If `vec(a) = 2 hat(i) - hat(j) + hat(k) and vec(b) = hat(i) - 2 hat(j) + hat(k) ` then projection of `vec(b)' on ' vec(a)` is

Options

A. `(5 sqrt(6))/(3)`
B. `(5)/(sqrt(6))`
C. `(5)/(sqrt(6))`
D. `5 sqrt(6)`

Answer

To find the projection of vector **b** on vector **a**, we can use the formula for the projection of one vector onto another. The projection of vector **b** onto vector **a** is given by:

\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a} \]

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

### Step 2: Calculate the dot product \(\vec{b} \cdot \vec{a}\)
The dot product is calculated as follows:
\[ \vec{b} \cdot \vec{a} = (1)(2) + (-2)(-1) + (1)(1) \]
Calculating each term:
- \(1 \cdot 2 = 2\)
- \(-2 \cdot -1 = 2\)
- \(1 \cdot 1 = 1\)

Now, summing these results:
\[ \vec{b} \cdot \vec{a} = 2 + 2 + 1 = 5 \]

### Step 3: Calculate the magnitude of \(\vec{a}\)
The magnitude of vector **a** is given by:
\[ |\vec{a}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} \]
Calculating each term:
- \(2^2 = 4\)
- \((-1)^2 = 1\)
- \(1^2 = 1\)

Now, summing these:
\[ |\vec{a}| = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Step 4: Calculate \(|\vec{a}|^2\)
\[ |\vec{a}|^2 = (\sqrt{6})^2 = 6 \]

### Step 5: Substitute into the projection formula
Now we can substitute into the projection formula:
\[ \text{proj}_{\vec{a}} \vec{b} = \left( \frac{5}{6} \right) \vec{a} \]

### Step 6: Write the final projection vector
Substituting \(\vec{a}\):
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{5}{6} (2\hat{i} - \hat{j} + \hat{k}) = \frac{10}{6}\hat{i} - \frac{5}{6}\hat{j} + \frac{5}{6}\hat{k} \]
This simplifies to:
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{5}{3}\hat{i} - \frac{5}{6}\hat{j} + \frac{5}{6}\hat{k} \]

### Final Answer
The projection of vector **b** on vector **a** is:
\[ \text{proj}_{\vec{a}} \vec{b} = \frac{5}{3}\hat{i} - \frac{5}{6}\hat{j} + \frac{5}{6}\hat{k} \]