To find the shortest distance between the two lines given by their symmetric equations, we will follow these steps:
### Step 1: Identify the points and direction ratios of the lines
The first line is given by:
\[
\frac{x-8}{3} = \frac{y+9}{-16} = \frac{z-10}{7}
\]
From this, we can identify:
- A point on the first line \( A_1 = (8, -9, 10) \)
- Direction ratios \( \mathbf{b_1} = (3, -16, 7) \)
The second line is given by:
\[
\frac{x-15}{3} = \frac{y-29}{8} = \frac{5-z}{5}
\]
From this, we can identify:
- A point on the second line \( A_2 = (15, 29, 5) \)
- Direction ratios \( \mathbf{b_2} = (3, 8, -5) \)
### Step 2: Find the vector between the two points
Calculate the vector \( \mathbf{A_2 - A_1} \):
\[
\mathbf{A_2 - A_1} = (15 - 8, 29 - (-9), 5 - 10) = (7, 38, -5)
\]
### Step 3: Calculate the cross product of the direction vectors
Now we compute the cross product \( \mathbf{b_2} \times \mathbf{b_1} \):
\[
\mathbf{b_1} = (3, -16, 7), \quad \mathbf{b_2} = (3, 8, -5)
\]
Using the determinant formula for the cross product:
\[
\mathbf{b_2} \times \mathbf{b_1} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 8 & -5 \\
3 & -16 & 7
\end{vmatrix}
\]
Calculating this determinant:
\[
= \mathbf{i} \begin{vmatrix} 8 & -5 \\ -16 & 7 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -5 \\ 3 & 7 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 8 \\ 3 & -16 \end{vmatrix}
\]
Calculating the minors:
\[
= \mathbf{i} (8 \cdot 7 - (-5) \cdot (-16)) - \mathbf{j} (3 \cdot 7 - (-5) \cdot 3) + \mathbf{k} (3 \cdot (-16) - 8 \cdot 3)
\]
\[
= \mathbf{i} (56 - 80) - \mathbf{j} (21 + 15) + \mathbf{k} (-48 - 24)
\]
\[
= \mathbf{i} (-24) - \mathbf{j} (36) + \mathbf{k} (-72)
\]
Thus,
\[
\mathbf{b_2} \times \mathbf{b_1} = (-24, -36, -72)
\]
### Step 4: Calculate the dot product
Now we calculate the dot product \( (\mathbf{A_2 - A_1}) \cdot (\mathbf{b_2} \times \mathbf{b_1}) \):
\[
(7, 38, -5) \cdot (-24, -36, -72) = 7 \cdot (-24) + 38 \cdot (-36) + (-5) \cdot (-72)
\]
Calculating this:
\[
= -168 - 1368 + 360 = -1176
\]
### Step 5: Calculate the magnitude of the cross product
Now, we find the magnitude of \( \mathbf{b_2} \times \mathbf{b_1} \):
\[
\|\mathbf{b_2} \times \mathbf{b_1}\| = \sqrt{(-24)^2 + (-36)^2 + (-72)^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84
\]
### Step 6: Calculate the shortest distance
Finally, we use the formula for the shortest distance \( d \):
\[
d = \frac{|(\mathbf{A_2 - A_1}) \cdot (\mathbf{b_2} \times \mathbf{b_1})|}{\|\mathbf{b_2} \times \mathbf{b_1}\|}
\]
Substituting the values we found:
\[
d = \frac{| -1176 |}{84} = \frac{1176}{84} = 14
\]
Thus, the shortest distance between the two lines is **14 units**.
