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.
AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
Q.A unit vector perpendicular to the vectors aˉ=2iˉ+3jˉ+4kˉ and bˉ=3jˉ+2kˉ is
(A) 223iˉ+2jˉ−2kˉ
(B) 223iˉ+2jˉ−3kˉ
(C) 223iˉ−2jˉ+3kˉ
(D) 223iˉ+2jˉ+3kˉ
›Reveal solutionSolution
The cross product aˉ×bˉ gives a vector perpendicular to both; normalizing it (either sign) gives ±223iˉ+2jˉ−3kˉ, matching option (B).
Concept and Intuition
Any vector perpendicular to both aˉ and bˉ must be parallel to aˉ×bˉ; normalizing that cross product (in either direction) gives all unit vectors perpendicular to the plane of aˉ,bˉ.
Q.If OA=2iˉ−jˉ+kˉ, OB=3iˉ−kˉ and OC=2jˉ+3kˉ are the position vectors of the points A, B and C, then a unit vector perpendicular to the plane containing A, B and C is
(A) 2218iˉ−4jˉ+2kˉ
(B) 76iˉ+2jˉ+3kˉ
(C) 119iˉ+2jˉ+6kˉ
(D) 938iˉ+2jˉ+5kˉ
›Reveal solutionSolution
The unit normal to a plane through three points comes from normalising the cross product of two vectors lying in that plane; here it is 938iˉ+2jˉ+5kˉ.
Concept and Intuition
Any two non-parallel vectors lying in the plane ABC (built from the three position vectors) span that plane, and their cross product is perpendicular to both — hence perpendicular to the whole plane. Normalising it gives the required unit vector.