Q8:

If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.

Theoretical Foundation & Definition

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin(\theta) \hat{n}$, where $\theta$ is the angle between the two vectors and $\hat{n}$ is a unit vector perpendicular to both. Given the condition $\vec{a} \times \vec{b} = \vec{0}$, we examine the mathematical requirements:

• $|\vec{a}| = 0$ (The zero vector case)
• $|\vec{b}| = 0$ (The zero vector case)
• $\sin(\theta) = 0 \implies \theta = 0^\circ$ or $180^\circ$ (The collinear vector case)

Analyzing the Converse Statement

The converse states: "If $\vec{a} \times \vec{b} = \vec{0}$, then either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$." To test the validity of this converse, we must determine if there exist non-zero vectors $\vec{a}$ and $\vec{b}$ such that their cross product is still the zero vector.

Counter-Example Formulation

Consider two non-zero vectors that are parallel (collinear). Let:
$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$
$\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k}$

Note that $\vec{b} = 2\vec{a}$, which confirms the vectors are linearly dependent and parallel. Thus, the angle $\theta$ between them is $0$.

Calculating the Cross Product

Using the determinant method for the cross product:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{vmatrix}$
$= \hat{i}(3 \times 8 - 4 \times 6) - \hat{j}(2 \times 8 - 4 \times 4) + \hat{k}(2 \times 6 - 3 \times 4)$
$= \hat{i}(24 - 24) - \hat{j}(16 - 16) + \hat{k}(12 - 12)$
$= 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$

Geometric Representation

Logical Conclusion

Because we have identified vectors $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$ such that $\vec{a} \times \vec{b} = \vec{0}$ (due to them being collinear), the condition "either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$" is not strictly necessary for the product to be zero. Therefore, the converse is false.

Final Solution: The converse is not true. If two non-zero vectors are parallel, their cross product is the zero vector, as demonstrated by $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = 4\hat{i} + 6\hat{j} + 8\hat{k}$.