Q.Find the unit vector in the direction of sum of vectors a=2i^−j^+k^ and b=2j^+k^.
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Concept understanding — Unit Vector Scaling
Unit Vector Scaling: From Intuition to Precision
Imagine you're drawing an arrow on graph paper. It has a direction and a length. Now suppose you want to keep the direction exactly the same, but make the arrow exactly one unit long. That's the core idea of unit vector scaling: take any vector and shrink or stretch it so its length becomes 1, without changing where it points.
The Intuition First
Think of a vector as a "directed step." A step of 3 metres north-east is a vector of length 3 in the north-east direction. To get a unit vector in the same direction, you'd take a step of exactly 1 metre north-east — scaling the original down by a factor of 3.
The key insight: direction is independent of length. A vector pointing north-east at length 5 and one at length 1 share the same direction. Unit vector scaling isolates that direction by forcing the length to be exactly 1.
The Precise Statement
v^=∥v∥v
Here v is any non-zero vector, ∥v∥ is its magnitude, and v^ ("v-hat") is the unit vector in the same direction. The operation: divide each component by the vector's length.
Example in 2D
Take v=(3,4). Its length is:
∥v∥=32+42=25=5
The unit vector is v^=(53,54).
Check: (3/5)2+(4/5)2=25/25=1. Direction unchanged — the ratio 3:4 is preserved.
Example in 3D
For v=(2,−1,2):
∥v∥=22+(−1)2+22=9=3
v^=(32,−31,32)
Why This Matters
Unit vectors are the building blocks of direction. In physics they represent pure directions for forces, velocities, or fields; in computer graphics, camera orientations and light directions. In mathematics they simplify dot products and projections — the dot product of a unit vector with another vector directly gives the component of that vector along the unit vector's direction.
Watch out
You cannot scale the zero vector to a unit vector — division by zero is undefined. The zero vector has no direction to preserve.
The One-Line Summary
Unit vector scaling takes any non-zero vector and divides it by its own length, producing a vector of length 1 that points exactly where the original pointed.
Normalising a vector into a unit vector is a routine computation throughout the NCERT Class 12 Vector Algebra chapter and appears constantly in CBSE board numericals and JEE Main problems. "Unit vector formula class 12 with examples" is a common search among students building up to direction-cosine and dot-product questions.
Add the two vectors, then divide the sum by its magnitude.
a=2i^−j^+k^, b=0i^+2j^+k^.
a+b=2i^+j^+2k^
∣a+b∣=22+12+22=9=3
Unit vector: 32i^+j^+2k^.
✓Final answer
32i^+31j^+32k^
a+b=2i^+j^+2k^ has magnitude 3, so the required unit vector is 31(2i^+j^+2k^).
The idea
A unit vector points the same way as a given vector but has length 1. To build one you divide the vector by its own magnitude. Here the given vector is the sum a+b, so first add, then normalise.
Add the vectors
Write b with its zero i^-component: b=0i^+2j^+k^.
a+b=(2+0)i^+(−1+2)j^+(1+1)k^=2i^+j^+2k^
Magnitude of the sum
∣a+b∣=22+12+22=4+1+4=9=3
Normalise
u^=∣a+b∣a+b=32i^+j^+2k^=32i^+31j^+32k^
Check:(32)2+(31)2+(32)2=94+1+4=1, confirming it is a unit vector.
✓Final answer
32i^+31j^+32k^
Method: Normalising a resultant into a unit vector
Use this whenever you need a unit vector in the direction of some combination of vectors (a sum, difference, or scalar multiple).
Steps
Step 1: Form the target vector first.
Before normalising, build the exact vector whose direction is wanted — here the sum a+b — by adding corresponding components. Do not normalise a and b separately.
Step 2: Find its magnitude.
∣v∣=x2+y2+z2.
Step 3: Divide the vector by its magnitude.
v^=∣v∣v.
As a check, the squares of the resulting components should add to 1.
Common Mistakes
Mistake 1: Normalising a and b separately, then adding the unit vectors.
Why it's wrong: the unit vector of a sum is not the sum of the unit vectors; you must add first, then normalise. Correct approach: compute a+b, then divide by ∣a+b∣.
Mistake 2: Forgetting the zero i^-component of b=2j^+k^.
Why it's wrong: leaving it out mis-sums the i^ term; b has i^-component 0. Correct approach: write b=0i^+2j^+k^ before adding.
Mistake 3: Stopping at the sum without dividing by the magnitude.
Why it's wrong: 2i^+j^+2k^ has length 3, so it is not yet a unit vector. Correct approach: divide by 3; the component squares should then sum to 1.