If `vec(a) = 2hat(i) + hat(j) - hat(k)` and `vec(b) = hat(i) - hat(k)`, then projection of `vec(a)` on `vec(b)` will be :

Options

A. `sqrt((3)/(2))`
B. `(3)/(sqrt(2))`
C. `(1)/(sqrt(2))`
D. `(3sqrt(6))/(2)`

Answer

To find the projection of vector **a** on vector **b**, we can follow these steps:

### Step 1: Write down the vectors
Given:
\[ \vec{a} = 2\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{k} \]

### Step 2: Find the unit vector of **b**
The unit vector **b** (denoted as **b̂**) is calculated by dividing vector **b** by its magnitude.

First, we need to find the magnitude of **b**:
\[ |\vec{b}| = \sqrt{(1)^2 + (0)^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

Now, we can find **b̂**:
\[ \hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{\hat{i} - \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{k} \]

### Step 3: Calculate the dot product of **a** and **b**
Next, we compute the dot product of **a** and **b**:
\[ \vec{a} \cdot \vec{b} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - \hat{k}) \]
Using the properties of dot product:
\[ = 2(1) + 1(0) + (-1)(-1) = 2 + 0 + 1 = 3 \]

### Step 4: Calculate the projection of **a** on **b**
The projection of **a** on **b** is given by the formula:
\[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \hat{b} \]
Substituting the values we found:
\[ \text{Projection} = \frac{3}{\sqrt{2}} \hat{b} \]

### Step 5: Substitute **b̂** into the projection formula
Now, substituting **b̂**:
\[ \text{Projection} = \frac{3}{\sqrt{2}} \left(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{k}\right) = \frac{3}{2}\hat{i} - \frac{3}{2}\hat{k} \]

### Final Answer
Thus, the projection of vector **a** on vector **b** is:
\[ \frac{3}{2}\hat{i} - \frac{3}{2}\hat{k} \]