If `vecA=2hati+3hatj-hatk` and `vecB=-hati+3hatj+4hatk` then projection of `vecA` on `vecB` will be

Options

A. `(3)/(sqrt(13))`
B. `(3)/(sqrt(26))`
C. `sqrt((3)/(26))`
D. `sqrt((3)/(13))`

Answer

To find the projection of vector \(\vec{A}\) on vector \(\vec{B}\), we can use the formula for projection:

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

Where:
- \(\vec{A} \cdot \vec{B}\) is the dot product of vectors \(\vec{A}\) and \(\vec{B}\).
- \(|\vec{B}|\) is the magnitude of vector \(\vec{B}\).

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

Given:
\[ \vec{A} = 2\hat{i} + 3\hat{j} - \hat{k} \]
\[ \vec{B} = -\hat{i} + 3\hat{j} + 4\hat{k} \]

The dot product is calculated as follows:
\[ \vec{A} \cdot \vec{B} = (2)(-1) + (3)(3) + (-1)(4) \]
\[ = -2 + 9 - 4 \]
\[ = 3 \]

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

The magnitude of vector \(\vec{B}\) is given by:
\[ |\vec{B}| = \sqrt{(-1)^2 + (3)^2 + (4)^2} \]
\[ = \sqrt{1 + 9 + 16} \]
\[ = \sqrt{26} \]

### Step 3: Calculate the projection of \(\vec{A}\) on \(\vec{B}\)

Using 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 to calculate \(|\vec{B}|^2\):
\[ |\vec{B}|^2 = 26 \]

Now substituting the values:
\[ \text{Projection of } \vec{A} \text{ on } \vec{B} = \frac{3}{26} \vec{B} \]
\[ = \frac{3}{26} (-\hat{i} + 3\hat{j} + 4\hat{k}) \]
\[ = -\frac{3}{26} \hat{i} + \frac{9}{26} \hat{j} + \frac{12}{26} \hat{k} \]

### Final Result

Thus, the projection of \(\vec{A}\) on \(\vec{B}\) is:
\[ -\frac{3}{26} \hat{i} + \frac{9}{26} \hat{j} + \frac{12}{26} \hat{k} \]