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.
KEAM 2026Set eng-2026-04204 marksMCQ
Q.Let a,b,c be three vectors such that ∣a∣=2, ∣b∣=3, ∣c∣=4 and a⋅b=a⋅c=0. If the angle between b and c is 3π, then a is equal to
(A) ±31(b×c)
(B) ±231(b×c)
(C) ±321(b×c)
(D) ±331(b×c)
(E) ±221(b×c)
›Reveal solutionSolution
a is perpendicular to both b and c, so it is along b×c; scale by ∣a∣/∣b×c∣.
Since a⋅b=a⋅c=0, a is perpendicular to the plane of b and c, hence parallel to b×c.
Q.If the points (1,0,0), (0,3,0) and (0,0,2) lie on a plane, then the unit normal vector n^ to the plane is
(A) 141(i^+3j^+2k^)
(B) 71(2i^+3j^+6k^)
(C) 141(2i^+3j^+k^)
(D) 71(3i^+2j^+6k^)
(E) 71(6i^+2j^+3k^)
›Reveal solutionSolution
n^=71(6i^+2j^+3k^).
Concept and Intuition
A plane with axis intercepts a,b,c is ax+by+cz=1; its normal comes from the cleared-denominator coefficients.