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 between two planes is parallel to the cross product of their normal vectors.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 obtained by computing the cross product of the normal vectors:\[\overrightarrow{d} = \overrightarrow{n_1} \times \overrightarrow{n_2}.\]Calculation of the cross product:\[\overrightarrow{d} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & -1 & 1 \\1 & 4 & -2\end{vmatrix}.\]Expanding the determinant yields:\[\overrightarrow{d} = \hat{i} \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix}- \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix}+ \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix}.\]\[\overrightarrow{d} = \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1)).\]Simplification leads to:\[\overrightarrow{d} = \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1).\]\[\overrightarrow{d} = -2\hat{i} + 7\hat{j} + 13\hat{k}.\]Therefore, the direction vector is:\[\boxed{-2\hat{i} + 7\hat{j} + 13\hat{k}}.\]