The projection of vector `veca=2hati+3hatj+2hatk`, on the vector `vecb=hati+2hatj+hatk` is

Options

A. `(5sqrt(6))/3`
B. `(sqrt(6))/3`
C. `5sqrt(6)`
D. None of the above

Answer

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

### Step 1: Calculate the dot product of \(\vec{A}\) and \(\vec{B}\)

The dot product \(\vec{A} \cdot \vec{B}\) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = (2\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) \]

Using the formula for the dot product:

\[ \vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10 \]

### Step 2: Calculate the magnitude of \(\vec{B}\)

The magnitude of \(\vec{B}\) is calculated using the formula:

\[ |\vec{B}| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

### Step 3: Use the projection formula

The formula for the projection of \(\vec{A}\) on \(\vec{B}\) is given by:

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} \]

Substituting the values we calculated:

\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{10}{\sqrt{6}} \]

### Step 4: Final result

Thus, the projection of \(\vec{A}\) on \(\vec{B}\) is:

\[ \frac{10}{\sqrt{6}} \]