The plane $x = 0$ is rotated through an angle $\alpha$ about its line of intersection with the plane $z = 0$. Then equation of the plane in new position is (are):
$x \pm \sqrt{\sec\alpha - 1} z = 0$
$x \pm \sqrt{\cos\alpha + 1} z = 0$
$x \pm \sqrt{\sec\alpha + 1} z = 0$
None of these
Step-by-Step Solution
Key Concept: The angle between two planes equals the angle between their normal vectors, found using the dot product formula.
The plane equation is $x+\lambda z=0$. For this plane to make angle $\alpha$ with the plane $x=0$, use the angle formula between planes. The normal to $x+\lambda z=0$ is $(1,0,\lambda)$ and to $x=0$ is $(1,0,0)$. The cosine of angle is $\frac{1}{\sqrt{1+\lambda^2}}=\cos\alpha$, which gives $\lambda=\pm\sqrt{\sec^2\alpha-1}$.
Correct Answer: Looking at the solution:
1. The rotated plane has equation $x + \lambda z = 0$
2. Using the angle formula between planes with normal vectors $(1,0,\lambda)$ and $(1,0,0)$:
$$\cos\alpha = \frac{|1|}{\sqrt{1+\lambda^2