The unit vector bisecting `vec(OY)` and `vec(OZ)` is

Options

A. `(veci+vecj+veck)/sqrt(3)`
B. `(veci-veck)/sqrt(2)`
C. `(vecj+veck)/sqrt(2)`
D. `(-vecj+veck)/sqrt(2)`

Explanation

To find the unit vector that bisects the vectors along the y-axis (denoted as \(\vec{OY}\)) and the z-axis (denoted as \(\vec{OZ}\)), we can follow these steps:

### Step 1: Identify the unit vectors along the y-axis and z-axis
The unit vector along the y-axis is represented as:
\[ \vec{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \]
The unit vector along the z-axis is represented as:
\[ \vec{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]

### Step 2: Find the sum of the unit vectors
To find the bisector of \(\vec{j}\) and \(\vec{k}\), we add these two vectors:
\[ \vec{b} = \vec{j} + \vec{k} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \]

### Step 3: Calculate the magnitude of the bisector vector
Next, we need to calculate the magnitude of the vector \(\vec{b}\):
\[ |\vec{b}| = \sqrt{(0)^2 + (1)^2 + (1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \]

### Step 4: Find the unit vector in the direction of the bisector
To find the unit vector, we divide the bisector vector \(\vec{b}\) by its magnitude:
\[ \text{Unit vector} = \frac{\vec{b}}{|\vec{b}|} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Conclusion
Thus, the unit vector bisecting \(\vec{OY}\) and \(\vec{OZ}\) is:
\[ \text{Unit vector} = \frac{1}{\sqrt{2}} \vec{j} + \frac{1}{\sqrt{2}} \vec{k} \]