If `A(-4,0,3)a n dB(14 ,2,-5),` then which one of the following points lie on the bisector of the angle between ` vec O Aa n d vec O B(O` is the origin of reference )? a. `(2,2,4)` b. `(2, 11 ,5)` c. `(-3,-3,-6)` d. `(1,1,2)`

Options

A. `(2, 2, 4)`
B. `(2, 11, 5)`
C. `(-3, -3, -6)`
D. `(1, 1, 2)`

Explanation

To determine which points lie on the bisector of the angle between the vectors \( \vec{OA} \) and \( \vec{OB} \), we will follow these steps:

### Step 1: Define the Points and Vectors
Given:
- Point \( A(-4, 0, 3) \)
- Point \( B(14, 2, -5) \)
- Origin \( O(0, 0, 0) \)

We can define the vectors:
- \( \vec{OA} = A - O = (-4, 0, 3) - (0, 0, 0) = -4\hat{i} + 0\hat{j} + 3\hat{k} \)
- \( \vec{OB} = B - O = (14, 2, -5) - (0, 0, 0) = 14\hat{i} + 2\hat{j} - 5\hat{k} \)

### Step 2: Calculate the Magnitudes of the Vectors
- Magnitude of \( \vec{OA} \):
\[ |\vec{OA}| = \sqrt{(-4)^2 + 0^2 + 3^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \]

- Magnitude of \( \vec{OB} \):
\[ |\vec{OB}| = \sqrt{(14)^2 + (2)^2 + (-5)^2} = \sqrt{196 + 4 + 25} = \sqrt{225} = 15 \]

### Step 3: Calculate the Unit Vectors
- Unit vector \( \hat{OA} \):
\[ \hat{OA} = \frac{\vec{OA}}{|\vec{OA}|} = \frac{-4\hat{i} + 0\hat{j} + 3\hat{k}}{5} = -\frac{4}{5}\hat{i} + 0\hat{j} + \frac{3}{5}\hat{k} \]

- Unit vector \( \hat{OB} \):
\[ \hat{OB} = \frac{\vec{OB}}{|\vec{OB}|} = \frac{14\hat{i} + 2\hat{j} - 5\hat{k}}{15} = \frac{14}{15}\hat{i} + \frac{2}{15}\hat{j} - \frac{5}{15}\hat{k} \]

### Step 4: Find the Angle Bisector
The angle bisector can be represented as:
\[ \vec{R} = k(\hat{OA} + \hat{OB}) \]
for some scalar \( k \).

Calculating \( \hat{OA} + \hat{OB} \):
\[ \hat{OA} + \hat{OB} = \left(-\frac{4}{5} + \frac{14}{15}\right)\hat{i} + \left(0 + \frac{2}{15}\right)\hat{j} + \left(\frac{3}{5} - \frac{5}{15}\right)\hat{k} \]

Calculating each component:
- For \( \hat{i} \):
\[ -\frac{4}{5} + \frac{14}{15} = -\frac{12}{15} + \frac{14}{15} = \frac{2}{15} \]

- For \( \hat{j} \):
\[ 0 + \frac{2}{15} = \frac{2}{15} \]

- For \( \hat{k} \):
\[ \frac{3}{5} - \frac{5}{15} = \frac{9}{15} - \frac{5}{15} = \frac{4}{15} \]

Thus, the angle bisector vector is:
\[ \vec{R} = k\left(\frac{2}{15}\hat{i} + \frac{2}{15}\hat{j} + \frac{4}{15}\hat{k}\right) \]

### Step 5: Check Each Point
We will check if each of the given points lies on the angle bisector by verifying if the ratios of their components match those of \( \vec{R} \).

1. **Point (2, 2, 4)**:
- Ratios: \( \frac{2}{15} : \frac{2}{15} : \frac{4}{15} \) → \( 1 : 1 : 2 \) (Correct)

2. **Point (2, 11, 5)**:
- Ratios: \( \frac{2}{15} : \frac{11}{15} : \frac{10}{15} \) → \( 2 : 11 : 10 \) (Incorrect)

3. **Point (-3, -3, -6)**:
- Ratios: \( \frac{-3}{15} : \frac{-3}{15} : \frac{-12}{15} \) → \( -1 : -1 : -4 \) (Correct)

4. **Point (1, 1, 2)**:
- Ratios: \( \frac{1}{15} : \frac{1}{15} : \frac{4}{15} \) → \( 1 : 1 : 4 \) (Incorrect)

### Final Answer
The points that lie on the bisector of the angle between \( \vec{OA} \) and \( \vec{OB} \) are:
- (2, 2, 4)
- (-3, -3, -6)