Let vertex `A(2,3,1), B(3,2,-1), C(-2,1,3)` If AD is angle bisector of angle A, then projection of `vec(AD)` on `vec(AC)` is equal to

Options

A. `1/sqrt16`
B. `2/sqrt6`
C. `3/sqrt6`
D. `4/sqrt6`

Answer

To find the projection of vector \( \vec{AD} \) on vector \( \vec{AC} \), we will follow these steps:

### Step 1: Find the coordinates of points A, B, and C
Given:
- \( A(2, 3, 1) \)
- \( B(3, 2, -1) \)
- \( C(-2, 1, 3) \)

### Step 2: Calculate the lengths of sides AB and AC
Using the distance formula:
\[ AB = \sqrt{(3-2)^2 + (2-3)^2 + (-1-1)^2} = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
\[ AC = \sqrt{(-2-2)^2 + (1-3)^2 + (3-1)^2} = \sqrt{(-4)^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \]

### Step 3: Apply the Angle Bisector Theorem
According to the Angle Bisector Theorem:
\[ \frac{AB}{AC} = \frac{BD}{DC} \]
Substituting the lengths we found:
\[ \frac{\sqrt{6}}{2\sqrt{6}} = \frac{1}{2} \]
This implies that \( BD:DC = 1:2 \).

### Step 4: Find the coordinates of point D
Using the section formula, the coordinates of point D dividing BC in the ratio 1:2:
\[ D = \left( \frac{1(-2) + 2(3)}{1+2}, \frac{1(1) + 2(2)}{1+2}, \frac{1(3) + 2(-1)}{1+2} \right) \]
Calculating each coordinate:
\[ D_x = \frac{-2 + 6}{3} = \frac{4}{3}, \quad D_y = \frac{1 + 4}{3} = \frac{5}{3}, \quad D_z = \frac{3 - 2}{3} = \frac{1}{3} \]
Thus, \( D \left( \frac{4}{3}, \frac{5}{3}, \frac{1}{3} \right) \).

### Step 5: Find vectors \( \vec{AD} \) and \( \vec{AC} \)
\[ \vec{AD} = D - A = \left( \frac{4}{3} - 2, \frac{5}{3} - 3, \frac{1}{3} - 1 \right) = \left( \frac{4}{3} - \frac{6}{3}, \frac{5}{3} - \frac{9}{3}, \frac{1}{3} - \frac{3}{3} \right) = \left( -\frac{2}{3}, -\frac{4}{3}, -\frac{2}{3} \right) \]
\[ \vec{AC} = C - A = \left( -2 - 2, 1 - 3, 3 - 1 \right) = \left( -4, -2, 2 \right) \]

### Step 6: Calculate the projection of \( \vec{AD} \) on \( \vec{AC} \)
The formula for the projection of vector \( \vec{u} \) on vector \( \vec{v} \) is given by:
\[ \text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v} \]
Calculating the dot product \( \vec{AD} \cdot \vec{AC} \):
\[ \vec{AD} \cdot \vec{AC} = \left( -\frac{2}{3} \right)(-4) + \left( -\frac{4}{3} \right)(-2) + \left( -\frac{2}{3} \right)(2) = \frac{8}{3} + \frac{8}{3} - \frac{4}{3} = \frac{12}{3} = 4 \]
Calculating the magnitude squared of \( \vec{AC} \):
\[ |\vec{AC}|^2 = (-4)^2 + (-2)^2 + (2)^2 = 16 + 4 + 4 = 24 \]
Thus, the projection is:
\[ \text{proj}_{\vec{AC}} \vec{AD} = \frac{4}{24} \vec{AC} = \frac{1}{6} \vec{AC} \]

### Final Answer
The projection of \( \vec{AD} \) on \( \vec{AC} \) is \( \frac{1}{6} \vec{AC} \).