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.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2023Set A-21 markMCQ
Q.If ∣a×b∣2+∣a⋅b∣2=144 and ∣a∣=4 then ∣b∣ is equal to
(A) 3
(B) 8
(C) 4
(D) 12
›Reveal solutionSolution
The key idea is to use the identity ∣a×b∣2+∣a⋅b∣2=∣a∣2∣b∣2. Substituting the given values gives ∣b∣=3.
The problem gives you a relation between the magnitudes of the cross product and dot product of two vectors, along with the magnitude of one vector, and asks for the magnitude of the other. The trick is to recognize that these two quantities are not independent — they are linked by a fundamental identity that comes straight from the definitions.
Recall that for any two vectors a and b, the magnitude of the cross product is ∣a×b∣=∣a∣∣b∣sinθ, where θ is the angle between them. The dot product is ∣a⋅b∣=∣a∣∣b∣cosθ. Now square both and add them:
Square the cross product magnitude:
∣a×b∣2=∣a∣2∣b∣2sin2θ
Square the dot product magnitude:
∣a⋅b∣2=∣a∣2∣b∣2cos2θ
Add them:
∣a×b∣2+∣a⋅b∣2=∣a∣2∣b∣2(sin2θ+cos2θ)
Since sin2θ+cos2θ=1, this simplifies beautifully to:
Q.The scalar components of a unit vector which is perpendicular to each of the vectors ^+2^−k^ and 3^−^+2k^ are
(A) −833,−835,837
(B) −3,−5,7
(C) 833,−835,−837
(D) 3,−5,−7
›Reveal solutionSolution
The perpendicular direction is the cross product (3,−5,−7); normalising by 83 gives the unit vector (833,−835,−837).
Q.A unit vector perpendicular to the planes containing the vectors i^−j^+k^ and −i^+j^+k^ is
(A) ±21(i^+j^)
(B) ±2(i^+j^)
(C) ±(i^+j^−k^)
(D) ±22(i^+j^)