Animated Solution for Mathematics - Three Dimensional Geometry: If the plane $2x+y-5z=0$ is rotated about its line of intersection with the plane $3x-y+4z-7=0$ by an angle of $\frac{\pi }{2}$, then the plane after the rotation passes through the point :

A 3D Adventure

Welcome, my dear student! Today, we are going to dive into the elegant world of 3D geometry. Imagine you are standing in a vast, empty space with two planes, $P_1:2x+y-5z=0$ and $P_2:3x-y+4z-7=0$, slicing through this space.

They meet at a line, a sharp edge where these two infinite sheets intersect. Our mission is to take $P_1$ and rotate it around this intersection line by exactly $\frac{\pi }{2}$ radians. It sounds daunting, but with the right tools, it becomes a beautiful dance of algebra.

How do we describe a plane that is constantly changing as it rotates, yet always stays anchored to that intersection line? We use the 'Family of Planes' equation. Any plane passing through the intersection of $P_1$ and $P_2$ can be written as $P_1+\lambda P_2=0$.

This equation is our anchor. It says, 'No matter what value $\lambda$ takes, this plane will always contain the line where $P_1$ and $P_2$ meet.' Let's write this out explicitly:

To make this useful, we need to group the terms by their variables $x,y,$ and $z$. This will reveal the normal vector of our new plane, which we'll call $P_3$:

Now, look at the coefficients of $x,y,$ and $z$. These form the normal vector of our new plane: $\overrightarrow{n_3}=(2+3\lambda ,1-\lambda ,-5+4\lambda )$.

Here is the crucial insight: the problem tells us the plane is rotated by $\frac{\pi }{2}$. In the language of geometry, this means the new plane $P_3$ is perpendicular to the original plane $P_1$.

If two planes are perpendicular, their normal vectors must also be perpendicular. The mathematical condition for two vectors to be perpendicular is that their dot product must be zero.

We know the normal vector of $P_1$ is $\overrightarrow{n_1}=(2,1,-5)$. So, we set the dot product $\overrightarrow{n_1}\cdot \overrightarrow{n_3}=0$:

Now, let's expand this carefully. I know algebra can sometimes feel like a chore, but stay with me—the numbers will fall into place beautifully:

Let's group the constants and the $\lambda$ terms. The constants are $4+1+25=30$, and the $\lambda$ terms are $6\lambda -\lambda -20\lambda =-15\lambda$. So, we have:

Solving for $\lambda$, we get $\lambda =2$. We have found the specific plane in the family that satisfies our rotation condition!

Now, we substitute $\lambda =2$ back into our grouped equation to find the final equation of the plane $P_3$:

This is the equation of our rotated plane. The final step is to see which of the given points lies on this plane. Let's test the point $(1,0,2)$:

It works! The point $(1,0,2)$ satisfies the equation, meaning it lies on the plane. You've navigated the geometry, mastered the family of planes, and used the dot product to find the solution.

Remember, every complex problem is just a series of simple, logical steps. Keep practicing, and you'll soon find that you don't just solve these problems—you understand them.