Q.Find the unit vector in the direction of the sum of the vectors, $\vec{a}=2\hat{i}+2\hat{j}-5\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$.

Concept: Direction Vectors — the unit vector in a given direction is found by dividing the vector by its magnitude.

First, find the sum:

Next, compute the magnitude:

The unit vector is the sum divided by its magnitude: …

Why Direction Vectors?

A unit vector in a given direction is simply a vector of length 1 that points exactly that way. To get it, you take any vector that already points in the desired direction and scale it down to length 1 — that is, divide by its own magnitude.

Here, the "desired direction" is the direction of the sum $\vec{a}+\vec{b}$. So the plan is straightforward:

Step 1: Add the vectors.

Step 2: Find the magnitude of the sum.

Step 3: Divide the sum by its magnitude.

Let's go.

1. Add the vectors component-wise

   $\vec{a} = 2\hat{i} + 2\hat{j} - 5\hat{k}$

   $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$

   Adding:

   • $\hat{i}$-components: $2 + 2 = 4$
   • $\hat{j}$-components: $2 + 1 = 3$
   • $\hat{k}$-components: $-5 + 3 = -2$

   So

2. Find the magnitude of this sum

   For a vector $x\hat{i}+y\hat{j}+z\hat{k}$, magnitude is $\sqrt{x^2+y^2+z^2}$.

Method: Unit vector in the direction of a sum (or combination) of vectors

Use this when the required direction is that of $\vec{a}+\vec{b}$ (or any linear combination), not of a single given vector.

Steps

Step 1: Form the resultant first.

Add component-wise:

Keep the signs of each component carefully as you combine.

Step 2: Compute the magnitude of the resultant.

$|\vec{a}+\vec{b}|=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2+(a_3+b_3)^2}.$ …

Common Mistakes

Mistake 1: Normalising $\vec{a}$ and $\vec{b}$ separately and adding the unit vectors.

Why it's wrong: $\hat{a}+\hat{b}$ points in a different direction than $\vec{a}+\vec{b}$ and generally is not even a unit vector. Correct approach: add the vectors first, then divide the resultant by its own magnitude.

Mistake 2: Sign error when adding the $\hat{k}$-components. …