Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is:

A Journey into Planes and Lines

Welcome, future engineers. Today, we are not just solving a problem; we are constructing a geometric reality. Imagine you are standing in a vast, empty 3D coordinate system.

At the origin, $(0,0,0)$, three distinct lines intersect, radiating outwards like the spokes of a wheel. We have been tasked with finding a specific plane—a flat, infinite sheet of glass—that contains one of these lines and stands perfectly perpendicular to a plane formed by the other two.

Decoding the Lines

First, let us look at the lines provided. They are given in symmetric form:

Notice something beautiful? There are no constants subtracted from $x,y,$ or $z$. This tells us immediately that all three lines pass through the origin.

Their direction vectors are simply the denominators: ${\vec{v}}_1=(2,3,4)$ ${\vec{v}}_2=(3,4,2)$ ${\vec{v}}_3=(4,2,3)$

The First Plane and the Normal Vector

Now, consider the plane $P_1$ that contains lines $L_2$ and $L_3$. To define this plane, we need a normal vector, ${\vec{n}}_1$. Think of this normal vector as a flagpole sticking straight out of the surface of the plane.

Because it is perpendicular to the plane, it must be perpendicular to every line lying on that plane, including $L_2$ and $L_3$. We calculate ${\vec{n}}_1={\vec{v}}_2\times {\vec{v}}_3$ using the determinant:

Expanding this, we get $\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-16)$, which simplifies to ${\vec{n}}_1=(8,-1,-10)$. This vector is the key to the orientation of our first plane.

The Required Plane

Now, we shift our focus to the plane we need to find, $P_{req}$. We know two things: it contains $L_1$ (so its normal must be perpendicular to ${\vec{v}}_1$) and it is perpendicular to $P_1$ (so its normal must be perpendicular to ${\vec{n}}_1$).

We need a vector ${\vec{n}}_{req}$ that is perpendicular to both ${\vec{v}}_1$ and ${\vec{n}}_1$. Again, we call upon the cross product:

The Final Elegance

Let us calculate this carefully. The $\hat{i}$ component is $(-30-(-4))=-26$. The $\hat{j}$ component is $-(-20-32)=52$. The $\hat{k}$ component is $(-2-24)=-26$.

So, our normal vector is $(-26,52,-26)$. As we discussed, the magnitude doesn't matter, only the direction. Dividing by $-26$, we get the simplified normal vector $(1,-2,1)$.

Since the plane passes through the origin $(0,0,0)$, the equation is simply $1(x-0)-2(y-0)+1(z-0)=0$. This gives us the final, elegant result:

You have successfully navigated the 3D space and defined a plane with precision. Keep this confidence; you are mastering the language of the universe.