Question

A vector parallel to the line of intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$ 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:

The direction vector of the line of intersection of two planes is always parallel to the cross product of their normal vectors.

Answer 1

The Correct Option is A

Step 1: Identify the normal vectors of the given planes.
The equation of a plane is $\vec{r} \cdot \vec{n} = d$, where $\vec{n}$ is the normal vector. For the first plane: \[ \vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1 \] \[ \vec{n_1} = 3\hat{i} - \hat{j} + \hat{k} \] For the second plane: \[ \vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \] \[ \vec{n_2} = \hat{i} + 4\hat{j} - 2\hat{k} \]
Step 2: Relationship with line of intersection.
The line of intersection of two planes is perpendicular to both normal vectors. Hence, a direction vector is: \[ \vec{n_1} \times \vec{n_2} \] 
Step 3: Calculate the cross product.
\[ \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} \] \[ = \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1)) \] \[ = \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1) \] \[ = -2\hat{i} + 7\hat{j} + 13\hat{k} \] 
Step 4: Final Answer.
\[ \boxed{-2\hat{i} + 7\hat{j} + 13\hat{k}} \]