Find the projection of `(8hat(i)+hat(j))` in the direction of `(hat(i)+2hat(j)-2hat(k))`

Answer

To find the projection of the vector \( \mathbf{A} = 8\hat{i} + \hat{j} \) in the direction of the vector \( \mathbf{B} = \hat{i} + 2\hat{j} - 2\hat{k} \), we can follow these steps:

### Step 1: Identify the vectors
Let:
\[ \mathbf{A} = 8\hat{i} + \hat{j} \]
\[ \mathbf{B} = \hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)
The dot product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (8\hat{i} + \hat{j}) \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) \]
Calculating the dot product:
\[ = 8 \cdot 1 + 1 \cdot 2 + 0 \cdot (-2) = 8 + 2 + 0 = 10 \]

### Step 3: Calculate the magnitude of vector \( \mathbf{B} \)
The magnitude of vector \( \mathbf{B} \) is given by:
\[ |\mathbf{B}| = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Calculate the projection of \( \mathbf{A} \) onto \( \mathbf{B} \)
The projection of vector \( \mathbf{A} \) onto vector \( \mathbf{B} \) is given by the formula:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|^2} \mathbf{B} \]
First, we need \( |\mathbf{B}|^2 \):
\[ |\mathbf{B}|^2 = 3^2 = 9 \]
Now substituting the values:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{10}{9} \mathbf{B} = \frac{10}{9} (\hat{i} + 2\hat{j} - 2\hat{k}) \]
Distributing \( \frac{10}{9} \):
\[ = \frac{10}{9} \hat{i} + \frac{20}{9} \hat{j} - \frac{20}{9} \hat{k} \]

### Final Answer
The projection of \( \mathbf{A} \) in the direction of \( \mathbf{B} \) is:
\[ \text{proj}_{\mathbf{B}} \mathbf{A} = \frac{10}{9} \hat{i} + \frac{20}{9} \hat{j} - \frac{20}{9} \hat{k} \]