Q.If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0.$ But the converse need not be true. Justify your answer with an example.

Concept: Vector Magnitude Properties — the dot product being zero does not imply either vector is zero; it only implies orthogonality.

Step 1: If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then by definition $\vec{a} \cdot \vec{b} = 0$ because the zero vector has zero magnitude.

Step 2: The converse says: if $\vec{a} \cdot \vec{b} = 0$, then $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$. This is false because the dot product is zero whenever the vectors are perpendicular, regardless of their magnitudes. …

The core idea here is about the meaning of the dot product. The dot product $\vec{a} \cdot \vec{b}$ measures how much two vectors point in the same direction. If either vector is zero, there's nothing to measure — the product is automatically zero. That part is true.

But the converse says: "If the dot product is zero, then at least one vector must be zero." That is not true. Why? Because the dot product can also be zero when two non-zero vectors are perpendicular (orthogonal). In that case, they have zero "overlap" in direction, even though both are perfectly fine non-zero vectors.

Let's walk through this carefully.

1. Recall the geometric definition of the dot product. For any two vectors $\vec{a}$ and $\vec{b}$,

where $\theta$ is the angle between them.

If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $|\vec{a}| = 0$ or $|\vec{b}| = 0$, so the product is $0$. That's the given "if" part — correct.

2. Now examine the converse.

   The converse claims: $\vec{a} \cdot \vec{b} = 0 \implies \vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.

   But from the formula, $\vec{a} \cdot \vec{b} = 0$ can also happen when $\cos \theta = 0$, i.e., when $\theta = 90^\circ$ (or $270^\circ$, etc.). That means the vectors are perpendicular, and neither needs to be zero.

3. Construct a concrete counterexample.

   Take any two non-zero perpendicular vectors in the plane. The simplest:

Compute the dot product:

Yet clearly $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$. …

Method: Disproving a Converse with a Counterexample

When asked whether the converse of a true 'if-then' statement holds, the reasoning pattern is: find the other way the conclusion of the original can occur, then exhibit one concrete case.

Steps

Step 1: State precisely what the converse claims.

Here the original is 'either vector zero $\Rightarrow$ dot product zero'; its converse is 'dot product zero $\Rightarrow$ some vector is zero'.

Step 2: Identify the missing case. …

Common Mistakes

Mistake 1: Believing the converse is true.

Why it's wrong: $\vec{a}\cdot\vec{b}=0$ also happens for perpendicular non-zero vectors, so a zero dot product does not force a zero vector. Correct approach: give a perpendicular non-zero counterexample.

Mistake 2: Choosing two parallel vectors as the counterexample. …