Q.Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Concept: Direction Vectors — a unit vector in the direction of $\vec{a}$ is $\frac{\vec{a}}{|\vec{a}|}$.

First, find the magnitude of $\vec{a}$:

Then, divide each component of $\vec{a}$ by this magnitude: …

Why Direction Vectors Work

A unit vector is a vector of length 1 that points in exactly the same direction as the original vector. Think of it as the "pure direction" of $\vec{a}$ — stripped of its magnitude, keeping only its orientation in space.

The key idea: if you have any non-zero vector $\vec{a}$, you can shrink or stretch it to length 1 by dividing it by its own magnitude. This works because:

• Multiplying a vector by a positive scalar changes its length but not its direction.
• Dividing by $|\vec{a}|$ scales the length to exactly 1.

So the formula is:

Where $\hat{a}$ (read "a-hat") is the unit vector in the direction of $\vec{a}$.

Step-by-Step Solution

1. Write down the given vector.

This means the components are: $a_x = 1$, $a_y = 1$, $a_z = 2$.

2. Find the magnitude of $\vec{a}$.

The magnitude (or length) of a vector in 3D space comes from the Pythagorean theorem extended to three dimensions:

Substitute the components:

3. Apply the unit vector formula.

Divide each component of $\vec{a}$ by $|\vec{a}|$: …

Method: Normalising a Vector to a Unit Vector

Use this whenever a question asks for "the unit vector in the direction of" a given vector — the technique is the same regardless of the numbers.

Steps

Step 1: Read off the components.

Write $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and identify $a_1,a_2,a_3$ (watch the signs).

Step 2: Compute the magnitude.

This length is what you must strip away to leave pure direction. Keep it as an exact surd, never rounded.

Step 3: Divide every component by the magnitude.

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

Common Mistakes

Mistake 1: Dividing by the sum of squares instead of its square root.

Why it's wrong: the magnitude is $\sqrt{a_1^2+a_2^2+a_3^2}$, not $a_1^2+a_2^2+a_3^2$. Dividing by $6$ instead of $\sqrt{6}$ produces a vector of length $\tfrac{\sqrt6}{6}\neq1$. Correct approach: take the square root before dividing.

Mistake 2: Not squaring the last coefficient.

Why it's wrong: $|\vec a|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$; the $2\hat{k}$ contributes $2^2=4$, not $2$. Correct approach: square every component fully, then add. …