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

Concept: Cross Product Normalization — a vector perpendicular to two given vectors is along their cross product; a unit vector is obtained by dividing by its magnitude.

Step 1: Compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$.

Step 2: Find a vector perpendicular to both by taking their cross product.

Step 3: Normalize this vector. Its magnitude is:

The required unit vector is:

The key idea is simple: if you need a vector perpendicular to two given vectors, the cross product is your direct tool. Here, the two vectors are $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$. Instead of computing these sums separately and then taking their cross product, we can use a neat property — the cross product simplifies to $2(\vec{b}\times\vec{a})$, which saves work.

Let’s go step by step.

1. Find $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$. Given $\vec{a} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}+3\hat{k}$,

2. Compute the cross product $(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})$. A vector perpendicular to both is given by their cross product. Let’s compute directly:

Expanding:

So $(\vec{a}+\vec{b})\times(\vec{a}-\vec{b}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.

3. Find the magnitude of this cross product.

4. Normalize to get the unit vector. A unit vector perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is:

Method: Unit vector perpendicular to two given vectors

The task splits into two independent jobs: the cross product supplies the perpendicular direction, and dividing by its magnitude fixes the length to $1$.

Steps

Step 1: Build the two vectors first, if given indirectly

If the vectors are described as combinations (e.g. $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$), compute them component-wise before anything else, keeping any zero components explicitly.

Step 2: Cross them for the perpendicular direction

$\vec{n}$ is automatically perpendicular to both $\vec{u}$ and $\vec{v}$, but its length is usually not $1$.

Step 3: Normalize by dividing by the magnitude

Dividing by the magnitude rescales the vector to unit length without changing its direction. Keep the $\pm$: there are exactly two opposite unit normals, and unless the question fixes a side, both are valid.

Step 4: Sanity note — a perpendicular direction exists only when $\vec{u}$ and $\vec{v}$ are non-parallel; if $\vec{n} = \vec{0}$ there is no single perpendicular to choose.

Common Mistakes

Mistake 1: Stopping at the cross product and reporting it as the answer.

Why it's wrong: the cross product has the right direction but the wrong length — the question asks for a unit vector. Correct approach: divide the cross product by its own magnitude to make its length exactly $1$.

Mistake 2: Dropping a zero component when setting up the determinant.

Why it's wrong: a vector like $\vec{a}-\vec{b} = 0\hat{i}-\hat{j}-2\hat{k}$ still has three components; writing only two shifts the columns and corrupts the determinant. Correct approach: always fill all three slots, using $0$ where a component is absent.

Mistake 3: Forgetting the $\pm$ (that there are two unit normals).

Why it's wrong: the negative of the answer is equally perpendicular and unit-length; presenting only one as "the" answer misses that both are correct. Correct approach: state $\pm\dfrac{\vec{n}}{|\vec{n}|}$ unless a specific orientation is required.