}_3-a_3{b}_2)\hat{i}-(a_1{b}_3-a_3{b}_1)\hat{j}+(a_1{b}_2-a_2{b}_1)\hat{k}$$

Compute $\vec{a}\times \vec{c}$: $$\vec{a}\times \vec{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ c_1& c_2& c_3\end{array}\right|=(a_2{c}_3-a_3{c}_2)\hat{i}-(a_1{c}_3-a_3{c}_1)\hat{j}+(a_1{c}_2-a_2{c}_1)\hat{k}$$

Add the two results: $$\text{RHS}=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$$ $$\text{RHS}=\left[(a_2{b}_3-a_3{b}_2)\hat{i}-(a_1{b}_3-a_3{b}_1)\hat{j}+(a_1{b}_2-a_2{b}_1)\hat{k}\right]+\left[(a_2{c}_3-a_3{c}_2)\hat{i}-(a_1{c}_3-a_3{c}_1)\hat{j}+(a_1{c}_2-a_2{c}_1)\hat{k}\right]$$

Step 3: Comparison

Comparing the expanded forms of LHS and RHS derived in Step 1 and Step 2, we observe that they are identical term by term.

Therefore, $$\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$$

Hence Proved.