Q.If $\vec{a}$ is any non-zero vector, then $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ equals ________.

The key idea is that the expression is the decomposition of $\vec{a}$ into its Cartesian components — essentially the vector triple product identity applied to basis vectors.

Step 1: Write $\vec{a}$ in component form:

$\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.

Step 2: Dot with each unit vector:

$\vec{a}\cdot\hat{i} = a_1$, $\vec{a}\cdot\hat{j} = a_2$, $\vec{a}\cdot\hat{k} = a_3$. …

The key here is to recognise what the expression $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ actually means.

When you take the dot product $\vec{a}\cdot\hat{i}$, you get the scalar component of $\vec{a}$ along the $x$-axis. Multiplying that scalar by the unit vector $\hat{i}$ gives you the vector component of $\vec{a}$ in the $x$-direction. The same holds for $\hat{j}$ and $\hat{k}$.

So the sum is simply the decomposition of $\vec{a}$ into its Cartesian components, added back together. That sum must equal $\vec{a}$ itself.

Let’s verify it step by step.

1. Write $\vec{a}$ in component form.

   Any vector in 3D can be written as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, where $a_1, a_2, a_3$ are real numbers.

2. Compute each dot product.

   • $\vec{a}\cdot\hat{i} = (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot\hat{i} = a_1$ (since $\hat{i}\cdot\hat{i}=1$, $\hat{j}\cdot\hat{i}=0$, $\hat{k}\cdot\hat{i}=0$).
   • Similarly, $\vec{a}\cdot\hat{j} = a_2$ and $\vec{a}\cdot\hat{k} = a_3$.

3. Multiply each scalar by its unit vector.

   • $(\vec{a}\cdot\hat{i})\hat{i} = a_1\hat{i}$
   • $(\vec{a}\cdot\hat{j})\hat{j} = a_2\hat{j}$
   • $(\vec{a}\cdot\hat{k})\hat{k} = a_3\hat{k}$

4. Add them up.

   $a_1\hat{i} + a_2\hat{j} + a_3\hat{k} = \vec{a}$ …

Method: Reconstructing a Vector from Its Projections onto $\hat{i},\hat{j},\hat{k}$

Use this for expressions of the form $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$.

Steps

Step 1: Interpret each term as a projection

$\vec{a}\cdot\hat{i}$ is the scalar component of $\vec{a}$ along the $x$-axis; multiplying by $\hat{i}$ turns it back into the vector component in that direction. The same holds for $\hat{j}$ and $\hat{k}$.

Step 2: Use orthonormality of the basis …

Common Mistakes

Mistake 1: Thinking the expression gives $|\vec{a}|$ or $|\vec{a}|^2$

Why it's wrong: each term is a scalar ($\vec{a}\cdot\hat{i}$) times a vector ($\hat{i}$), so the sum is a vector, not a length. Correct approach: it reconstructs $\vec{a}$ itself.

Mistake 2: Not using the orthonormality of $\hat{i},\hat{j},\hat{k}$

Why it's wrong: only because the basis is orthonormal do the dot products reduce to the plain coordinates $a_1,a_2,a_3$. Correct approach: use $\hat{i}\cdot\hat{i}=1$, $\hat{i}\cdot\hat{j}=0$, etc., to pick off the components. …