Animated Solution for Mathematics - Three Dimensional Geometry: The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$ is :

Mastering the Family of Planes

Imagine you are standing in a room, holding two large sheets of paper. You hold them at an angle, letting them intersect. Where they meet, they form a sharp, distinct line.

In the world of 3D geometry, this is the line of intersection. Our goal today is to find a third plane that passes through this exact line, but with a twist: it must be perfectly perpendicular to a third, given plane. This is a classic JEE Advanced challenge that tests your ability to visualize space and manipulate algebraic constraints.

The Family of Planes

We start with two planes:

We need a plane that passes through their intersection. Instead of trying to find the line itself, we use the elegant concept of the 'family of planes.'

We define this family as $P_1+\lambda P_2=0$. Think of $\lambda$ as a steering wheel. As you turn this wheel (change the value of $\lambda$), the plane rotates around the line of intersection like a door on a hinge. Every value of $\lambda$ gives us a unique plane, but all of them share that same, fixed line of intersection.

The DNA of the Plane

To find the specific plane we need, we must understand its orientation. The orientation of any plane is defined by its normal vector—the vector perpendicular to its surface.

Let's substitute our equations into the family formula:

By grouping the terms, we get:

The coefficients of $x$, $y$, and $z$ are the components of our normal vector:

This vector is the 'DNA' of our new plane; it tells us exactly how the plane is tilted in space.

The Perpendicularity Constraint

Now, we introduce the constraint. Our new plane must be perpendicular to $P_3:x-y+z=0$. The normal vector of $P_3$ is ${\vec{n}}_3=\hat{i}-\hat{j}+\hat{k}$.

For two planes to be perpendicular, their normal vectors must be orthogonal. We apply the dot product test: $\vec{n}\cdot {\vec{n}}_3=0$.

Substituting our components, we get:

Expanding this, we get:

The constants $1-1+1$ leave us with $1$, and the $\lambda$ terms $2\lambda -3\lambda +4\lambda$ give us $3\lambda$. Thus, $1+3\lambda =0$, leading us to the elegant result:

The Final Transformation

With $\lambda =-\frac{1}{3}$ in hand, we substitute it back into our family equation:

Multiplying by $3$ to clear the fraction, we get:

Expanding this, we find:

Notice the beauty of the cancellation: the $y$ terms vanish, leaving us with $x-z+2=0$.

Finally, we convert this to vector form. Since $x=\vec{r}\cdot \hat{i}$, $y=\vec{r}\cdot \hat{j}$, and $z=\vec{r}\cdot \hat{k}$, our equation becomes:

You have successfully navigated the geometry, constrained the orientation, and arrived at the solution. Keep this logic in your toolkit—it is the key to unlocking any problem involving intersecting planes.