Cross Product Normalization (Unit Vector Perpendicular to Two Vectors)
A frequent job in vector algebra is: given two vectors, find a vector of length 1 that is perpendicular to both of them. The cross product does the perpendicular part; normalization does the length part. Putting them together is what this idea is about.
Step 1 — the cross product gives the direction
For two non-parallel vectors a and b, the cross product
a×b=∣a∣∣b∣sinθn^
is a vector that is perpendicular to botha and b. Its direction is fixed by the right-hand rule. So a×b already points the way we want — but its length is ∣a∣∣b∣sinθ, which is usually not1.
Step 2 — normalize to get unit length
To normalize any non-zero vector means to divide it by its own magnitude, producing a vector of length 1 in the same direction. Applying that to the cross product:
n^=±∣a×b∣a×b
This n^ is a unit vector perpendicular to botha and b. The ± matters: there are exactly two such unit normals, pointing in opposite directions. The plus sign gives the right-hand-rule direction of a×b; the minus sign gives the other side.
Why the division works
Dividing by the magnitude only rescales the vector — it never changes its direction. So n^ keeps the perpendicularity that the cross product built in, while its length becomes ∣a×b∣∣a×b∣=1.
Quick example
Let a=i^+j^ and b=j^+k^. Then
a×b=i^−j^+k^,∣a×b∣=1+1+1=3.
So a unit vector perpendicular to both is
n^=±31(i^−j^+k^).
Watch out
Normalization needs a non-zero cross product. If a and b are parallel, a×b=0 and ∣a×b∣=0 — you cannot divide by zero, and geometrically there is no single perpendicular direction to pick.
Takeaway: cross product for the perpendicular direction, then divide by its magnitude for unit length — that two-step recipe delivers the unit normal n^=±(a×b)/∣a×b∣.
Students preparing for boards search "unit vector perpendicular to two vectors formula" and "cross product normalization class 12 maths," both of which are covered in the Vector Algebra chapter of the NCERT/CBSE Class 12 Mathematics curriculum. This two-step cross-product-then-normalize technique is also a frequent JEE Main and state CET question type.
A vector perpendicular to two vectors is their cross product; divide by its length to get a unit vector.
Step 1 — Form the two vectors.
a+b=4i^+4j^+0k^,a−b=2i^+0j^+4k^.
Step 2 — Cross product.
(a+b)×(a−b)=i^42j^40k^04=16i^−16j^−8k^.
Step 3 — Normalise.
16i^−16j^−8k^=256+256+64=576=24,
so the unit vector is 241(16i^−16j^−8k^)=31(2i^−2j^−k^).
✓Final answer
The required unit vector is ±31(2i^−2j^−k^) (both directions are perpendicular to the two given vectors).
The cross product (a+b)×(a−b)=16i^−16j^−8k^ has length 24, so a unit vector perpendicular to both is ±31(2i^−2j^−k^).
The idea
The cross product of two vectors is always perpendicular to both of them. So to find something perpendicular to a+b and a−b at the same time, cross those two vectors, then shrink the result to length 1 by dividing by its magnitude. Because the opposite direction is perpendicular too, the answer carries a ±.
Step-by-step
1. Build the two vectors. With a=3i^+2j^+2k^ and b=i^+2j^−2k^,
a+b=(3+1)i^+(2+2)j^+(2−2)k^=4i^+4j^,
a−b=(3−1)i^+(2−2)j^+(2+2)k^=2i^+4k^.
2. Cross them.
(a+b)×(a−b)=i^42j^40k^04.
i^: (4)(4)−(0)(0)=16
j^: −[(4)(4)−(0)(2)]=−16
k^: (4)(0)−(4)(2)=−8
⇒c=16i^−16j^−8k^=8(2i^−2j^−k^).
3. Find the magnitude.
∣c∣=162+(−16)2+(−8)2=256+256+64=576=24.
4. Normalise.
c^=∣c∣c=248(2i^−2j^−k^)=31(2i^−2j^−k^).
The negative of this is equally valid, since it is also perpendicular to both given vectors.
✓Final answer
The required unit vector is ±31(2i^−2j^−k^).
Method: A Unit Vector Perpendicular to Two Given Vectors
The cross product of two vectors is perpendicular to both — normalise it to get a perpendicular unit vector.
Steps
Step 1: Assemble the two vectors, then cross them.
After forming the required vectors (e.g. a+b and a−b), compute their cross product via the determinant. The result is automatically perpendicular to each.
Step 2: Find its magnitude.
∣c∣=c12+c22+c32
Step 3: Divide to normalise, and include ±.
c^=±∣c∣c
Both directions are perpendicular to the two given vectors, so both signs are valid answers.
Common Mistakes
Mistake 1: Crossing a and b directly.
Why it's wrong: the answer must be perpendicular to a+b and a−b, so those are the two vectors to cross — not a and b themselves. Correct approach: first form a+b and a−b, then cross them.
Mistake 2: Forgetting to normalise.
Why it's wrong: the raw cross product is perpendicular but not of length 1. Correct approach: divide by its magnitude (24 here) to get a unit vector.
Mistake 3: Omitting the ±.
Why it's wrong: the opposite direction is equally perpendicular to both vectors. Correct approach: report ±31(2i^−2j^−k^).