let `vec a = 2hat i +3hat j` and `vec b = hat i +4hat j` then find projection of `vec a` on `vec b`

Answer

To find the projection of vector \(\vec{a}\) on vector \(\vec{b}\), we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 2\hat{i} + 3\hat{j} \]
\[ \vec{b} = \hat{i} + 4\hat{j} \]

### 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:
\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 \]
For our vectors:
- \(a_1 = 2\), \(a_2 = 3\)
- \(b_1 = 1\), \(b_2 = 4\)

Calculating the dot product:
\[ \vec{a} \cdot \vec{b} = (2)(1) + (3)(4) = 2 + 12 = 14 \]

### 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 \(\vec{b}\):
\[ |\vec{b}| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \]

### Step 4: Calculate the projection of \(\vec{a}\) on \(\vec{b}\)
The projection of vector \(\vec{a}\) on vector \(\vec{b}\) is given by the formula:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]
We already calculated \(\vec{a} \cdot \vec{b} = 14\) and \(|\vec{b}| = \sqrt{17}\), so:
\[ |\vec{b}|^2 = 17 \]
Thus,
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{14}{17} \vec{b} \]
Substituting \(\vec{b}\):
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{14}{17} (\hat{i} + 4\hat{j}) = \frac{14}{17}\hat{i} + \frac{56}{17}\hat{j} \]

### Final Answer
The projection of vector \(\vec{a}\) on vector \(\vec{b}\) is:
\[ \text{proj}_{\vec{b}} \vec{a} = \frac{14}{17}\hat{i} + \frac{56}{17}\hat{j} \]