Animated Solution for Mathematics - Three Dimensional Geometry: 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

Welcome, my dear student, to the fascinating world of 3D analytical geometry. Today, we are not just solving a problem; we are constructing a plane in space.

Imagine you are standing in a vast, empty room. You have three lines, all originating from a single point: the origin $(0,0,0)$. This is our anchor.

The lines are given by:

Because none of these equations have constant terms, we immediately know that all three lines intersect at the origin. This is our first victory.

Defining $P_2$

Now, let us focus on the plane $P_2$, which contains the second and third lines. To define a plane, we need a normal vector, ${\vec{n}}_2$, which is a vector pointing straight out of the surface, perpendicular to everything on it.

Since the lines with direction vectors ${\vec{d}}_2=(3,4,2)$ and ${\vec{d}}_3=(4,2,3)$ lie on this plane, ${\vec{n}}_2$ must be perpendicular to both. We summon the power of the cross product: ${\vec{n}}_2={\vec{d}}_2\times {\vec{d}}_3$.

We set up our determinant:

Expanding this, we get:

This vector, $(8,-1,-10)$, is the backbone of our green plane $P_2$.

The Perpendicular Handshake

Now, we turn to our target plane, $P_1$. We know it contains the first line with direction ${\vec{d}}_1=(2,3,4)$.

But there is a twist: $P_1$ is perpendicular to $P_2$. This means their normal vectors must also be perpendicular. So, the normal ${\vec{n}}_1$ of our target plane must be perpendicular to both ${\vec{d}}_1$ and ${\vec{n}}_2$.

Again, the cross product is our hero: ${\vec{n}}_1={\vec{d}}_1\times {\vec{n}}_2$. We set up the second determinant:

Expanding this, we get:

We have our normal vector, but it looks a bit heavy. In geometry, the magnitude of a normal vector doesn't change the plane's orientation, only its direction ratios do.

So, we can scale ${\vec{n}}_1$ by dividing by $-26$, giving us the simplified direction ratios $(1,-2,1)$.

Since our plane passes through the origin $(0,0,0)$, the equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ becomes:

This simplifies beautifully to the final answer: $x-2y+z=0$

You have done it! You have navigated the 3D space and found the equation of the plane. Keep this intuition alive, and no problem will ever be too complex for you.