Find the projection of `hati+3hatj+7hatk` on the vector `2hati-3hatj+6hatk`

Answer

To find the projection of the vector \( \mathbf{A} = \hat{i} + 3\hat{j} + 7\hat{k} \) on the vector \( \mathbf{B} = 2\hat{i} - 3\hat{j} + 6\hat{k} \), we will follow these steps:

### Step 1: Identify the vectors
Let:
\[ \mathbf{A} = \hat{i} + 3\hat{j} + 7\hat{k} \]
\[ \mathbf{B} = 2\hat{i} - 3\hat{j} + 6\hat{k} \]

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

Substituting the components of \( \mathbf{A} \) and \( \mathbf{B} \):
\[ \mathbf{A} \cdot \mathbf{B} = (1)(2) + (3)(-3) + (7)(6) \]
Calculating each term:
\[ = 2 - 9 + 42 = 35 \]

### Step 3: Calculate the magnitude of vector \( \mathbf{B} \)
The magnitude of vector \( \mathbf{B} \) is calculated as:
\[ |\mathbf{B}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \]
Substituting the components of \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 4: Calculate the projection of \( \mathbf{A} \) on \( \mathbf{B} \)
The formula for the projection of vector \( \mathbf{A} \) on vector \( \mathbf{B} \) is given by:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} \mathbf{B} \]
First, calculate \( |\mathbf{B}|^2 \):
\[ |\mathbf{B}|^2 = 7^2 = 49 \]
Now substitute the values into the projection formula:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{35}{49} \mathbf{B} \]
Simplifying \( \frac{35}{49} \):
\[ \frac{35}{49} = \frac{5}{7} \]
Thus, the projection becomes:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{5}{7} (2\hat{i} - 3\hat{j} + 6\hat{k}) = \frac{10}{7}\hat{i} - \frac{15}{7}\hat{j} + \frac{30}{7}\hat{k} \]

### Final Answer
The projection of \( \mathbf{A} \) on \( \mathbf{B} \) is:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{10}{7}\hat{i} - \frac{15}{7}\hat{j} + \frac{30}{7}\hat{k} \]