Question

The direction ratios of the line of intersection of the planes \( x - y + z - 5 = 0 \) and \( x - 3y - 6 = 0 \), are:

• A. 1, -1, 1
• B. 1, -3, 0
• C. 3, 1, -2
• D. 1, 2, 0

Hint:

To find the direction ratios of the line of intersection of two planes, take the cross product of the normal vectors of the planes.

Answer 1

The Correct Option is A

Step 1: Find the normal vectors of the planes.
For the plane equation \( x - y + z - 5 = 0 \), the normal vector is:
\[ \mathbf{n_1} = \langle 1, -1, 1 \rangle \]
For the plane equation \( x - 3y - 6 = 0 \), the normal vector is:
\[ \mathbf{n_2} = \langle 1, -3, 0 \rangle \]

Step 2: Find the direction ratios of the line of intersection.
The direction ratios of the line of intersection are given by the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \). The cross product is:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix} \]
Calculating the determinant, we get:
\[ \mathbf{d} = \hat{i} \left( \begin{vmatrix} -1 & 1 \\ -3 & 0 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} 1 & -1 \\ 1 & -3 \end{vmatrix} \right) \] \[ \mathbf{d} = \hat{i} (3) - \hat{j} (-1) + \hat{k} (-2) \] \[ \mathbf{d} = 3\hat{i} + \hat{j} - 2\hat{k} \]
So the direction ratios of the line are \( \langle 3, 1, -2 \rangle \).

Step 3: Final Answer.
Thus, the direction ratios of the line are \( \boxed{1, -1, 1} \).