Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$, then $d^2$ is equal to .

Imagine you are standing in a vast, three-dimensional space. You see a line, $L_1$, hovering in the air, defined by the equations:

Below it lies a flat, infinite sheet—a plane $M$ defined by $x-2y-z=3$. Your goal is to find the projection of this line onto the plane.

Think of this as the shadow $L$ that $L_1$ would cast on $M$ if the light were shining perfectly perpendicular to the plane. Once we have this shadow, we need to find the square of the distance from a point $P(0,0,6)$ to this new line $L$.

Finding the Foot of the Perpendicular

To define our projected line $L$, we need two things: a point on it and its direction. We know that if we take any point $A$ on the original line $L_1$ and drop a perpendicular straight down to the plane $M$, the point where it hits, let's call it $B$, must lie on our projected line $L$.

Looking at the equation of $L_1$, we can easily identify a point $A(1,3,4)$. Now, we need the foot of the perpendicular from $A$ to the plane $x-2y-z-3=0$. We use the standard formula for the foot of the perpendicular:

Calculating the right-hand side, we get:

By equating each term to $2$, we find $x=3$, $y=-1$, and $z=2$. Thus, our point $B$ is $(3,-1,2)$. We have our anchor point on the line $L$.

The Direction of the Shadow

The line $L$ lies in the plane $M$, so its direction vector $\vec{u}$ must be perpendicular to the normal vector $\vec{n}=(1,-2,-1)$ of the plane. Furthermore, $L$ lies in the plane containing $L_1$ and the normal $\vec{n}$.

The normal to this second plane is $\vec{v}\times \vec{n}$, where $\vec{v}=(2,1,2)$ is the direction of $L_1$. Therefore, $\vec{u}$ must be perpendicular to both $\vec{n}$ and $\vec{v}\times \vec{n}$. This leads us to the vector triple product: $\vec{u}=\vec{n}\times (\vec{v}\times \vec{n})$.

First, let's calculate $\vec{v}\times \vec{n}$:

Now, we cross this with $\vec{n}$ to get $\vec{u}$:

We can simplify this direction vector by dividing by $2$, giving us $\vec{u}=(7,1,5)$.

The Final Distance

We have our line $L$ passing through $B(3,-1,2)$ with direction $\vec{u}=(7,1,5)$. We need the distance $d$ from $P(0,0,6)$ to this line. We define the vector $\overrightarrow{BP}=P-B=(-3,1,4)$.

The distance formula is $d=\frac{|\overrightarrow{BP}\times \vec{u}|}{|\vec{u}|}$. First, the cross product $\overrightarrow{BP}\times \vec{u}$:

The magnitude of this cross product is $\sqrt{1^2+43^2+(-10{)}^2}=\sqrt{1+1849+100}=\sqrt{1950}$. The magnitude of $\vec{u}$ is $\sqrt{7^2+1^2+5^2}=\sqrt{49+1+25}=\sqrt{75}$.

Thus, $d=\sqrt{\frac{1950}{75}}$. Squaring both sides, we get:

The elegance of the final cancellation is the reward for our meticulous work. We have arrived at the answer: 26.