The position vectors of the vertices A, B and C of a $\Delta ABC$ are respectively $\hat{i}-\hat{j}-3\hat{k}$, $2\hat{i}+\hat{j}-2\hat{k}$ and $-5\hat{i}+2\hat{j}-6\hat{k}$. The length of the bisector AD of $\angle BAC$, where D is on the line segment BC, is:

Question

Accepted Answer

To determine the length of the angle bisector AD of $\angle BAC$ in $\Delta ABC$, we start by using the given position vectors of the vertices:

- $ \vec{A} = \hat{i} - \hat{j} - 3\hat{k} $
- $ \vec{B} = 2\hat{i} + \hat{j} - 2\hat{k} $
- $ \vec{C} = -5\hat{i} + 2\hat{j} - 6\hat{k} $

The angle bisector theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides. Therefore, the point D on segment BC divides BC in the ratio of the lengths of sides AB and AC.

First, we calculate the vectors $\vec{AB}$ and $\vec{AC}$:

$$
\vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + \hat{j} - 2\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) = \hat{i} + 2\hat{j} + \hat{k}
$$

$$
\vec{AC} = \vec{C} - \vec{A} = (-5\hat{i} + 2\hat{j} - 6\hat{k}) - (\hat{i} - \hat{j} - 3\hat{k}) = -6\hat{i} + 3\hat{j} - 3\hat{k}
$$

Next, we find the magnitudes of $\vec{AB}$ and $\vec{AC}$:

$$
|\vec{AB}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}
$$

$$
|\vec{AC}| = \sqrt{(-6)^2 + 3^2 + (-3)^2} = \sqrt{54} = 3\sqrt{6}
$$

The ratio in which D divides BC is $|\vec{AC}| : |\vec{AB}| = 3\sqrt{6} : \sqrt{6} = 3:1$.

Using the section formula, the position vector of D, $\vec{D}$, is given by:

$$
\vec{D} = \frac{3\vec{B} + 1\vec{C}}{3 + 1} = \frac{3(2\hat{i} + \hat{j} - 2\hat{k}) + 1(-5\hat{i} + 2\hat{j} - 6\hat{k})}{4}
$$

$$
= \frac{(6\hat{i} + 3\hat{j} - 6\hat{k}) + (-5\hat{i} + 2\hat{j} - 6\hat{k})}{4}
$$

$$
= \frac{(1\hat{i} + 5\hat{j} - 12\hat{k})}{4} = \frac{1}{4}\hat{i} + \frac{5}{4}\hat{j} - 3\hat{k}
$$

Now, we calculate the vector $\vec{AD}$:

$$
\vec{AD} = \vec{D} - \vec{A} = \left(\frac{1}{4}\hat{i} + \frac{5}{4}\hat{j} - 3\hat{k}\right) - (\hat{i} - \hat{j} - 3\hat{k})
$$

$$
= \left(\frac{1}{4} - 1\right)\hat{i} + \left(\frac{5}{4} + 1\right)\hat{j} + (0)\hat{k}
$$

$$
= -\frac{3}{4}\hat{i} + \frac{9}{4}\hat{j}
$$

Finally, the length of $\vec{AD}$ is:

$$
|\vec{AD}| = \sqrt{\left(-\frac{3}{4}\right)^2 + \left(\frac{9}{4}\right)^2} = \sqrt{\frac{9}{16} + \frac{81}{16}} = \sqrt{\frac{90}{16}} = \sqrt{\frac{45}{8}} = \frac{3\sqrt{10}}{4}
$$

Thus, the length of the bisector AD is $\frac{3\sqrt{10}}{4}$, which corresponds to Option 4.

**Explanation for Incorrect Options:**

1. **Option 1: $\frac{15}{2}$** - This value is significantly larger than the calculated length of the bisector, indicating a miscalculation or misunderstanding of the problem.

2. **Option 2: $\frac{\sqrt{11}}{2}$** - This value does not match the derived length from the correct application of the angle bisector theorem and vector calculations.

3. **Option 3: $\frac{1}{4}$** - This value is too small and does not reflect the actual length of the bisector as calculated.

Therefore, Option 4 is the correct answer, as it matches the calculated length of the angle bisector AD.