Q.Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, $c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$. Then show that $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$.

Concept: Distributive property of the cross product over vector addition.

Reasoning:

1. Write $\vec{b}+\vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k}$.

2. Compute $\vec{a}\times(\vec{b}+\vec{c})$ using the determinant form:

3. Expand the determinant along the first row:

4. Separate terms containing $\vec{b}$ and $\vec{c}$: …

The distributive property of the cross product is not something to memorise blindly — it follows from the way the cross product is defined. When you see $\vec{a} \times (\vec{b} + \vec{c})$, think: "I am taking the cross product of $\vec{a}$ with the sum of two vectors." The cross product itself is built from the components using determinants, and determinants are linear in each row. That linearity is the real reason distribution works.

Let’s walk through it step by step.

1. Write the vectors in component form

We have:

The sum $\vec{b} + \vec{c}$ is simply:

2. Compute $\vec{a} \times (\vec{b} + \vec{c})$ using the determinant formula

The cross product of $\vec{a}$ with any vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ is:

So for $\vec{v} = \vec{b} + \vec{c}$, we get:

Expanding the determinant along the first row:

3. Use the linearity of the determinant in each row

Each $2 \times 2$ determinant splits because the entries in the second row are sums. For example:

That is exactly:

The same holds for the other two determinants. So the whole expression becomes:

4. Separate into two cross products

Group the terms involving $\vec{b}$ together and those involving $\vec{c}$ together: …

Method: Proving the Cross-Product Distributive Law by Components

To prove a cross-product identity that must hold for all vectors, expand both sides in components and use the linearity of the determinant.

Steps

Step 1: Write everything in components.

Express $\vec{a},\vec{b},\vec{c}$ by their components and form $\vec{b}+\vec{c}$ componentwise.

Step 2: Set up the determinant with the summed row. …

Common Mistakes

Mistake 1: Assuming the property instead of proving it.

Why it's wrong: the question asks to show distributivity, so quoting it is not enough. Correct approach: expand both sides in components (or via the determinant) and compare.

Mistake 2: Swapping the order while 'distributing'. …