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Let A(-2, 2, 3) and B(13, -3, 13) be to ...

Let A(-2, 2, 3) and B(13, -3, 13) be to two points and `angle`
be a line passing through the point A.
Direction ratios of the normal to the plane passing
throuth the origin and the points A and B are

A

`15, -5, 10`

B

`11, -1, 16`

C

`3, 13, 2`

D

`7, 13, -4`

Text Solution

AI Generated Solution

The correct Answer is:
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 can follow these steps: ### Step 1: Define the Points We have two points: - Point A: \( A(-2, 2, 3) \) - Point B: \( B(13, -3, 13) \) - The origin: \( O(0, 0, 0) \) ### Step 2: Find the Position Vectors The position vectors from the origin to points A and B can be defined as: - \( \vec{OA} = -2 \hat{i} + 2 \hat{j} + 3 \hat{k} \) - \( \vec{OB} = 13 \hat{i} - 3 \hat{j} + 13 \hat{k} \) ### Step 3: Calculate the Cross Product To find the direction ratios of the normal to the plane, we need to compute the cross product of the vectors \( \vec{OA} \) and \( \vec{OB} \). The cross product \( \vec{OA} \times \vec{OB} \) can be calculated using the determinant of a matrix: \[ \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 3 \\ 13 & -3 & 13 \end{vmatrix} \] ### Step 4: Calculate the Determinant Expanding the determinant, we have: \[ \vec{OA} \times \vec{OB} = \hat{i} \begin{vmatrix} 2 & 3 \\ -3 & 13 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 3 \\ 13 & 13 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 2 \\ 13 & -3 \end{vmatrix} \] Calculating each of these 2x2 determinants: 1. For \( \hat{i} \): \[ 2 \cdot 13 - 3 \cdot (-3) = 26 + 9 = 35 \] 2. For \( \hat{j} \): \[ -(-2 \cdot 13 - 3 \cdot 13) = 26 - 39 = -13 \quad \text{(thus, it becomes +13)} \] 3. For \( \hat{k} \): \[ -2 \cdot (-3) - 2 \cdot 13 = 6 - 26 = -20 \] Putting it all together, we get: \[ \vec{OA} \times \vec{OB} = 35 \hat{i} + 13 \hat{j} - 20 \hat{k} \] ### Step 5: Direction Ratios The direction ratios of the normal vector to the plane are the coefficients of \( \hat{i}, \hat{j}, \hat{k} \): - Direction ratios are \( 35, 13, -20 \). ### Step 6: Simplifying the Direction Ratios We can simplify the direction ratios by taking out the common factor of 5: \[ \text{Direction ratios} = 7, 13, -4 \] ### Final Answer The direction ratios of the normal to the plane passing through the origin and the points A and B are \( 7, 13, -4 \). ---
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