Magnitude of a Vector

The magnitude (or modulus) of a vector measures its length. For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$ in 2D, the components form the two shorter sides of a right triangle, and the vector itself is the hypotenuse. By Pythagoras:

For a 3D vector $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$, we apply Pythagoras twice (first in the $xy$-plane, then lift to height $a_3$):

Example (2D): $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$

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

So the hook estimate: it is exactly 5, a classic 3-4-5 Pythagorean triple. Check your estimate from card 01!

Example (3D): $\mathbf{u} = 2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}$

$|\mathbf{u}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

The magnitude (or modulus) of a vector measures its length. For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$ in 2D, the components form the two shorter sides of a right triangle, and the vector itself is the hypotenuse. By Pythagoras:

Pause, copy the magnitude formulas: 2D $|\vec{a}|=\sqrt{a_1^2+a_2^2}$, 3D $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$, with the Pythagorean derivation into your book.

We just saw that $|\vec{a}|=\sqrt{a_1^2+a_2^2}$ in 2D and $|\vec{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$ in 3D, from the Pythagorean theorem. That raises a question: for a displacement vector $\overrightarrow{AB}$, you first compute components, then apply the magnitude formula, what is the common error at step one, and how does it propagate to the final answer? This card answers it → computing $B-A$ in the wrong order gives $-\overrightarrow{AB}$, but since magnitude is always positive, the error cancels in the final answer ($|\overrightarrow{BA}|=|\overrightarrow{AB}|$).

A reliable method for any magnitude calculation:

1. Read off the components identify $a_1$, $a_2$ (and $a_3$). Watch signs carefully.
2. Square each component squaring removes any negatives.
3. Sum the squares, then take the square root.
4. Simplify factor the expression under the root if possible; look for perfect squares.

For displacement vectors $\overrightarrow{AB}$, first compute $\overrightarrow{AB} = B - A$ (subtract position vectors), then apply the formula.

For displacement vectors $\overrightarrow{AB}$, first compute $\overrightarrow{AB} = B - A$ (subtract position vectors), then apply the formula.

Pause, copy the two-step magnitude procedure: (1) compute components using end-minus-start; (2) apply the magnitude formula; include a worked 3D example into your book.