Finding Magnitude and Direction of a Vector from Its Components
Physics · Motion In A Plane · NEET
If a vector has components Ax and Ay, its magnitude is A = sqrt(Ax^2 + Ay^2) (Pythagoras) and its direction is the angle theta = tan-1(Ay/Ax) measured from the x-axis. Memory hook: components make a right triangle, so the vector is the hypotenuse (magnitude) and the angle is "y over x, then inverse tan". This is the reverse of resolution: resolution splits a vector into parts, this joins the parts back into one arrow.
The x-component (Ax) and y-component (Ay) form the two sides of a right triangle; the vector A is the hypotenuse. Its length is A = sqrt(Ax^2 + Ay^2) and its direction is theta = tan-1(Ay/Ax) measured from the x-axis.
Your doubts, answered
How do I get the magnitude of a vector when I only know Ax and Ay?
Use the Pythagoras rule: A = sqrt(Ax^2 + Ay^2). The components form the two perpendicular sides of a right triangle and the vector is the hypotenuse. Example: if Ax = 3 and Ay = 4, then A = sqrt(9 + 16) = sqrt(25) = 5. For a 3-D vector A = Ax i + Ay j + Az k, extend it to A = sqrt(Ax^2 + Ay^2 + Az^2).
Which formula gives the direction: sin, cos, or tan?
For the angle theta measured from the x-axis, use tan: theta = tan-1(Ay/Ax). This is because tan theta = opposite/adjacent = Ay/Ax. You use sin and cos when you go the other way (resolution: Ax = A cos theta, Ay = A sin theta). So: components to angle uses tan-inverse; angle to components uses sin and cos.
Why is tan-1(Ay/Ax) sometimes the wrong angle?
A calculator only returns values between -90 and +90 degrees, so it cannot tell which quadrant the vector is in. Always look at the SIGNS of Ax and Ay first. If Ax is negative (vector points left), add 180 degrees to the calculator answer. Sketch the vector so your final angle matches the real direction.
How do I find magnitude and direction from a position or velocity vector like v = 4i + 4j?
Read off the components directly: here Ax = 4 and Ay = 4. Magnitude = sqrt(4^2 + 4^2) = sqrt(32) = 4 sqrt(2) m/s. Direction = tan-1(4/4) = tan-1(1) = 45 degrees from the x-axis. The unit vectors i and j simply label the x and y components.
If two perpendicular forces act, how do I get the resultant's size and angle?
Perpendicular forces are already the two components. Magnitude = sqrt(F1^2 + F2^2). The angle with F1 is tan-1(F2/F1). Example: 8 N along x and 6 N along y give resultant sqrt(64+36) = 10 N, at tan-1(6/8) = tan-1(3/4) from the 8 N force.
⚠️ The NEET trap ✗ Using theta = tan-1(Ax/Ay) or picking sin/cos to find the angle from components. ✓ From the x-axis, the angle is theta = tan-1(Ay/Ax) (y-component over x-component). sin and cos are for the reverse step (finding components from a known angle). 🧠 Angle FROM components = tan-inverse of (y over x). Flipping the ratio flips your answer to the complement (90 - theta).
Real NEET questions
2023
The position of a particle is given by r(t) = 4t i + 2t^2 j + 5k, where t is in seconds and r in metres. Find the magnitude and direction (with respect to the x-axis) of the velocity v(t) at t = 1 s.
A · 3 sqrt(2) m/s, 30 degrees
B · 3 sqrt(2) m/s, 45 degrees
C · 4 sqrt(2) m/s, 45 degrees ✓
D · 4 sqrt(2) m/s, 60 degrees
Solution: Step 1: Velocity is the time-derivative of position. v = dr/dt = 4 i + 4t j (the 5k term is constant, so it gives 0). Step 2: At t = 1 s, v = 4 i + 4 j, so vx = 4 and vy = 4. Step 3: Magnitude = sqrt(vx^2 + vy^2) = sqrt(16 + 16) = sqrt(32) = 4 sqrt(2) m/s. Step 4: Direction from x-axis = tan-1(vy/vx) = tan-1(4/4) = tan-1(1) = 45 degrees. Answer: C.
2026
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it are respectively:
A · 20 m/s^2; tan-1(4/3) with the 8 N force
B · 2 m/s^2; tan-1(3/4) with the 6 N force
C · 2 m/s^2; tan-1(4/3) with the 8 N force
D · 2 m/s^2; tan-1(3/4) with the 8 N force ✓
Solution: Step 1: The two forces are perpendicular, so treat them as x and y components. Resultant force = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 N. Step 2: Acceleration magnitude = F/m = 10/5 = 2 m/s^2. Step 3: Direction with respect to the 8 N force = tan-1(6/8) = tan-1(3/4). (The 8 N force is the 'x' side, the 6 N force is the 'y' side.) Answer: D.
Solved Motion In A Plane NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the magnitude of a vector from its components?
For a 2-D vector, A = sqrt(Ax^2 + Ay^2). For a 3-D vector, A = sqrt(Ax^2 + Ay^2 + Az^2). It comes straight from the Pythagoras theorem.
How do you find the direction of a vector from its components?
The angle with the x-axis is theta = tan-1(Ay/Ax). Always check the signs of Ax and Ay to place the vector in the correct quadrant and adjust the calculator angle if needed.
Is this the same as resolution of a vector?
It is the exact reverse. Resolution splits one vector into Ax = A cos theta and Ay = A sin theta. Here you already have Ax and Ay and rebuild the single vector's magnitude and angle.
Why do these questions appear so often in NEET?
Almost every 2-D problem (projectiles, forces, velocity from a position vector, relative velocity) ends by asking for a resultant's magnitude or angle. Mastering sqrt(Ax^2 + Ay^2) and tan-1(Ay/Ax) lets you finish many chapters' final steps quickly and without error.
When do I need to add 180 degrees to the angle?
When the x-component Ax is negative (the vector points to the left half). The calculator's tan-1 gives an angle in the wrong half, so add 180 degrees to get the true direction. A quick sketch confirms it.
