Question

The direction cosines of the straight line given by the planes $x=0$ and $z=0$ are:

• A. $1, 0, 0$
• B. $0, 0, 1$
• C. $1, 1, 0$
• D. $0, 1, 0$
• E. $0, 1, 1$

Hint:

Geometrically, the intersection of the $yz$-plane ($x=0$) and the $xy$-plane ($z=0$) is the $y$-axis itself. Any line along a coordinate axis has direction cosines where only one component is 1 and the others are 0.

Answer 1

The Correct Option is D

Concept: A straight line defined by the intersection of two planes is parallel to the cross product of the normal vectors of those two planes. Once the direction vector of the line is found, the direction cosines are obtained by normalizing that vector.

Step 1: Identify the normal vectors of the given planes.
The first plane is $x=0$. Its normal vector is $\vec{n_1} = (1, 0, 0)$. The second plane is $z=0$. Its normal vector is $\vec{n_2} = (0, 0, 1)$.

Step 2: Find the direction vector of the line of intersection.
The direction vector $\vec{v}$ of the line is perpendicular to both normals: \[ \vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 1 & 0 & 0 0 & 0 & 1 \end{vmatrix} \] Expanding the determinant: \[ \vec{v} = \hat{i}(0-0) - \hat{j}(1-0) + \hat{k}(0-0) = -\hat{j} \] The direction vector is $(0, -1, 0)$, which is equivalent to $(0, 1, 0)$ for direction purposes.

Step 3: Determine the direction cosines.
The direction vector represents the $y$-axis. The direction cosines $(l, m, n)$ for a vector along the $y$-axis are: \[ l = \cos(90^\circ) = 0, \quad m = \cos(0^\circ) = 1, \quad n = \cos(90^\circ) = 0 \] Thus, the direction cosines are $(0, 1, 0)$.