The plane $3y + 4z = 0$ is rotated about its line of intersection with the plane $x = 0$ through an angle $60°$. The equation of the plane in its new position is:
Step-by-Step Solution
Key Concept: A plane through a line of intersection is found using a parameter, and the angle condition determines that parameter uniquely via the normal vector formula.
The plane through the line of intersection of $3y + 4z = 0$ and $x = 0$ has the form $3y + 4z + \lambda x = 0$, or $3y + 4z + \lambda x = 0$. For a $60°$ angle with plane $3y + 4z = 0$, we use $\cos 60° = \frac{3^2 + 4^2}{\sqrt{3^2+4^2}\sqrt{3^2+4^2+\lambda^2}} = \frac{1}{2}$. This gives $\frac{25}{25+\lambda^2} = \frac{1}{4}$, so $\lambda = \pm 5\sqrt{3}$. The required plane is $3y + 4z \pm 5\sqrt{3}x = 0$.
Correct Answer: 1,2