Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive,

a) when the three vectors are coplanar;

b) in the general case

Griffiths problem

Pairoaj

In this problem use the following relationships

$$A \cdot B=AB \cos{\theta}$$

$$A \times B = AB \sin{\theta} \,\hat{n}$$

From the drawn diagram above,

$$\left\lvert B+C \right\rvert \,\cos{\theta_3}=\left\lvert B \right\rvert \cos{\theta_1}+\left\lvert C \right\rvert \cos{\theta_2}$$

$$\rightarrow \left\lvert A \right\rvert \left\lvert B+C \right\rvert \,\cos{\theta_3}=\left\lvert A \right\rvert \left\lvert B \right\rvert \cos{\theta_1}+\left\lvert A \right\rvert \left\lvert C \right\rvert \cos{\theta_2}$$

$$\rightarrow A\cdot(B+C)=A\cdot B+A \cdot C$$

hence the **dot product is distributive**. Similarly,

$$\left\lvert B+C \right\rvert \,\sin{\theta_3}=\left\lvert B \right\rvert \sin{\theta_1}+\left\lvert C \right\rvert \sin{\theta_2}$$

$$\rightarrow \left\lvert A \right\rvert \left\lvert B+C \right\rvert \,\sin{\theta_3}\,\hat{n}=\left\lvert A \right\rvert \left\lvert B \right\rvert \sin{\theta_1}\,\hat{n}+\left\lvert A \right\rvert \left\lvert C \right\rvert \sin{\theta_2}\,\hat{n}$$

$$\rightarrow A \times (B+C)=(A\times B) + (A\times C)$$

and hence the **cross product is distributive**

for the **general case** just calculate the value of

$$A \cdot (B+C)$$

where

$$A = A_x \,\hat{i}+A_y \,\hat{j} + A_z \,\hat{k}$$ $$B = B_x \,\hat{i}+B_y \,\hat{j} + B_z \,\hat{k}$$ $$C = C_x \,\hat{i}+C_y \,\hat{j} + C_z \,\hat{k}$$

for instance,

$$A \cdot (B+C) = (A_x,A_y,A_z) \cdot (B_x+C_x,B_y+C_y+B_z+C_z)$$

$$ = A_x(B_x+C_x)\,\hat{x}+A_y(B_y+C_y)\,\hat{y}+A_z(B_z+C_z)\,\hat{z}$$

$$ = (A_xB_x+A_xC_x)\,\hat{x}+(A_yB_y+A_yC_y)\,\hat{y}+(A_zB_z+A_zC_z)\,\hat{z}$$

now check that it is equal to $A \cdot B+A\cdot C$

the same can be done for the cross product $A \times (B+C) = (A\times B)+(A \times C)$

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