Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point (1,4,0) along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

Imagine you are standing in a vast, three-dimensional space at the point $P(1,4,0)$. Before you lies a target, a line $L_1$ defined by the equation:

Your goal is to reach this line, but you are bound by a specific direction dictated by a second line, $L_2$, given by:

Defining the Path

To reach our destination, we must first define our path. We are starting at $P(1,4,0)$ and moving in the direction of $L_2$. The direction ratios of $L_2$ are $(1,2,3)$.

Therefore, our path, which we shall call $L_3$, is a line passing through $P$ with the direction vector $\vec{d}=\hat{i}+2\hat{j}+3\hat{k}$. Using the point-direction form, the equation of our path $L_3$ is:

Here, $t$ is our parameter. Any point $Q$ on this path can be expressed in terms of $t$ as:

The Intersection

We seek the point where our path $L_3$ intersects the target line $L_1$. For this to happen, the point $Q$ must satisfy the equation of $L_1$. We substitute the coordinates of $Q$ into the equation of $L_1$:

Simplifying the numerators, we obtain:

To solve for $t$, we equate the first two parts:

Cross-multiplying yields $3(t-1)=4(t-1)$, which implies $t-1=0$, or $t=1$. This confirms that at $t=1$, our path intersects the target line.

The Final Distance

With $t=1$, we find the exact coordinates of our intersection point $Q$ by substituting $t=1$ back into our expression for $Q$:

Thus, the intersection point is $Q(2,6,3)$. We now calculate the distance between our starting point $P(1,4,0)$ and our destination $Q(2,6,3)$ using the 3D distance formula:

The final distance is $\sqrt{14}$. Through the power of parametric equations, we have navigated the 3D space to arrive at the solution.