Q.Find a vector of magnitude 6, which is perpendicular to both the vectors $2\hat{i}-\hat{j}+2\hat{k}$ and $4\hat{i}-\hat{j}+3\hat{k}$.

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 $\vec{a}$ and $\vec{b}$, the cross product

is a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$. Its direction is fixed by the right-hand rule. So $\vec{a}\times\vec{b}$ already points the way we want — but its length is $|\vec{a}||\vec{b}|\sin\theta$, which is usually not $1$.

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:

This $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. The $\pm$ matters: there are exactly two such unit normals, pointing in opposite directions. The plus sign gives the right-hand-rule direction of $\vec{a}\times\vec{b}$; 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 $\hat n$ keeps the perpendicularity that the cross product built in, while its length becomes $\dfrac{|\vec a\times\vec b|}{|\vec a\times\vec b|}=1$.

Quick example

Let $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=\hat{j}+\hat{k}$. Then

So a unit vector perpendicular to both is …