Giving directions like "walk 3 km north" has two parts: a distance (3 km) and a direction (north). A vector carries both. A unit vector keeps only the direction part — it has magnitude exactly 1, like a signpost that points the way without telling you how far to go.
We write unit vectors with a hat: v^ (read "v-hat").
Note
"Unit" comes from "unity" — one. A unit vector is simply a vector of length one.
The Idea Behind Verification
If someone hands you a vector and claims it is a unit vector, how do you check? You measure its length. Length 1 means yes; any other length means no. That is the whole idea:
v is a unit vector ⟺∣v∣=1.
The magnitude is computed from the components:
∣v∣=x2+y2(2D),∣v∣=x2+y2+z2(3D).
So verification is a two-step routine: compute the magnitude, then compare it with 1.
A Quick Check
Is b=(21,21) a unit vector?
∣b∣=21+21=1=1.
Yes — it is the unit vector pointing at 45∘. By contrast, (3,4) has magnitude 25=5, so it is not a unit vector.
Watch out
Do not assume a vector is "unit" just because every component is less than 1. For example (0.5,0.5) has magnitude 0.5≈0.707=1. Only the magnitude decides.
Why it's wrong: −b negates all three components of b; changing only one is a frequent slip. Correct approach: distribute the minus sign across every component before adding.
Mistake 2: Multiplying only one component by the coefficient …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
Q.Let O be the origin and r be the position vector of a point P. If OP makes angles 6π and 3π with i and j respectively, then a vector along OP with magnitude 2 units is
(A) i+3j
(B) j+3k
(C) 3i+j
(D) 3j+k
›Reveal solutionSolution
Direction cosines (23,21,0) scaled to length 2 give 3i+j. Correct option: (C).
Q.If a and b are two vectors such that a=2i+2j+pk, ∣b∣=7, a⋅b=4 and ∣a×b∣=517 then p=
(A) ±5
(B) ±6
(C) ±1
(D) ±3
›Reveal solutionSolution
Use the identity ∣a×b∣2=∣a∣2∣b∣2−(a⋅b)2 to find ∣a∣, then compute p from ∣a∣2=8+p2. The result is p=±5.
The core idea here is that the magnitude of the cross product and the dot product are not independent — they are linked by a fundamental identity that involves the magnitudes of the two vectors. You are given a⋅b, ∣b∣, and ∣a×b∣, but a itself has an unknown component p. The plan is to first find ∣a∣ using the identity, then solve for p from the expression of ∣a∣ in terms of p.
Recall the key identity. For any two vectors a and b,
∣a×b∣2=∣a∣2∣b∣2−(a⋅b)2.
This is a direct consequence of ∣a×b∣=∣a∣∣b∣sinθ and a⋅b=∣a∣∣b∣cosθ, combined with sin2θ+cos2θ=1. It lets you find ∣a∣ without knowing the angle between them.