If `veca=hati+hatj+hatk` & `vecB=hati-hatj+hatk` are the sides of AB & AC of `/_\ABC` if AD is the angle bisector of `angleBAC` where D lies on BC, then length of AD is

Options

A. 1
B. `sqrt2`
C. `sqrt3`
D. 2

Explanation

To find the length of the angle bisector \( AD \) in triangle \( ABC \) where \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{B} = \hat{i} - \hat{j} + \hat{k} \), we will follow these steps:

### Step 1: Identify the vectors
We have:
- \( \vec{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \vec{B} = \hat{i} - \hat{j} + \hat{k} \)

### Step 2: Calculate the lengths of the vectors
To find the lengths of \( \vec{A} \) and \( \vec{B} \):
\[ |\vec{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
\[ |\vec{B}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]

### Step 3: Find the cosine of the angle between the vectors
The cosine of the angle \( \theta \) between the vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Calculating the dot product:
\[ \vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1 \]
Thus,
\[ \cos \theta = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3} \]

### Step 4: Find \( \theta \)
Using the cosine value, we can find \( \theta \) using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \]

### Step 5: Find \( \cos \frac{\theta}{2} \)
Using the half-angle formula:
\[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 + \frac{1}{3}}{2}} = \sqrt{\frac{\frac{4}{3}}{2}} = \sqrt{\frac{2}{3}} \]

### Step 6: Calculate the length of the angle bisector \( AD \)
The length of the angle bisector \( AD \) can be calculated using the formula:
\[ AD = \frac{2 |\vec{A}| |\vec{B}|}{|\vec{A}| + |\vec{B}|} \cos \frac{\theta}{2} \]
Substituting the values:
\[ AD = \frac{2 \cdot \sqrt{3} \cdot \sqrt{3}}{\sqrt{3} + \sqrt{3}} \cdot \sqrt{\frac{2}{3}} = \frac{2 \cdot 3}{2\sqrt{3}} \cdot \sqrt{\frac{2}{3}} = \frac{3}{\sqrt{3}} \cdot \sqrt{\frac{2}{3}} = \sqrt{3} \cdot \sqrt{\frac{2}{3}} = \sqrt{2} \]

### Final Answer
The length of \( AD \) is \( \sqrt{2} \).

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