A unit vector is a vector whose magnitude is exactly 1 (dimensionless). Its only job is to specify direction. Any vector A can be decomposed into its magnitude |A| and a unit vector â along its direction:
â = A / |A|
This is the definition given in NCERT Class 11 Physics Chapter 3, page 30. The operation is called "normalisation" — divide the vector by its own magnitude to strip the size and keep only the direction.
Three standard unit vectors form the Cartesian coordinate system: î (along +x), ĵ (along +y), k̂ (along +z). Any vector in 3D space can be written as A = Aₓ î + Aᵧ ĵ + A_z k̂, where Aₓ, Aᵧ, A_z are the scalar components.
The trap that costs marks: Students confuse the unit vector with the vector's components. When a question says "find the unit vector along A = 3î + 4ĵ," the answer is NOT 3î + 4ĵ — that has magnitude 5, not 1. You must divide each component by 5 to get (3/5)î + (4/5)ĵ.
A second common confusion: treating the unit vector as having units. A unit vector is dimensionless. If F = 10 N along â, the unit "newton" belongs to the magnitude 10, not to â.
Watch-out for NEET: Questions on unit vectors typically test whether you can normalise a given vector, identify properties (magnitude = 1, dimensionless), or distinguish between a vector, its magnitude, and its unit vector. These appear as quick recall or single-step application items — free marks if the definition is sharp, lost marks if the normalisation step is forgotten.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the magnitude of a unit vector?
Answer: B. By definition (NCERT Class 11 Physics Chapter 3, page 30), a unit vector has magnitude exactly equal to 1. This is the defining property — it carries direction only.
Why A is wrong: A is wrong because a vector with magnitude 0 is the null vector, not a unit vector.
Why C is wrong: C is wrong because the magnitude of a unit vector is always 1 regardless of which vector it was derived from — the normalisation process guarantees this.
Why D is wrong: D is wrong because the unit vector's magnitude is 1 by definition; the original vector's magnitude was divided out during normalisation.
A unit vector is:
Answer: A. A unit vector is dimensionless and has magnitude 1. Its sole purpose is to indicate direction (NCERT Class 11 Physics Chapter 3, page 30).
Why B is wrong: B is wrong because 'unit' in unit vector refers to magnitude 1, not to physical units like metres or seconds. A unit vector is dimensionless.
Why C is wrong: C is wrong because a vector with all components equal to 1 (e.g. î + ĵ + k̂) has magnitude √3, not 1. It is not a unit vector.
Why D is wrong: D is wrong because there is no concept of a 'smallest possible vector' — vectors can have arbitrarily small magnitudes. A unit vector specifically has magnitude exactly 1.
If A = 3î + 4ĵ, the unit vector along A is:
Answer: D. |A| = √(3² + 4²) = √(9 + 16) = √25 = 5. The unit vector â = A/|A| = (3/5)î + (4/5)ĵ. Verify: magnitude = √(9/25 + 16/25) = √(25/25) = 1. ✓
Why A is wrong: A is wrong because 3î + 4ĵ has magnitude 5, not 1. Forgetting to divide by the magnitude is the most common error in unit-vector problems.
Why B is wrong: B is wrong because 7 is the sum of the components (3 + 4), not the magnitude. The magnitude requires √(3² + 4²) = 5, not simple addition.
Why C is wrong: C is wrong because dividing each component by itself (3 by 3, 4 by 4) gives î + ĵ, which has magnitude √2 ≠ 1. The correct operation is to divide each component by the same number: the vector's magnitude.
The unit vector along B = 2î − 2ĵ + k̂ is:
Answer: A. |B| = √(4 + 4 + 1) = √9 = 3. So b̂ = (2/3)î − (2/3)ĵ + (1/3)k̂. Verify: √(4/9 + 4/9 + 1/9) = √(9/9) = 1. ✓
Why B is wrong: B is wrong because it divides by 5 = 2 + 2 + 1, the sum of the component sizes, instead of the magnitude √(4 + 4 + 1) = 3.
Why C is wrong: C is wrong because it divides every component by 2 instead of by the magnitude 3; the result has magnitude √(1 + 1 + 1/4) = 1.5, not 1.
Why D is wrong: D is wrong because 9 is the square of the magnitude (9 = 3²), not the magnitude itself. You must divide by √9 = 3, not by 9.
Which of the following is true about the standard unit vectors î, ĵ, k̂?
Answer: C. By definition, î, ĵ, k̂ are mutually perpendicular unit vectors along the +x, +y, +z axes respectively. Each has magnitude 1 (NCERT Class 11 Physics Chapter 3, page 30).
Why A is wrong: A is wrong because î, ĵ, k̂ point along three mutually perpendicular directions (+x, +y, +z). They are orthogonal, not parallel.
Why B is wrong: B is wrong because each of î, ĵ, k̂ has magnitude 1 by definition, not √3. The magnitude √3 belongs to the vector î + ĵ + k̂, not to the individual basis vectors.
