To find the direction cosines of the normal to the plane containing the lines given by the equations \(\frac{x}{2} = \frac{y}{3} = \frac{z}{5}\) and \(\frac{x-1}{2} = \frac{y-1}{3} = \frac{z-1}{5}\), we can follow these steps:
### Step 1: Identify the direction ratios of the lines
The first line is given by:
\[
\frac{x}{2} = \frac{y}{3} = \frac{z}{5}
\]
From this, we can see that the direction ratios of the first line are \( (2, 3, 5) \).
The second line is given by:
\[
\frac{x-1}{2} = \frac{y-1}{3} = \frac{z-1}{5}
\]
This line can be rewritten as:
\[
x = 1 + 2t, \quad y = 1 + 3t, \quad z = 1 + 5t
\]
From this, we can see that the direction ratios of the second line are also \( (2, 3, 5) \).
### Step 2: Confirm that the lines are parallel
Since both lines have the same direction ratios, they are parallel.
### Step 3: Find a vector normal to the plane
To find a normal vector to the plane containing these two parallel lines, we need to consider a vector connecting points on both lines.
Let’s take a point on the first line (when \( t = 0 \)):
\[
P_1 = (0, 0, 0)
\]
And a point on the second line (when \( t = 0 \)):
\[
P_2 = (1, 1, 1)
\]
Now, we can find the vector \( \vec{P_1P_2} \):
\[
\vec{P_1P_2} = P_2 - P_1 = (1 - 0, 1 - 0, 1 - 0) = (1, 1, 1)
\]
### Step 4: Calculate the normal vector using the cross product
Let \( \vec{a} = (2, 3, 5) \) (direction ratios of the lines) and \( \vec{b} = (1, 1, 1) \) (the vector we found).
Now we compute the cross product \( \vec{n} = \vec{a} \times \vec{b} \):
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 5 \\
1 & 1 & 1
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i} \begin{vmatrix}
3 & 5 \\
1 & 1
\end{vmatrix} - \hat{j} \begin{vmatrix}
2 & 5 \\
1 & 1
\end{vmatrix} + \hat{k} \begin{vmatrix}
2 & 3 \\
1 & 1
\end{vmatrix}
\]
Calculating each of the 2x2 determinants:
- For \( \hat{i} \): \( (3 \cdot 1 - 5 \cdot 1) = 3 - 5 = -2 \)
- For \( \hat{j} \): \( (2 \cdot 1 - 5 \cdot 1) = 2 - 5 = -3 \) (note the negative sign in front)
- For \( \hat{k} \): \( (2 \cdot 1 - 3 \cdot 1) = 2 - 3 = -1 \)
So, we have:
\[
\vec{n} = (-2) \hat{i} + 3 \hat{j} - 1 \hat{k} = (-2, 3, -1)
\]
### Step 5: Find the direction cosines
The direction cosines are given by:
\[
\cos \alpha = \frac{a_1}{\sqrt{a_1^2 + a_2^2 + a_3^2}}, \quad \cos \beta = \frac{a_2}{\sqrt{a_1^2 + a_2^2 + a_3^2}}, \quad \cos \gamma = \frac{a_3}{\sqrt{a_1^2 + a_2^2 + a_3^2}}
\]
where \( \vec{n} = (a_1, a_2, a_3) = (-2, 3, -1) \).
Calculating the magnitude:
\[
\sqrt{(-2)^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}
\]
Now substituting into the formulas:
\[
\cos \alpha = \frac{-2}{\sqrt{14}}, \quad \cos \beta = \frac{3}{\sqrt{14}}, \quad \cos \gamma = \frac{-1}{\sqrt{14}}
\]
### Final Answer
The direction cosines of the normal to the plane are:
\[
\left( \frac{-2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \right)
\]
