Find the projection of `vec(b)+ vec(c )` on `vec(a)` where `vec(a)= 2 hat(i) -2hat(j) + hat(k), vec(b)= hat(i) + 2hat(j)- 2hat(k), vec(c ) = 2hat(i) - hat(j) + 4hat(k)`

Answer

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

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

### Step 2: Calculate \(\vec{b} + \vec{c}\)
We add the vectors \(\vec{b}\) and \(\vec{c}\):
\[ \vec{b} + \vec{c} = (\hat{i} + 2\hat{j} - 2\hat{k}) + (2\hat{i} - \hat{j} + 4\hat{k}) \]
Combine like terms:
\[ = (1 + 2)\hat{i} + (2 - 1)\hat{j} + (-2 + 4)\hat{k} \]
\[ = 3\hat{i} + 1\hat{j} + 2\hat{k} \]

### Step 3: Find the projection of \(\vec{b} + \vec{c}\) on \(\vec{a}\)
The formula for the projection of vector \(\vec{x}\) on vector \(\vec{y}\) is given by:
\[ \text{Projection of } \vec{x} \text{ on } \vec{y} = \frac{\vec{x} \cdot \vec{y}}{|\vec{y}|} \]
Here, \(\vec{x} = \vec{b} + \vec{c} = 3\hat{i} + 1\hat{j} + 2\hat{k}\) and \(\vec{y} = \vec{a} = 2\hat{i} - 2\hat{j} + \hat{k}\).

### Step 4: Calculate the dot product \((\vec{b} + \vec{c}) \cdot \vec{a}\)
\[ \vec{b} + \vec{c} \cdot \vec{a} = (3\hat{i} + 1\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 2\hat{j} + \hat{k}) \]
Calculating the dot product:
\[ = 3 \cdot 2 + 1 \cdot (-2) + 2 \cdot 1 \]
\[ = 6 - 2 + 2 = 6 \]

### Step 5: Calculate the magnitude of \(\vec{a}\)
\[ |\vec{a}| = \sqrt{(2)^2 + (-2)^2 + (1)^2} \]
\[ = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 6: Find the projection
Now we can find the projection:
\[ \text{Projection of } (\vec{b} + \vec{c}) \text{ on } \vec{a} = \frac{6}{3} = 2 \]

### Final Answer
The projection of \(\vec{b} + \vec{c}\) on \(\vec{a}\) is \(2\).

---