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 use the formula for the projection of vector **a** onto vector **b**:

\[ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]

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

### Step 2: Calculate the dot product \(\vec{a} \cdot \vec{b}\)
The dot product of two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is given by:
\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

For our vectors:
- \(a_1 = 2\), \(a_2 = 1\), \(a_3 = -1\)
- \(b_1 = 1\), \(b_2 = 0\), \(b_3 = -1\)

Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (2)(1) + (1)(0) + (-1)(-1) = 2 + 0 + 1 = 3 \]

### Step 3: Calculate the magnitude of vector \(\vec{b}\)
The magnitude of vector \(\vec{b}\) is given by:
\[ |\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \]
Calculating it:
\[ |\vec{b}| = \sqrt{(1)^2 + (0)^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

### Step 4: Calculate the projection
Using the projection formula:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]
First, we need \(|\vec{b}|^2\):
\[ |\vec{b}|^2 = (\sqrt{2})^2 = 2 \]

Now substituting the values:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{3}{2} \vec{b} \]
Substituting \(\vec{b}\):
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{3}{2} (\hat{i} - \hat{k}) = \frac{3}{2} \hat{i} - \frac{3}{2} \hat{k} \]

### Final Answer:
The projection of vector **a** on vector **b** is:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{3}{2} \hat{i} - \frac{3}{2} \hat{k} \]