Q.If $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$, find the unit vector in the direction of

A unit vector in the direction of $\vec{v}$ is $\dfrac{\vec{v}}{|\vec{v}|}$. Note $6\vec{b}$ points the same way as $\vec{b}$.

$\vec{a}=\hat{i}+\hat{j}+2\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-2\hat{k}$.

(i) $6\vec{b}=12\hat{i}+6\hat{j}-12\hat{k}$, $|6\vec{b}|=\sqrt{144+36+144}=18$.

Unit vector $=\dfrac{12\hat{i}+6\hat{j}-12\hat{k}}{18}=\dfrac13(2\hat{i}+\hat{j}-2\hat{k})$.

(ii) $2\vec{a}-\vec{b}=(2-2)\hat{i}+(2-1)\hat{j}+(4+2)\hat{k}=\hat{j}+6\hat{k}$, $|2\vec{a}-\vec{b}|=\sqrt{0+1+36}=\sqrt{37}$.

Unit vector $=\dfrac{\hat{j}+6\hat{k}}{\sqrt{37}}$.

The idea

A unit vector in the direction of a non-zero vector $\vec{v}$ is $\hat{v}=\dfrac{\vec{v}}{|\vec{v}|}$. Multiplying a vector by a positive scalar (like $6$) does not change its direction, only its length — so $6\vec{b}$ and $\vec{b}$ share the same unit vector.

Part (i): direction of $6\vec{b}$

Part (ii): direction of $2\vec{a}-\vec{b}$

Mind the sign on the $\hat{k}$ term: $4-(-2)=6$.

Method: Unit vectors of scaled and combined vectors

Use this for "find the unit vector in the direction of $k\vec{b}$ / $m\vec{a}+n\vec{b}$" type parts.

Steps

Step 1: Exploit that a positive scalar does not change direction.

A vector like $6\vec{b}$ points the same way as $\vec{b}$, so it has the same unit vector as $\vec{b}$ — you may normalise $\vec{b}$ directly and skip multiplying by $6$. This shortcut applies only to a single positive multiple, not to a genuine combination.

Step 2: Form each target vector by component arithmetic, watching signs.

For a combination such as $2\vec{a}-\vec{b}$, compute component by component and be careful subtracting a negative coordinate (e.g. $4-(-2)=6$).

Step 3: Divide each target vector by its own magnitude.

Common Mistakes

Mistake 1: Computing the unit vector of $6\vec{b}$ as $\dfrac{6\vec{b}}{|\vec{b}|}$.

Why it's wrong: you must divide by the magnitude of the same vector, $|6\vec{b}|=6|\vec{b}|=18$; dividing by $|\vec{b}|$ leaves a length-6 vector. Correct approach: divide $6\vec{b}$ by $|6\vec{b}|$ — or just note $6\vec{b}$ shares $\vec{b}$'s unit vector.

Mistake 2: Sign slip in $2\vec{a}-\vec{b}$ on the $\hat{k}$ term.

Why it's wrong: $4-(-2)=6$, not $2$; subtracting a negative adds. Correct approach: substitute the sign explicitly before subtracting.

Mistake 3: Writing $\hat{j}+6\hat{k}$ as the final answer for part (ii).

Why it's wrong: that is the direction vector, not yet a unit vector. Correct approach: divide by $|2\vec{a}-\vec{b}|=\sqrt{37}$.