Q.Find unit vector in the direction of vector $\vec{a}=2\hat{i}+3\hat{j}+\hat{k}$.

The key idea is that a unit vector in the direction of a given vector is found by dividing the vector by its magnitude.

Step 1: Write the vector:

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.

Step 2: Compute its magnitude:

$|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.

Step 3: The unit vector $\hat{a}$ is: …

Concept and Intuition

A unit vector is a vector with magnitude exactly 1. It tells you only the direction, not the length. Think of it as a pure direction pointer.

If you have any non-zero vector $\vec{a}$, you can think of it as:

$\vec{a} = (\text{magnitude of }\vec{a}) \times (\text{unit vector in the direction of }\vec{a})$

So to isolate just the direction, you divide the vector by its own length. That's the entire idea:

This works because dividing by the magnitude scales the vector down to length 1, while keeping its direction unchanged.

Step-by-step solution

1. Write the vector clearly

   $\vec{a} = 2\hat{i} + 3\hat{j} + 1\hat{k}$

   The components are: $a_x = 2$, $a_y = 3$, $a_z = 1$.

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

   The magnitude (or length) of a vector in 3D is given by the square root of the sum of squares of its components:

Compute:

3. Apply the unit vector formula

This means each component gets divided by $\sqrt{14}$:

4. Verify the result …

Method: Finding the unit vector in the direction of a given vector

Use this to "normalise" any non-zero vector to length $1$ while preserving its direction.

Steps

Step 1: Compute the magnitude.

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

Take the square root of the sum of squares — never just add the components.

Step 2: Divide the vector by its 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: Using $|\vec{a}|=2+3+1=6$ instead of the square-root formula.

Why it's wrong: magnitude is $\sqrt{2^2+3^2+1^2}=\sqrt{14}$, not the sum of components; dividing by $6$ gives a vector that is not a unit vector. Correct approach: square the components, add, then take the square root.

Mistake 2: Forgetting the $\hat{k}$-component's coefficient of $1$. …