To find the direction ratios of the normal to the plane passing through the origin and the points A(-2, 2, 3) and B(13, -3, 13), we will follow these steps:
### Step 1: Define the Points
Let the points be defined as follows:
- Point O (the origin) = (0, 0, 0)
- Point A = (-2, 2, 3)
- Point B = (13, -3, 13)
### Step 2: Find the Position Vectors
The position vectors from the origin to points A and B are:
- OA = A - O = (-2 - 0, 2 - 0, 3 - 0) = (-2, 2, 3)
- OB = B - O = (13 - 0, -3 - 0, 13 - 0) = (13, -3, 13)
### Step 3: Write the Vectors
We can express these vectors in terms of unit vectors:
- OA = -2i + 2j + 3k
- OB = 13i - 3j + 13k
### Step 4: Calculate the Cross Product
To find the normal vector to the plane formed by OA and OB, we need to calculate the cross product OA × OB.
Using the determinant method:
\[
\text{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 2 & 3 \\
13 & -3 & 13
\end{vmatrix}
\]
### Step 5: Compute the Determinant
Calculating the determinant:
\[
\text{n} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 13 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 3 \\ 13 & 13 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 2 \\ 13 & -3 \end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. For \( \hat{i} \):
\[
2 \cdot 13 - 3 \cdot (-3) = 26 + 9 = 35
\]
2. For \( \hat{j} \):
\[
-2 \cdot 13 - 3 \cdot 13 = -26 - 39 = -65 \quad \text{(remember to change the sign for j)}
\]
So it becomes \( +65 \hat{j} \).
3. For \( \hat{k} \):
\[
-2 \cdot (-3) - 2 \cdot 13 = 6 - 26 = -20
\]
Putting it all together:
\[
\text{n} = 35 \hat{i} + 65 \hat{j} - 20 \hat{k}
\]
### Step 6: Direction Ratios
The direction ratios of the normal vector are:
\[
(35, 65, -20)
\]
### Step 7: Simplification
To express the direction ratios in a simpler form, we can divide by 5:
\[
(7, 13, -4)
\]
### Final Answer
The direction ratios of the normal to the plane passing through the origin and the points A and B are:
\[
\boxed{(7, 13, -4)}
\]
