Question

A vector parallel to the line of intersection of the planes \[ \overrightarrow{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1 \quad \text{and} \quad \overrightarrow{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \] is:

• A. \( -2\hat{i} + 7\hat{j} + 13\hat{k} \)
• B. \( 2\hat{i} - 7\hat{j} + 13\hat{k} \)
• C. \( -\hat{i} + 4\hat{j} + 7\hat{k} \)
• D. \( \hat{i} - 4\hat{j} + 7\hat{k} \)

Hint:

To find the direction vector of the line of intersection of two planes, calculate the cross product of their normal vectors: \( \overrightarrow{n_1} \times \overrightarrow{n_2} \).

Answer 1

The Correct Option is A

The line of intersection of the two planes is parallel to the cross product of the normal vectors of the planes. The normal vectors are: \[ \overrightarrow{n_1} = 3\hat{i} - \hat{j} + \hat{k}, \quad \overrightarrow{n_2} = \hat{i} + 4\hat{j} - 2\hat{k}. \] The direction vector of the line is given by: \[ \overrightarrow{d} = \overrightarrow{n_1} \times \overrightarrow{n_2}. \] Compute the cross product: \[ \overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix}. \] Expand the determinant: \[ \overrightarrow{d} = \hat{i} \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix}. \] \[ \overrightarrow{d} = \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1)). \] Simplify: \[ \overrightarrow{d} = \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1). \] \[ \overrightarrow{d} = -2\hat{i} + 7\hat{j} + 13\hat{k}. \] Thus, the direction vector is: \[ \boxed{-2\hat{i} + 7\hat{j} + 13\hat{k}}. \]