Q.Compare the components for the following vector equations.

Vector Component Extraction: The Intuition

Imagine pushing a heavy box across the floor at an angle — not straight forward, but slightly downward. Some of your effort moves the box forward, and some presses it into the floor. The force you apply is a single vector, but its effect splits into two independent directions: horizontal and vertical.

That splitting is vector component extraction. Any vector can be seen as the sum of two (or three) simpler vectors pointing along chosen reference directions — usually the coordinate axes. Each of those simpler vectors is a component.

The Precise Statement

Given a vector $\vec{v}$ in a plane, and perpendicular axes $x$ and $y$, the components of $\vec{v}$ are its projections onto those axes:

where $\hat{i}$ and $\hat{j}$ are unit vectors along the $x$ and $y$ axes, and $v_x$, $v_y$ are scalar components (numbers, possibly negative).

If $\vec{v}$ makes an angle $\theta$ from the positive $x$-axis, then:

Why This Works: The Geometry

Draw a vector from the origin. Drop a perpendicular from its tip to the $x$-axis — that gives $v_x$. Drop another to the $y$-axis — that gives $v_y$. The original vector is the diagonal of the rectangle formed by $v_x$ and $v_y$. This is the Pythagorean theorem in reverse: if you know the hypotenuse and one angle, trigonometry gives you the legs.

A Concrete Example

A force of 10 N acts at $30^\circ$ above the horizontal.

• $F_x = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$ N
• $F_y = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5$ N

So the force vector is $8.66\hat{i} + 5\hat{j}$ N.

Why This Matters …