The projection of `bar(A) = 2hati +2hatj +5hatk` on to `bar(B) = hati -hatj` is:

Options

A. Zero
B. `(4)/(sqrt(2))`
C. `-(4)/(sqrt(2))`
D. 4

Explanation

To solve the problem of finding the projection of vector **A** onto vector **B**, we can follow these steps:

### Step 1: Identify the vectors
Given:
- **A** = \( \bar{A} = 2\hat{i} + 2\hat{j} + 5\hat{k} \)
- **B** = \( \bar{B} = \hat{i} - \hat{j} \)

### Step 2: Calculate the dot product of **A** and **B**
The dot product \( \bar{A} \cdot \bar{B} \) is calculated as follows:
\[ \bar{A} \cdot \bar{B} = (2\hat{i} + 2\hat{j} + 5\hat{k}) \cdot (\hat{i} - \hat{j}) \]
Using the properties of dot product:
\[ = 2(\hat{i} \cdot \hat{i}) + 2(\hat{j} \cdot (-\hat{j})) + 5(\hat{k} \cdot \hat{i}) + 5(\hat{k} \cdot (-\hat{j})) \]
\[ = 2(1) + 2(-1) + 5(0) + 5(0) \]
\[ = 2 - 2 + 0 + 0 = 0 \]

### Step 3: Calculate the magnitude of vector **B**
The magnitude of vector **B** is given by:
\[ |\bar{B}| = \sqrt{(\hat{i})^2 + (-\hat{j})^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Calculate the projection of **A** onto **B**
The projection of vector **A** onto vector **B** is given by the formula:
\[ \text{Projection of } \bar{A} \text{ onto } \bar{B} = \frac{\bar{A} \cdot \bar{B}}{|\bar{B}|^2} \bar{B} \]
Since we found that \( \bar{A} \cdot \bar{B} = 0 \):
\[ \text{Projection of } \bar{A} \text{ onto } \bar{B} = \frac{0}{|\bar{B}|^2} \bar{B} = 0 \cdot \bar{B} = \bar{0} \]

### Final Answer
The projection of vector **A** onto vector **B** is \( \bar{0} \) (the zero vector).

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