Walk 3 km east, then 4 km north. You have walked 7 km, but you are only 5 km from where you started — and that 5 is the magnitude of your displacement vector. $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$, read aloud the magnitude of v equals the square root of v-x squared plus v-y squared. Letter by letter: $v_x$ is the component along x, $v_y$ the component along y, both in the story's own unit, and $|\vec{v}|$ is the arrow's length in that same unit. Whichever of the three the question leaves blank is the one you solve for.

It is Pythagoras and nothing more: the components are the two legs of a right triangle and the vector is its hypotenuse. Which buys you a free sanity rail for the rest of the chapter — the magnitude is always at least as big as either component, and always smaller than the two added together. Run the ladder in order: square, add, root. Read backwards for a missing leg, $v_x = \sqrt{|\vec{v}|^2 - v_y^2}$, and the subtraction happens under the root, between the SQUARES — never between the sides themselves.