Question

If L is a line common to the planes \(3x + 4y + 7z = 1\), \(x - y + z = 5\) then the direction ratios of the line L are

• A. \((16, 0, -1)\)
• B. \((11, 4, -7)\)
• C. \((2, 5, 1)\)
• D. \((4, -7, 11)\)

Hint:

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

Answer 1

The Correct Option is B

Step 1: Identify Normals:
Normal to plane 1: \(\vec{n_1} = (3, 4, 7)\). Normal to plane 2: \(\vec{n_2} = (1, -1, 1)\).
Step 2: Find Direction Vector of Line:
The line of intersection is perpendicular to both normals. Thus, its direction vector \(\vec{d}\) is given by the cross product \(\vec{n_1} \times \vec{n_2}\). \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{vmatrix} \] \[ \vec{d} = \hat{i}(4(1) - 7(-1)) - \hat{j}(3(1) - 7(1)) + \hat{k}(3(-1) - 4(1)) \] \[ \vec{d} = \hat{i}(4 + 7) - \hat{j}(3 - 7) + \hat{k}(-3 - 4) \] \[ \vec{d} = 11\hat{i} + 4\hat{j} - 7\hat{k} \] Direction ratios are \((11, 4, -7)\).