Q.If 𝑎⃗ = 3𝑖̂ + 2𝑗̂ + 4𝑘̂ , 𝑏⃗⃗ = 𝑖̂ + 𝑗̂ − 3𝑘̂ and 𝑐⃗ = 6𝑖̂ − 𝑗̂ + 2𝑘̂ are three given vectors, then (2𝑎⃗. 𝑖̂)𝑖̂ − (𝑏⃗⃗. 𝑗̂)𝑗̂ + (𝑐⃗. 𝑘̂)𝑘̂ is same as the vector
(A) 𝑎⃗
(B) 𝑏⃗⃗ + 𝑐⃗
(C) 𝑎⃗ − 𝑏⃗⃗
(D) 𝑐⃗

Concept: Vector Component Extraction — each dot product with a unit vector isolates the corresponding component of the vector.

Step 1: Compute each dot product.

$2\vec{a} \cdot \hat{i} = 2(3) = 6$

$\vec{b} \cdot \hat{j} = 1$

$\vec{c} \cdot \hat{k} = 2$

Step 2: Multiply each scalar by its unit vector.

$(6)\hat{i} - (1)\hat{j} + (2)\hat{k} = 6\hat{i} - \hat{j} + 2\hat{k}$ …

The key idea here is vector component extraction. When you dot a vector with a unit vector like $\hat{i}$, you get the scalar component of that vector along the $x$-axis. Multiplying that scalar back by $\hat{i}$ gives you the vector component along $x$ — essentially, you’re picking out just the $x$-part of the original vector.

So the expression $(2\vec{a} \cdot \hat{i})\hat{i} - (\vec{b} \cdot \hat{j})\hat{j} + (\vec{c} \cdot \hat{k})\hat{k}$ is doing exactly this: it takes the $x$-component of $2\vec{a}$, the $y$-component of $\vec{b}$ (with a minus sign), and the $z$-component of $\vec{c}$, and assembles them into a new vector. No mixing of axes happens — each term lives on its own coordinate axis.

Let’s work it out.

1. First term: $(2\vec{a} \cdot \hat{i})\hat{i}$

   $\vec{a} = 3\hat{i} + 2\hat{j} + 4\hat{k}$, so $2\vec{a} = 6\hat{i} + 4\hat{j} + 8\hat{k}$.

   Dotting with $\hat{i}$ picks out the $x$-component: $2\vec{a} \cdot \hat{i} = 6$.

   Multiplying back by $\hat{i}$ gives $6\hat{i}$.

2. Second term: $-(\vec{b} \cdot \hat{j})\hat{j}$

   $\vec{b} = \hat{i} + \hat{j} - 3\hat{k}$, so $\vec{b} \cdot \hat{j} = 1$.

   With the minus sign, this becomes $-1 \cdot \hat{j} = -\hat{j}$.

3. Third term: $+(\vec{c} \cdot \hat{k})\hat{k}$

   $\vec{c} = 6\hat{i} - \hat{j} + 2\hat{k}$, so $\vec{c} \cdot \hat{k} = 2$.

   Multiplying by $\hat{k}$ gives $2\hat{k}$.

4. Combine them:

   $6\hat{i} - \hat{j} + 2\hat{k}$.

Now compare with the given options:

• $\vec{a} = 3\hat{i} + 2\hat{j} + 4\hat{k}$ — not a match. …

Method: Isolating a component with the dot-then-scale pattern

Use this whenever an expression is built from pieces of the form $(\vec{v}\cdot\hat{i})\hat{i}$, $(\vec{v}\cdot\hat{j})\hat{j}$ or $(\vec{v}\cdot\hat{k})\hat{k}$ and you must simplify it to a single vector.

Steps

Step 1: Recognise what $(\vec{v}\cdot\hat{u})\hat{u}$ does.

Dotting a vector with a unit axis vector returns the scalar component along that axis; multiplying that scalar back by the same unit vector rebuilds only the piece of $\vec{v}$ on that axis. So $(\vec{v}\cdot\hat{i})\hat{i}$ is nothing but the $\hat{i}$-part of $\vec{v}$ — the axes never mix.

Step 2: Read each scalar component off directly.

For any $\vec{v}=x\hat{i}+y\hat{j}+z\hat{k}$,

$\vec{v}\cdot\hat{i}=x,\qquad \vec{v}\cdot\hat{j}=y,\qquad \vec{v}\cdot\hat{k}=z.$ …

Common Mistakes

Mistake 1: Dropping the scalar $2$ inside $(2\vec{a}\cdot\hat{i})\hat{i}$.

Why it's wrong: the factor $2$ scales the whole term, so $2\vec{a}\cdot\hat{i}=2(3)=6$, not $3$. Correct approach: apply the coefficient to the component you extract before comparing with the options.

Mistake 2: Ignoring the minus sign in front of $(\vec{b}\cdot\hat{j})\hat{j}$.

Why it's wrong: the expression subtracts the $\hat{j}$ piece, so the middle term is $-\hat{j}$, not $+\hat{j}$. Correct approach: carry each term's own sign into the assembled vector.

Mistake 3: Mixing components across vectors. …