To find the shortest distance between the two given lines, we can use the formula for the distance \( d \) between two skew lines represented in vector form. The lines are given as:
1. \(\frac{x-3}{1} = \frac{y-8}{4} = \frac{z-3}{22}\)
2. \(\frac{x+3}{1} = \frac{y+7}{1} = \frac{z-6}{7}\)
### Step 1: Identify the direction vectors and points on the lines
From the first line, we can express it in vector form:
- Point on Line 1: \( A(3, 8, 3) \)
- Direction vector of Line 1: \( \mathbf{b_1} = (1, 4, 22) \)
From the second line, we can express it in vector form:
- Point on Line 2: \( B(-3, -7, 6) \)
- Direction vector of Line 2: \( \mathbf{b_2} = (1, 1, 7) \)
### Step 2: Find the vector connecting the two points
The vector connecting points \( A \) and \( B \) is given by:
\[
\mathbf{a_2 - a_1} = B - A = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3)
\]
### Step 3: Calculate the cross product of the direction vectors
Now, we need to find the cross product \( \mathbf{b_1} \times \mathbf{b_2} \):
\[
\mathbf{b_1} = (1, 4, 22), \quad \mathbf{b_2} = (1, 1, 7)
\]
Using the determinant method for the cross product:
\[
\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 4 & 22 \\
1 & 1 & 7
\end{vmatrix}
\]
Calculating the determinant:
\[
= \mathbf{i}(4 \cdot 7 - 22 \cdot 1) - \mathbf{j}(1 \cdot 7 - 22 \cdot 1) + \mathbf{k}(1 \cdot 1 - 4 \cdot 1)
\]
\[
= \mathbf{i}(28 - 22) - \mathbf{j}(7 - 22) + \mathbf{k}(1 - 4)
\]
\[
= 6\mathbf{i} + 15\mathbf{j} - 3\mathbf{k}
\]
### Step 4: Calculate the magnitude of the cross product
Now we calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):
\[
|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{6^2 + 15^2 + (-3)^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}
\]
### Step 5: Calculate the distance using the formula
The distance \( d \) between the two lines is given by:
\[
d = \frac{|(\mathbf{a_2 - a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|}
\]
Calculating the dot product \( (\mathbf{a_2 - a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) \):
\[
\mathbf{a_2 - a_1} = (-6, -15, 3)
\]
\[
\mathbf{b_1} \times \mathbf{b_2} = (6, 15, -3)
\]
\[
(-6, -15, 3) \cdot (6, 15, -3) = (-6 \cdot 6) + (-15 \cdot 15) + (3 \cdot -3) = -36 - 225 - 9 = -270
\]
Now substituting back into the distance formula:
\[
d = \frac{|-270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30}
\]
### Final Answer:
The shortest distance between the two lines is \( 3\sqrt{30} \).
