Define vector product of two vectors. Also prove using vector method Sin (A+B) = SinA CosB + CosA SinB . [6]

Definition of Vector Product

The vector product (or cross product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as a vector $\vec{C}=\vec{A}\times \vec{B}$ whose magnitude is given by: $$|\vec{C}|=|\vec{A}||\vec{B}|\sin \theta$$ where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$.

The direction of $\vec{C}$ is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$, determined by the right-hand rule: if the fingers of the right hand curl from $\vec{A}$ to $\vec{B}$ through the smaller angle, the thumb points in the direction of $\vec{A}\times \vec{B}$.

In component form, if $\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$ and $\vec{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$, then: $$\vec{A}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$$

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Proof: $\sin (A+B)=\sin A\cos B+\cos A\sin B$

Given: Two vectors $\vec{u}$ and $\vec{v}$ with magnitudes $|\vec{u}|=1$ and $|\vec{v}|=1$.

Assumption: Let $\vec{u}$ make an angle $A$ with the positive x-axis, and $\vec{v}$ make an angle $B$ with the positive x-axis.

Step 1: Express vectors in component form

Since $|\vec{u}|=1$: $$\vec{u}=\cos A\hat{i}+\sin A\hat{j}$$

Since $|\vec{v}|=1$: $$\vec{v}=\cos B\hat{i}+\sin B\hat{j}$$

Step 2: Compute the cross product $\vec{u}\times \vec{v}$

$$\vec{u}\times \vec{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \cos A& \sin A& 0\\ \cos B& \sin B& 0\end{array}\right|$$

Expanding along the third row (or using the determinant formula):

$$\vec{u}\times \vec{v}=\hat{k}\left(\cos A\sin B-\sin A\cos B\right)$$

So the magnitude of the cross product is: $$|\vec{u}\times \vec{v}|=|\cos A\sin B-\sin A\cos B|$$

Step 3: Compute the cross product $\vec{v}\times \vec{u}$

$$\vec{v}\times \vec{u}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \cos B& \sin B& 0\\ \cos A& \sin A& 0\end{array}\right|$$

$$\vec{v}\times \vec{u}=\hat{k}\left(\cos B\sin A-\sin B\cos A\right)$$

So the magnitude is: $$|\vec{v}\times \vec{u}|=|\sin A\cos B-\cos A\sin B|$$

Step 4: Use the geometric definition of cross product

The angle between $\vec{u}$ and $\vec{v}$ is $(A-B)$ if $A>B$, or more generally, the angle between them is $|A-B|$. However, we need the angle for $\sin (A+B)$.

Let us instead consider two unit vectors:

• $\vec{p}$ making angle $A$ with the x-axis: $\vec{p}=\cos A\hat{i}+\sin A\hat{j}$
• $\vec{q}$ making angle $-(B)$ with the x-axis: $\vec{q}=\cos B\hat{i}-\sin B\hat{j}$

The angle between $\vec{p}$ and $\vec{q}$ is $A+B$.

Compute $\vec{p}\times \vec{q}$:

$$\vec{p}\times \vec{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \cos A& \sin A& 0\\ \cos B& -\sin B& 0\end{array}\right|$$

$$\vec{p}\times \vec{q}=\hat{k}\left(\cos A(-\sin B)-\sin A\cos B\right)$$

$$\vec{p}\times \vec{q}=\hat{k}\left(-\cos A\sin B-\sin A\cos B\right)$$

$$\vec{p}\times \vec{q}=-\hat{k}\left(\sin A\cos B+\cos A\sin B\right)$$

Step 5: Equate with geometric definition

By the geometric definition: $$|\vec{p}\times \vec{q}|=|\vec{p}||\vec{q}|\sin (A+B)=1\cdot 1\cdot \sin (A+B)=\sin (A+B)$$

From the component calculation: $$|\vec{p}\times \vec{q}|=|\sin A\cos B+\cos A\sin B|$$

Since $A$ and $B$ are angles in the standard range where $\sin (A+B)\ge 0$:

$$\sin (A+B)=\sin A\cos B+\cos A\sin B$$

Step 6: Verification with a figure

The vector method confirms the identity by showing that the magnitude of the cross product of two unit vectors separated by angle $(A+B)$ equals $\sin (A+B)$, while the component calculation yields $\sin A\cos B+\cos A\sin B$.

Therefore, $\sin (A+B)=\sin A\cos B+\cos A\sin B$.