If the position vector of a point p with respect to the origin O is `hat(i) + 3hat(j) -2hat(k)` and that of a point Q is `3hat(i) + hat(j)- 2hat(k)`, then what is the position vector of the bisector of the `anglePOQ` ?

Options

A. A. `hat(i) -hat(j)- hat(k)`
B. B. `hat(i) + hat(j)- hat(k)`
C. C. `hat(i) + hat(j) + hat(k)`
D. D. None of these

Answer

To find the position vector of the bisector of the angle \( \angle POQ \), we will follow these steps:

### Step 1: Identify the position vectors of points P and Q
The position vector of point P is given as:
\[ \vec{OP} = \hat{i} + 3\hat{j} - 2\hat{k} \]
The position vector of point Q is given as:
\[ \vec{OQ} = 3\hat{i} + \hat{j} - 2\hat{k} \]

### Step 2: Calculate the magnitudes of the position vectors
To find the magnitudes, we use the formula:
\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]

For \( \vec{OP} \):
\[ |\vec{OP}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

For \( \vec{OQ} \):
\[ |\vec{OQ}| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

### Step 3: Determine the type of triangle formed by points O, P, and Q
Since both magnitudes are equal, we conclude that triangle \( POQ \) is an isosceles triangle.

### Step 4: Find the position vector of the midpoint of PQ
The midpoint \( E \) of line segment \( PQ \) can be found using the formula:
\[ \vec{E} = \frac{\vec{P} + \vec{Q}}{2} \]

Calculating \( \vec{P} + \vec{Q} \):
\[ \vec{P} + \vec{Q} = (\hat{i} + 3\hat{j} - 2\hat{k}) + (3\hat{i} + \hat{j} - 2\hat{k}) = (1 + 3)\hat{i} + (3 + 1)\hat{j} + (-2 - 2)\hat{k} = 4\hat{i} + 4\hat{j} - 4\hat{k} \]

Now divide by 2:
\[ \vec{E} = \frac{4\hat{i} + 4\hat{j} - 4\hat{k}}{2} = 2\hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 5: Write the position vector of the angle bisector
In an isosceles triangle, the angle bisector also bisects the opposite side. Since the midpoint \( E \) is at \( 2\hat{i} + 2\hat{j} - 2\hat{k} \), the position vector of the angle bisector can be expressed in the direction of this vector.

Thus, the position vector of the angle bisector can be represented as:
\[ \vec{r} = k(2\hat{i} + 2\hat{j} - 2\hat{k}) \quad \text{for some scalar } k \]

### Final Answer:
The position vector of the bisector of angle \( POQ \) is:
\[ \vec{r} = \hat{i} + \hat{j} - \hat{k} \]