Why D is wrong: D is wrong because |î + ĵ + k̂| = √(1 + 1 + 1) = √3 ≈ 1.73, which is not 1. The sum of unit vectors is not generally a unit vector.
A force F has magnitude 10 N and acts along the direction of the vector 6î + 8ĵ. The force vector in component form is:
Answer: C. C is correct. First find the unit vector: |6î + 8ĵ| = √(36 + 64) = 10. So F̂ = (6/10)î + (8/10)ĵ = 0.6î + 0.8ĵ. Then F = 10 × (0.6î + 0.8ĵ) = 6î + 8ĵ N. Note that the direction vector was itself given with magnitude 10, so dividing by 10 and then multiplying by the 10 N magnitude returns the same components — a coincidence of the numbers chosen here, not a shortcut that generalises. Check the answer: |6î + 8ĵ| = 10 N, as required.
Why A is wrong: A is wrong because 0.6î + 0.8ĵ is the UNIT vector F̂, not the force. It is dimensionless and has magnitude 1, so it cannot carry newtons. This is the half-finished answer: the normalisation step was done and the multiplication by the 10 N magnitude was left out.
Why B is wrong: B is wrong because multiplying the direction vector directly by 10 without first normalising yields 60î + 80ĵ, which has magnitude √(3600 + 6400) = 100 N — ten times too large.
Why D is wrong: D is wrong because splitting the 10 N equally between two components gives each component 10 N, yielding a resultant of 10√2 ≈ 14.1 N and the wrong direction (45° instead of 53.1°).
Two vectors P and Q have the same unit vector. Which statement must be true?
Answer: D. If P̂ = Q̂, then both vectors point in the same direction (their directions are identical after normalisation). However, their magnitudes |P| and |Q| can be different. For example, 3î and 7î both have unit vector î but different magnitudes.
Why A is wrong: A is wrong because sharing a unit vector means sharing a direction, not a magnitude. Vectors 2î and 5î share the unit vector î but have magnitudes 2 and 5.
Why B is wrong: B is wrong because same unit vector means same direction, which is the opposite of perpendicular (perpendicular vectors have orthogonal unit vectors, not identical ones).
Why C is wrong: C is wrong because equality of vectors requires BOTH same magnitude AND same direction. Same unit vector guarantees only same direction.
A student writes: "The unit vector along a displacement of 5 m in the +x direction is 5î m." What error has the student made?
Answer: B. The displacement vector is 5î m. To get the unit vector, divide by the magnitude: û = (5î m)/(5 m) = î. The metres cancel — a unit vector is dimensionless and has magnitude 1. The student failed to normalise and incorrectly attached a physical unit to the unit vector.
Why A is wrong: A is wrong because the problem states the displacement is in the +x direction, which correctly corresponds to +î. The direction is not the issue; the magnitude and dimensionality are.
Why C is wrong: C is wrong because ĵ points along +y, not +x. This option changes both the direction and the axis, compounding the error.
Why D is wrong: D is wrong because 5î m has magnitude 5 (not 1) and carries units of metres. A unit vector must have magnitude 1 and be dimensionless.
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Given
A particle has a position vector r = 2î − 3ĵ + 6k̂ (in metres) relative to the origin.
Required
Find the unit vector along r.
Concept
A unit vector along any vector A is obtained by dividing the vector by its magnitude: â = A / |A| (NCERT Class 11 Physics Chapter 3, page 30). The result is dimensionless and has magnitude 1.
Formula
r̂ = r / |r|, where |r| = √(rₓ² + rᵧ² + r_z²).
Substitution
|r| = √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49
Calculation
|r| = 7
Therefore: r̂ = (2/7)î + (−3/7)ĵ + (6/7)k̂
Note: the integers 2, −3, 6, and 7 are exact (counting/derived integers). They do not limit significant figures in this context.
Final answer
r̂ = (2/7)î − (3/7)ĵ + (6/7)k̂
Verification: |r̂| = √(4/49 + 9/49 + 36/49) = √(49/49) = 1 ✓. The result is dimensionless. ✓
Common trap
The most common error is forgetting to compute the magnitude and dividing each component by different numbers (e.g. dividing 2 by 2, 3 by 3, 6 by 6 — yielding î − ĵ + k̂, which has magnitude √3, not 1). All components must be divided by the same number: the total magnitude.
Similar NEET-style question
"If A = î + 2ĵ + 2k̂, find the unit vector in the direction of A."
(Answer: |A| = 3; â = (1/3)î + (2/3)ĵ + (2/3)k̂.)
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A vector A in two dimensions can be expressed as A = Ax î + Ay ĵ where Ax = A cos θ and Ay = A sin θ are the rectangular components along the x and y axes; î and ĵ are unit vectors along x and y respectively. The magnitude is A = √(Ax² + Ay²); the direction tan θ = Ay/Ax.
-- NCERT Class 11 Physics, Ch. 3, p. 32More in Kinematics: 14 exam traps and mistakes · 6 formulas · 10 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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