Animated Solution for Mathematics - Three Dimensional Geometry: The plane passing through the points $(1,2,1)$, $(2,1,2)$ and parallel to the line, $2x=3y,z=1$ also passes through the point:

A Journey into 3D Space

Imagine you are standing in a vast, empty room. You have two points floating in the air, $P_1(1,2,1)$ and $P_2(2,1,2)$.

You want to construct a flat, infinite sheet of paper—a plane—that passes perfectly through these two points. But there is a catch: this plane must also be perfectly parallel to a specific line, $L$, defined by $2x=3y$ and $z=1$.

This is not just an algebra problem; it is a challenge of spatial visualization. Let us break it down, step by step, and uncover the elegance hidden within the coordinates.

Decoding the Line

First, we must understand our constraints. The line $L$ is given by $2x=3y$ and $z=1$.

To find the direction of this line, we need to bring it into the symmetric form:

By rearranging $2x=3y$, we get $\frac{x}{3}=\frac{y}{2}$. Since $z$ is fixed at $1$, the line has no movement in the $z$-direction, meaning its $z$-component is $0$.

Thus, the direction vector of our line is $\vec{b}=3\hat{i}+2\hat{j}+0\hat{k}$. This vector is our compass; it tells us the orientation of the line that our plane must respect.

The Vector Bridge

Now, let us look at the two points $P_1$ and $P_2$. If we draw a line segment connecting them, that segment lies entirely within our plane.

We can represent this segment as a vector, $\vec{a}$, by subtracting the coordinates of $P_1$ from $P_2$:

We now have two vectors that define the orientation of our plane: $\vec{a}$, which lies on the plane, and $\vec{b}$, which is parallel to the plane.

The Normal Hero

To write the equation of a plane, we need a normal vector, $\vec{n}$. This is the vector that stands perfectly perpendicular to the surface.

How do we find a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$? The answer is the cross product!

By calculating $\vec{n}=\vec{a}\times \vec{b}$, we create a vector that is orthogonal to both, effectively defining the 'tilt' of our plane. We set up the determinant:

Expanding this, we get $\hat{i}((-1)(0)-(1)(2))-\hat{j}((1)(0)-(1)(3))+\hat{k}((1)(2)-(-1)(3))$. This simplifies beautifully to $\vec{n}=-2\hat{i}+3\hat{j}+5\hat{k}$.

This vector, $(-2,3,5)$, is the key to our equation.

The Final Construction

With our normal vector $\vec{n}=(-2,3,5)$ and a point $P_1(1,2,1)$, we can use the standard point-normal form:

Substituting our values, we get $-2(x-1)+3(y-2)+5(z-1)=0$. Expanding this, we arrive at $-2x+2+3y-6+5z-5=0$, which simplifies to $-2x+3y+5z-9=0$.

Multiplying by $-1$ for elegance, we get the final equation:

Finally, we test our result. Plugging in the point $(-2,0,1)$, we see $2(-2)-3(0)-5(1)+9=-4-0-5+9=0$. It fits perfectly! You have successfully navigated the 3D landscape and found the plane.