Projection of the vector `veca=2hati+3hatj-2hatk` on the vector `vecb=hati+2hatj+3hatk` is

Options

A. `2/sqrt14`
B. `1/sqrt14`
C. `3/sqrt14`
D. none of these

Explanation

To find the projection of the vector \(\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}\) on the vector \(\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}\), we can use the formula for the projection of vector \(\vec{a}\) onto vector \(\vec{b}\):

\[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]

### Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

The dot product \(\vec{a} \cdot \vec{b}\) is calculated as follows:

\[ \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j} - 2\hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) \]

Calculating the dot product:

\[ = 2 \cdot 1 + 3 \cdot 2 + (-2) \cdot 3 \]
\[ = 2 + 6 - 6 \]
\[ = 2 \]

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

The magnitude of vector \(\vec{b}\) is given by:

\[ |\vec{b}| = \sqrt{1^2 + 2^2 + 3^2} \]
\[ = \sqrt{1 + 4 + 9} \]
\[ = \sqrt{14} \]

### Step 3: Calculate the projection

Now we can substitute the values into the projection formula:

\[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]

First, we need \( |\vec{b}|^2 \):

\[ |\vec{b}|^2 = 14 \]

Now substituting back:

\[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{2}{14} \vec{b} \]
\[ = \frac{1}{7} \vec{b} \]

Now substituting \(\vec{b}\):

\[ = \frac{1}{7} (\hat{i} + 2\hat{j} + 3\hat{k}) \]
\[ = \frac{1}{7} \hat{i} + \frac{2}{7} \hat{j} + \frac{3}{7} \hat{k} \]

### Final Answer

The projection of the vector \(\vec{a}\) on the vector \(\vec{b}\) is:

\[ \frac{1}{7} \hat{i} + \frac{2}{7} \hat{j} + \frac{3}{7} \hat{k} \]