Q.Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$.

A vector perpendicular to two vectors is their cross product; divide by its length to get a unit vector.

Step 1 — Form the two vectors.

Step 2 — Cross product.

Step 3 — Normalise.

so the unit vector is $\dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k}) = \dfrac{1}{3}(2\hat{i}-2\hat{j}-\hat{k}).$

The idea

The cross product of two vectors is always perpendicular to both of them. So to find something perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ at the same time, cross those two vectors, then shrink the result to length $1$ by dividing by its magnitude. Because the opposite direction is perpendicular too, the answer carries a $\pm$.

Step-by-step

1. Build the two vectors. With $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$,

2. Cross them.

• $\hat{i}$: $(4)(4)-(0)(0) = 16$
• $\hat{j}$: $-\big[(4)(4)-(0)(2)\big] = -16$
• $\hat{k}$: $(4)(0)-(4)(2) = -8$

3. Find the magnitude.

4. Normalise.

The negative of this is equally valid, since it is also perpendicular to both given vectors.

Method: A Unit Vector Perpendicular to Two Given Vectors

The cross product of two vectors is perpendicular to both — normalise it to get a perpendicular unit vector.

Steps

Step 1: Assemble the two vectors, then cross them.

After forming the required vectors (e.g. $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$), compute their cross product via the determinant. The result is automatically perpendicular to each.

Step 2: Find its magnitude.

Step 3: Divide to normalise, and include $\pm$.

Both directions are perpendicular to the two given vectors, so both signs are valid answers.

Common Mistakes

Mistake 1: Crossing $\vec{a}$ and $\vec{b}$ directly.

Why it's wrong: the answer must be perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, so those are the two vectors to cross — not $\vec{a}$ and $\vec{b}$ themselves. Correct approach: first form $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, then cross them.

Mistake 2: Forgetting to normalise.

Why it's wrong: the raw cross product is perpendicular but not of length $1$. Correct approach: divide by its magnitude ($24$ here) to get a unit vector.

Mistake 3: Omitting the $\pm$.

Why it's wrong: the opposite direction is equally perpendicular to both vectors. Correct approach: report $\pm\frac13(2\hat{i}-2\hat{j}-\hat{k})$.