Animated Solution for Mathematics - Three Dimensional Geometry: The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is :

Welcome, future engineers. Today, we are going to demystify a classic JEE Advanced problem. Imagine you are standing in a vast, empty 3D room with a fixed point $P(2,-1,6)$ floating in the air.

Somewhere in that room, there is a "ghost line." It is defined as the intersection of two planes:

This is a classic trap designed to test if you can see the underlying structure of space.

To find the distance from a point to a line, we need two things: the line's direction and a point on the line. Geometrically, if a line lies on a plane, it must be perpendicular to that plane's normal vector.

Since our line lies on both planes, it must be perpendicular to both normal vectors, ${\vec{n}}_1$ and ${\vec{n}}_2$. We extract these normals from the plane equations: For $P_1$, ${\vec{n}}_1=0\hat{i}+3\hat{j}-2\hat{k}$. For $P_2$, ${\vec{n}}_2=3\hat{i}+0\hat{j}-1\hat{k}$.

The direction vector $\vec{b}$ of our line is the cross product: $\vec{b}={\vec{n}}_1\times {\vec{n}}_2$. Calculating this determinant:

To simplify, we can use the direction ratios $(1,2,3)$ by dividing by $-3$. Remember, direction is about the ratio, not the magnitude.

Now that we have the direction, we need an anchor point $A$ on the line. Since the line is infinite, we have the freedom to choose. Let's set $z=4$.

Plugging $z=4$ into $P_1$:

Plugging $z=4$ into $P_2$:

Our anchor point $A$ is $(0,3,4)$.

We now have point $A(0,3,4)$ on the line and point $P(2,-1,6)$ in space. The vector $\overrightarrow{AP}$ is:

The shortest distance $d$ from a point to a line is given by:

Think of the numerator $|\overrightarrow{AP}\times \vec{b}|$ as the area of a parallelogram formed by $\overrightarrow{AP}$ and $\vec{b}$. Since the area is base $\times$ height, and the base is $|\vec{b}|$, dividing by $|\vec{b}|$ gives us the height—the perpendicular distance.

We compute the cross product $\overrightarrow{AP}\times \vec{b}$ using the direction vector $\vec{b}=(1,2,3)$:

The magnitude of the cross product is:

The magnitude of $\vec{b}$ is:

Finally, the distance $d$ is:

There it is. The complexity of the planes collapses into a simple, elegant result. Keep practicing this visualization, and you will master 3D geometry.