A vector has component along the X-axis equal to 25 unit and along the Y-axis equal to 60 unit. Find the magnitude and direction of the vector.

1. Decoding the problem

The vector components are basically the shadows of the vector on the X-axis and Y-axis.

• $A_x = 25$ units (shadow on X-axis)
• $A_y = 60$ units (shadow on Y-axis)

2. Magnitude (length) of a vector

When a vector has components $A_x$ and $A_y$, the magnitude $|\vec{A}|$ follows directly from the Pythagorean theorem because $A_x$ and $A_y$ form the legs of a right-angled triangle:

3. Direction (angle with X-axis)

The direction angle $\theta$ is defined here as the anticlockwise angle from +X-axis to the vector. Using basic trigonometry in the same right-angled triangle,

From this equation, we get $\theta = \tan^{-1}\!\left(\dfrac{A_y}{A_x}\right)$. The arctangent (inverse tangent) directly gives us the required angle, provided we know the signs of $A_x$ and $A_y$. Here both are positive, so the vector sits in the first quadrant.

4. Why these steps?

Pythagoras tells us the shortest distance (the hypotenuse) in a right-angled triangle. Trigonometry provides the relation between an angle and the side lengths, hence it gives the direction of the vector. No other algebra is needed.

Imagine a ladder placed on the ground

• One end of the ladder sits 25 steps along the floor in the X‐direction.
• The other end reaches 60 steps upward in the Y‐direction.

The total length of the ladder is the straight line from the start to the end—this is the magnitude of the vector.

To get the length, we use the same rule you learnt for right-angled triangles (Pythagoras):

Next, the direction is simply the angle the ladder makes with the X-axis. For a right-angled triangle, that angle $\theta$ is found by

This gives the direction of our vector.

So, we have both the size (65 units) and the tilt (about $67^\circ$ upward from the X-axis).

Step-by-Step Solution

1. Write the given data
   $A_x = 25$ units, $A_y = 60$ units

2. Magnitude

   $$|\vec{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65 \text{ units}$$

3. Direction angle with +X-axis

   $$\tan \theta = \frac{A_y}{A_x} = \frac{60}{25} = \frac{12}{5}$$ $$\theta = \tan^{-1}\!\left(\frac{12}{5}\right)$$

   Using a calculator (or log tables),

   $$\theta \approx 67.38^\circ$$

4. State the final answer
   The vector magnitude is 65 units, and its direction is about $67^\circ$ above the +X-axis (first quadrant).