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.
KCET 2020Set A-11 markMCQ
Q.The two vectors i^+j^+k^ and i^+3j^+5k^ represent the two sides AB and AC respectively of a △ABC. The length of the median through A is
(A) 214
(B) 14
(C) 7
(D) 14
›Reveal solutionSolution
The median from A goes to the midpoint of BC. The vector from A to that midpoint is the average of the two side vectors, and its magnitude gives the median length: 14.
The median through A joins A to the midpoint of the opposite side BC. If we know the vectors representing sides AB and AC, the vector from A to the midpoint of BC is simply the average of those two vectors. That’s because the midpoint’s position vector (with A as origin) is the average of the position vectors of B and C. So the median length is just the magnitude of that average.
Let’s work it out.
Let AB=i^+j^+k^ and AC=i^+3j^+5k^.
Taking A as the origin, the position vectors of B and C are exactly these vectors.
The midpoint M of BC has position vector
AM=2AB+AC
because the midpoint’s coordinates are the average of the coordinates of B and C.