Animated Solution for Mathematics - Three Dimensional Geometry: The length of the perpendicular drawn from the point $(2,1,4)$ to the plane containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{j}-2\hat{k})$ is :

Welcome, future engineers. Today, we are not just solving a problem; we are constructing a 3D landscape in our minds. Imagine you are standing in a vast, empty void.

In front of you, there is a flat, infinite sheet of paper—a plane—floating in space. On this plane, two lines are drawn, intersecting at a single point. And hovering above this plane, like a star in the night sky, is a point $P(2,1,4)$.

Our mission is to find the shortest distance from that star to the sheet of paper. This is the essence of 3D coordinate geometry: turning abstract vectors into a tangible, physical reality.

Before we can measure the distance, we must define the plane. A plane is like a rigid board; to fix it in space, we need an anchor point and an orientation. Look at the equations of our two lines:

Do you see it? The position vector part, $(\hat{i}+\hat{j})$, is identical in both. This is our anchor!

It tells us that both lines pass through the point $A(1,1,0)$. Since both lines lie on the plane, this point $A$ must also lie on the plane. We have our first piece of the puzzle.

Now, we need the orientation. A plane is defined by its 'tilt,' and that tilt is dictated by its normal vector—a vector that stands perfectly perpendicular to the surface. To find this normal, we need the 'skeleton' of the plane: the direction vectors of the lines lying upon it.

From our equations, we extract:

These two vectors, $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$, define the directions of our lines. If we want a vector that is perpendicular to both, we must invoke the power of the cross product. The cross product $\vec{n}=\overrightarrow{b_1}\times \overrightarrow{b_2}$ will give us the normal vector, the arrow that points straight out of the plane.

Let us calculate this normal vector with precision. We set up our determinant:

Expanding this, we get:

Here is a pro-tip for your JEE journey: vectors are about direction, not magnitude. We can scale this vector to make our lives easier. Dividing by $-3$, we get a simplified normal vector $\overrightarrow{n^{\prime }}=\hat{i}-\hat{j}-\hat{k}$. The direction ratios are $(1,-1,-1)$.

With our anchor point $A(1,1,0)$ and our normal vector $(1,-1,-1)$, we can write the equation of the plane using the point-normal form:

Substituting our values:

Expanding this, we get $x-1-y+1-z=0$. The constants $-1$ and $+1$ cancel out beautifully, leaving us with the elegant equation:

We have arrived at the final act. We need the perpendicular distance $d$ from the point $P(2,1,4)$ to the plane $x-y-z=0$. We use the standard distance formula:

Plugging in our coordinates $(2,1,4)$ and the coefficients $(1,-1,-1)$:

Since distance is always positive, $|-3|=3$. Thus, $d=\frac{3}{\sqrt{3}}$. Rationalizing this, we get the final result: