Find the scalar projection of :
`vec(a)=hat(i)-hat(j)` on `vec(b)=hat(i)+hat(j)`

Answer

To find the scalar projection of vector **a** on vector **b**, we can use the formula:

\[ \text{Scalar Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

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

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

For our vectors:
- \(a_1 = 1\) (coefficient of \(\hat{i}\) in \(\vec{a}\))
- \(a_2 = -1\) (coefficient of \(\hat{j}\) in \(\vec{a}\))
- \(b_1 = 1\) (coefficient of \(\hat{i}\) in \(\vec{b}\))
- \(b_2 = 1\) (coefficient of \(\hat{j}\) in \(\vec{b}\))

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

### Step 3: Calculate the magnitude of vector \(\vec{b}\)
The magnitude of a vector \(\vec{b} = b_1 \hat{i} + b_2 \hat{j}\) is given by:
\[ |\vec{b}| = \sqrt{b_1^2 + b_2^2} \]

For our vector \(\vec{b}\):
\[ |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Calculate the scalar projection
Now we can substitute the values into the scalar projection formula:
\[ \text{Scalar Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{0}{\sqrt{2}} = 0 \]

### Final Answer
The scalar projection of vector **a** on vector **b** is:
\[ \boxed{0} \]