Calculate |a⃗| = √(3² + 4²) = 5 and understand unit vectors.

Find Magnitude of Vector 3î + 4ĵ

Question

Find the magnitude of the vector $\vec{a} = 3\hat{i} + 4\hat{j}$ .

Also find the unit vector in the direction of $\vec{a}$ .

Solution — Step by Step

For any vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ , the magnitude is:

|\vec{a}| = \sqrt{x^2 + y^2 + z^2}

This comes from the 3D extension of the Pythagorean theorem — the magnitude is just the length of the vector treated as a hypotenuse.

Here, $\vec{a} = 3\hat{i} + 4\hat{j}$ , so $x = 3$ , $y = 4$ , and $z = 0$ (no $\hat{k}$ component, meaning the vector lies in the XY-plane).

|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5}

The 3-4-5 Pythagorean triplet makes this a clean answer — and that’s exactly why NCERT chose it.

A unit vector has magnitude 1. We get it by dividing $\vec{a}$ by its magnitude:

\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{3\hat{i} + 4\hat{j}}{5} = \frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}

We can verify: $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$ ✓

Why This Works

The magnitude formula works because $\hat{i}$ , $\hat{j}$ , $\hat{k}$ are mutually perpendicular unit vectors. When you write $3\hat{i} + 4\hat{j}$ , you’re moving 3 units along the x-axis and 4 units along the y-axis — two perpendicular directions.

The resultant length, by Pythagoras, is $\sqrt{3^2 + 4^2}$ . This is the same geometry you used in Class 10 for right triangles — vectors just generalize it to three dimensions.

The unit vector $\hat{a}$ preserves the direction of $\vec{a}$ but scales the length to exactly 1. This is useful whenever you need “direction only” — for instance, in force problems where you want direction separate from magnitude.

Alternative Method — Geometrical Interpretation

Draw $\vec{a} = 3\hat{i} + 4\hat{j}$ on a coordinate plane. Starting from the origin, go 3 units along the x-axis and 4 units along the y-axis. The vector is the straight line from the origin to that point $(3, 4)$ .

The magnitude is simply the distance from $(0, 0)$ to $(3, 4)$ :

d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5

Same answer, but now you see why the formula works — it’s literally the distance formula from coordinate geometry. The two approaches are identical in structure.

Common Mistake

Forgetting to square root — or squaring incorrectly.

Many students write $|\vec{a}| = 3^2 + 4^2 = 25$ and stop there, skipping the square root. The magnitude is $\sqrt{25} = 5$ , not 25.

A subtler error: writing $|3\hat{i} + 4\hat{j}| = 3 + 4 = 7$ . You cannot add the components directly. Magnitudes do NOT add linearly — only their squares do, and only when the components are perpendicular.

Memorize common Pythagorean triplets: (3, 4, 5), (5, 12, 13), (8, 15, 17). When you spot these as vector components in an exam, the magnitude is instantly the third number — no calculation needed. This saves 30 seconds per question in JEE Main.

Question

Solution — Step by Step

Why This Works

Alternative Method — Geometrical Interpretation

Common Mistake

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