Unit vectors and Resolution of Vectors

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We live in a three-dimensional world. So when an object moves, its direction can change in three independent ways: along x, along y, and along z. To handle such motion neatly, physics uses unit vectors and the resolution of vectors.

Topic: Unit Vectors & Resolution of Vectors (3D)

What is a Unit Vector?

A unit vector is a vector whose magnitude is exactly 1. Its job is to show direction only. That’s why unit vectors are like “compass arrows”: they tell you the direction but not the distance.

Standard unit vectors in 3D

In a Cartesian (x–y–z) coordinate system, we use three mutually perpendicular unit vectors:

$$\hat{i}\ \rightarrow \text{along } +x \text{ axis}$$ $$\hat{j}\ \rightarrow \text{along } +y \text{ axis}$$ $$\hat{k}\ \rightarrow \text{along } +z \text{ axis}$$ $$|\hat{i}|=|\hat{j}|=|\hat{k}|=1$$
Key idea: Once you know \(\hat{i},\hat{j},\hat{k}\), you can describe any vector in 3D as a combination of these three directions.

How to Create a Unit Vector from Any Vector (Derivation)

Suppose you have a vector \(\vec{A}\) (not necessarily a unit vector). The unit vector along the direction of \(\vec{A}\) is written as \(\hat{A}\).

$$\hat{A}=\frac{\vec{A}}{|\vec{A}|}$$

Why does this work?

Dividing a vector by its magnitude keeps the direction same but scales the length to 1.

$$|\hat{A}|=\left|\frac{\vec{A}}{|\vec{A}|}\right|=\frac{|\vec{A}|}{|\vec{A}|}=1$$
Common mistake: Students sometimes divide by components \((A_x, A_y, A_z)\) instead of the magnitude \(|\vec{A}|\). Always divide by the magnitude.

Resolution of a Vector in 3D (Main Idea)

Consider a vector \(\vec{V}\) pointing in any direction in space. We can “break” it into three perpendicular components:

  • \(V_x\) along x-axis
  • \(V_y\) along y-axis
  • \(V_z\) along z-axis

Then the vector can be written using unit vectors as:

$$\vec{V}=V_x\hat{i}+V_y\hat{j}+V_z\hat{k}$$
Vector V resolved into Vx, Vy, Vz components
Figure 2: Resolving a 3D vector into components along x, y, z.
Once you find \(V_x, V_y, V_z\), the problem becomes much easier because you can treat each direction separately.

Magnitude of a Vector in Component Form (Derivation)

If \(\vec{V}=V_x\hat{i}+V_y\hat{j}+V_z\hat{k}\), then the magnitude of \(\vec{V}\) is given by 3D Pythagoras:

$$|\vec{V}|=\sqrt{V_x^2+V_y^2+V_z^2}$$

Why?

Because \(V_x, V_y, V_z\) are perpendicular components. First combine \(V_x\) and \(V_y\) in the xy-plane, then include \(V_z\).

3D Pythagoras derivation: resultant in xy-plane then include z
Figure 3: Magnitude of \(\vec{V}\) using stepwise Pythagoras in 3D.

Direction Cosines and Direction Angles (Important Derivation)

Let \(\vec{V}\) make angles \(\alpha\) with x-axis, \(\beta\) with y-axis, and \(\gamma\) with z-axis. Then:

$$\cos\alpha=\frac{V_x}{|\vec{V}|},\quad \cos\beta=\frac{V_y}{|\vec{V}|},\quad \cos\gamma=\frac{V_z}{|\vec{V}|}$$ $$V_x=|\vec{V}|\cos\alpha,\quad V_y=|\vec{V}|\cos\beta,\quad V_z=|\vec{V}|\cos\gamma$$

Direction cosine identity (must-know)

Square and add the three equations:

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$
Vector V making angles alpha beta gamma with x y z axes
Figure 4: Direction angles \((\alpha,\beta,\gamma)\) of a vector with coordinate axes.
This identity helps you check if your angles are correct. If the sum is not 1, something is wrong.

Unit Vector Along a Given Vector in 3D (With Components)

If \(\vec{V}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}\), then:

$$|\vec{V}|=\sqrt{a_x^2+a_y^2+a_z^2}$$ $$\hat{V}=\frac{\vec{V}}{|\vec{V}|}=\frac{a_x\hat{i}+a_y\hat{j}+a_z\hat{k}}{\sqrt{a_x^2+a_y^2+a_z^2}}$$
This is extremely common in physics: electric field direction, force direction, velocity direction, etc.

