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.
Concept: Cross Product Normalization — a vector perpendicular to two given vectors is along their cross product; a unit vector is obtained by dividing by its magnitude.
Step 1: Compute a+b and a−b.
a+b=(1+1)i^+(1+2)j^+(1+3)k^=2i^+3j^+4k^
a−b=(1−1)i^+(1−2)j^+(1−3)k^=0i^−j^−2k^
Step 2: Find a vector perpendicular to both by taking their cross product.
The unit vector is 6−i^+2j^−k^ (its negative is also perpendicular).
The cross product (a+b)×(a−b) gives a vector perpendicular to both, and normalizing it yields the unit vector 61(−i^+2j^−k^).
The key idea is simple: if you need a vector perpendicular to two given vectors, the cross product is your direct tool. Here, the two vectors are a+b and a−b. Instead of computing these sums separately and then taking their cross product, we can use a neat property — the cross product simplifies to 2(b×a), which saves work.
Let’s go step by step.
Find a+b and a−b.
Given a=i^+j^+k^ and b=i^+2j^+3k^,
Compute the cross product (a+b)×(a−b).
A vector perpendicular to both is given by their cross product. Let’s compute directly:
(a+b)×(a−b)=i^20j^3−1k^4−2
Expanding:
=i^3−14−2−j^204−2+k^203−1
=i^[3(−2)−4(−1)]−j^[2(−2)−4(0)]+k^[2(−1)−3(0)]
=i^[−6+4]−j^[−4−0]+k^[−2−0]
=i^(−2)−j^(−4)+k^(−2)=−2i^+4j^−2k^
So (a+b)×(a−b)=−2i^+4j^−2k^.
Tip
You could also use the identity (a+b)×(a−b)=a×a−a×b+b×a−b×b=0−a×b+b×a−0=2(b×a).
Computing b×a directly gives −i^+2j^−k^, and doubling it yields the same result. This shortcut avoids the determinant of the sum/difference vectors.
Find the magnitude of this cross product.
∣−2i^+4j^−2k^∣=(−2)2+42+(−2)2=4+16+4=24=26
Normalize to get the unit vector.
A unit vector perpendicular to both a+b and a−b is:
26−2i^+4j^−2k^=61(−i^+2j^−k^)
Watch out
The negative of this vector, 61(i^−2j^+k^), is also a unit vector perpendicular to both. Both are correct; the problem likely expects one of them. Always check if the question asks for "a" unit vector (any one) or "the" unit vector (often the one with a specific sign convention).
✓Final answer
The required unit vector is 61(−i^+2j^−k^) (or its negative).
Method: Unit vector perpendicular to two given vectors
The task splits into two independent jobs: the cross product supplies the perpendicular direction, and dividing by its magnitude fixes the length to 1.
Steps
Step 1: Build the two vectors first, if given indirectly
If the vectors are described as combinations (e.g. a+b and a−b), compute them component-wise before anything else, keeping any zero components explicitly.
Step 2: Cross them for the perpendicular direction
n=u×v=i^u1v1j^u2v2k^u3v3.
n is automatically perpendicular to both u and v, but its length is usually not 1.
Step 3: Normalize by dividing by the magnitude
n^=±∣n∣n.
Dividing by the magnitude rescales the vector to unit length without changing its direction. Keep the ±: there are exactly two opposite unit normals, and unless the question fixes a side, both are valid.
Step 4: Sanity note — a perpendicular direction exists only when u and v are non-parallel; if n=0 there is no single perpendicular to choose.
Common Mistakes
Mistake 1: Stopping at the cross product and reporting it as the answer.
Why it's wrong: the cross product has the right direction but the wrong length — the question asks for a unit vector. Correct approach: divide the cross product by its own magnitude to make its length exactly 1.
Mistake 2: Dropping a zero component when setting up the determinant.
Why it's wrong: a vector like a−b=0i^−j^−2k^ still has three components; writing only two shifts the columns and corrupts the determinant. Correct approach: always fill all three slots, using 0 where a component is absent.
Mistake 3: Forgetting the ± (that there are two unit normals).
Why it's wrong: the negative of the answer is equally perpendicular and unit-length; presenting only one as "the" answer misses that both are correct. Correct approach: state ±∣n∣n unless a specific orientation is required.