Q.If $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0,$ then what can be concluded about the vector $\vec{b}?$

The key idea is that the dot product of a vector with itself equals the square of its magnitude.

Step 1:

Given $\vec{a} \cdot \vec{a} = 0$. Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, we have $|\vec{a}|^2 = 0$, so $|\vec{a}| = 0$.

Thus $\vec{a}$ is the zero vector.

Step 2: …

The key here is to read the conditions carefully — they look like two separate dot product equations, but the first one is actually a statement about the vector $\vec{a}$ itself.

Why the first condition is special.

The dot product of a vector with itself gives the square of its magnitude:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2$.

If this equals zero, then $|\vec{a}|^2 = 0$, which forces $|\vec{a}| = 0$. The only vector with zero magnitude is the zero vector. So $\vec{a} = \vec{0}$.

Now the second condition.

If $\vec{a} = \vec{0}$, then $\vec{a} \cdot \vec{b} = \vec{0} \cdot \vec{b} = 0$ is automatically true for any vector $\vec{b}$. There is no restriction on $\vec{b}$ at all. …

Method: Reading $\vec{a}\cdot\vec{a}=0$ as a Magnitude Statement

This is a reasoning problem, not a computation. The key is recognising that a vector dotted with itself is a length condition, not an angle condition.

Steps

Step 1: Translate a self-dot-product into a magnitude.

So $\vec{a}\cdot\vec{a}=0$ forces $|\vec{a}|=0$, i.e. $\vec{a}=\vec{0}$ — the only vector of zero length.

Step 2: Feed that conclusion into the remaining condition. …

Common Mistakes

Mistake 1: Reading $\vec{a}\cdot\vec{a}=0$ as a perpendicularity condition.

Why it's wrong: $\vec{a}\cdot\vec{a}=|\vec{a}|^2$ is a length statement, forcing $\vec{a}=\vec{0}$ — it is not about an angle. Correct approach: conclude $\vec{a}$ is the zero vector.

Mistake 2: Concluding $\vec{b}$ is perpendicular to $\vec{a}$. …