Solved Numerical Examples

Example 1: Magnitude of a 3D vector

A vector is \(\vec{V}=3\hat{i}+4\hat{j}+12\hat{k}\). Find its magnitude.

Solution:
$$|\vec{V}|=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$$

So, the magnitude is 13 units.

Example 2: Unit vector along a given vector

Find the unit vector along \(\vec{A}=2\hat{i}-2\hat{j}+1\hat{k}\).

Solution:
$$|\vec{A}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$$ $$\hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}=\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{1}{3}\hat{k}$$
Example 3: Using direction cosines

A vector has magnitude 20 and makes angles \(\alpha=60^\circ\), \(\beta=45^\circ\) with x and y axes. Find \(V_x\) and \(V_y\). (Use \(\cos60^\circ=0.5\), \(\cos45^\circ=0.707\))

Solution:
$$V_x=|\vec{V}|\cos\alpha=20\times 0.5=10$$ $$V_y=|\vec{V}|\cos\beta=20\times 0.707\approx 14.14$$

(To find \(V_z\), you would need \(\gamma\) or use the identity if enough information is given.)

Example 4: Find direction cosines from components

For \(\vec{V}=6\hat{i}+2\hat{j}+3\hat{k}\), find \(\cos\alpha,\cos\beta,\cos\gamma\).

Solution:
$$|\vec{V}|=\sqrt{6^2+2^2+3^2}=\sqrt{36+4+9}=\sqrt{49}=7$$ $$\cos\alpha=\frac{V_x}{|\vec{V}|}=\frac{6}{7},\quad \cos\beta=\frac{2}{7},\quad \cos\gamma=\frac{3}{7}$$

Check: \((6/7)^2+(2/7)^2+(3/7)^2=(36+4+9)/49=1\) ✅

Practice Questions (For Students)

A) Conceptual

  1. What is a unit vector? Why is its magnitude always 1?
  2. Why do we use \(\hat{i},\hat{j},\hat{k}\) to describe vectors?
  3. What do direction cosines represent physically?
  4. Write the identity involving \(\cos\alpha,\cos\beta,\cos\gamma\). Why is it useful?
  5. Can two different vectors have the same unit vector? Explain.

B) Numerical

  1. Find \(|\vec{A}|\) if \(\vec{A}=5\hat{i}-12\hat{j}+0\hat{k}\).
  2. Find the unit vector along \(\vec{B}=4\hat{i}+0\hat{j}-3\hat{k}\).
  3. If \(|\vec{V}|=50\) and \(\cos\alpha=0.6\), find \(V_x\).
  4. For \(\vec{V}=2\hat{i}+3\hat{j}+6\hat{k}\), find \(\cos\gamma\).
  5. A vector has direction cosines \((0.2,0.8,k)\). Find \(k\).
  6. A drone’s velocity is \(\vec{V}=12\hat{i}+5\hat{j}+9\hat{k}\ \text{m/s}\). Find its speed.
Answers (Click to reveal)
$$|\vec{A}|=\sqrt{5^2+(-12)^2+0^2}=\sqrt{169}=13$$ $$|\vec{B}|=\sqrt{4^2+0^2+(-3)^2}=5,\quad \hat{B}=\frac{4}{5}\hat{i}-\frac{3}{5}\hat{k}$$ $$V_x=|\vec{V}|\cos\alpha=50\times 0.6=30$$ $$|\vec{V}|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7,\quad \cos\gamma=\frac{6}{7}$$ $$0.2^2+0.8^2+k^2=1\Rightarrow k^2=0.32\Rightarrow k\approx 0.566$$ $$|\vec{V}|=\sqrt{12^2+5^2+9^2}=\sqrt{250}\approx 15.81\ \text{m/s}$$

Summary (Takeaway)

  • Unit vectors \(\hat{i},\hat{j},\hat{k}\) give standard directions along x, y, z axes.
  • Any vector in 3D can be written as \(\vec{V}=V_x\hat{i}+V_y\hat{j}+V_z\hat{k}\).
  • Magnitude: \( |\vec{V}|=\sqrt{V_x^2+V_y^2+V_z^2} \)
  • Direction cosines:
    $$\cos\alpha=\frac{V_x}{|\vec{V}|},\quad \cos\beta=\frac{V_y}{|\vec{V}|},\quad \cos\gamma=\frac{V_z}{|\vec{V}|}$$ $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$
  • Unit vector along \(\vec{V}\): \(\hat{V}=\vec{V}/|\vec{V}|\)
© Eduvigyan.com • Topic: Unit Vectors & Resolution of Vectors (3D) • Back to Mechanics
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