Animated Solution for Mathematics - Three Dimensional Geometry: The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is :

To find the shortest distance from the point $P(2,-1,4)$ to the line $L$ defined by

we must identify the perpendicular distance from the point to the line. This is equivalent to finding the length of the segment $PM$, where $M$ is the foot of the perpendicular dropped from $P$ onto $L$.

A line in 3D space is a trajectory. We represent this trajectory using a parameter $\lambda$. By setting the symmetric equation equal to $\lambda$, we create a bridge between the abstract equation and concrete coordinates:

This allows us to express any point $M$ on the line as a function of $\lambda$:

Thus, the general point $M$ is $(10\lambda -3,-7\lambda +2,\lambda )$. For a specific value of $\lambda$, this point $M$ represents the foot of the perpendicular.

We define the vector $\overrightarrow{PM}$, which connects our fixed point $P$ to our variable point $M$. We find this by subtracting the coordinates of $P$ from $M$:

Simplifying the components, we obtain:

The direction vector of the line $L$, extracted from the denominators of the symmetric equation, is $\vec{d}=10\hat{i}-7\hat{j}+1\hat{k}$.

For $\overrightarrow{PM}$ to be the shortest distance, it must be perpendicular to the line $L$. Mathematically, this requires the dot product of $\overrightarrow{PM}$ and the direction vector $\vec{d}$ to be zero:

Substituting our components into the dot product equation:

Expanding the terms carefully:

Grouping the terms leads to:

This simplifies to the parameter value $\lambda =0.5$.

Using $\lambda =0.5$, we find the coordinates of the foot of the perpendicular $M$:

Thus, $M=(2,-1.5,0.5)$. We now calculate the vector $\overrightarrow{PM}$ using these coordinates:

The shortest distance is the magnitude of this vector:

The final shortest distance is $\sqrt{12.5}$ (or approximately $3.536$ units).