Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0,2x-y+z+3=0$ is:

The goal is to find the shortest distance between two lines in 3D space. We are given one line, $L_1$, and a second line, $L_2$, defined by the intersection of two planes.

The equation for $L_1$ is given in symmetric form:

From this, we identify a point $A(1,-1,0)$ on the line. The direction vector is extracted from the denominators:

Line $L_2$ is the intersection of planes $P_1:x+y+z+1=0$ and $P_2:2x-y+z+3=0$. The direction vector ${\vec{b}}_2$ must be perpendicular to the normals ${\vec{n}}_1=\hat{i}+\hat{j}+\hat{k}$ and ${\vec{n}}_2=2\hat{i}-\hat{j}+\hat{k}$.

We calculate the cross product ${\vec{b}}_2={\vec{n}}_1\times {\vec{n}}_2$:

To find a point $B$ on $L_2$, we set $y=0$ in the plane equations:

Subtracting the first from the second yields $x=-2$. Substituting $x=-2$ into $x+z=-1$ gives $z=1$. Thus, our point is $B(-2,0,1)$.

We use the shortest distance formula for skew lines:

The vector connecting the points is ${\vec{a}}_2-{\vec{a}}_1=-3\hat{i}+\hat{j}+\hat{k}$. Next, we compute the cross product of the direction vectors:

The magnitude of this cross product is:

Finally, the dot product is $(-3)(2)+(1)(2)+(1)(2)=-2$. Taking the absolute value and dividing by the magnitude:

The shortest distance between the two lines is $\frac{1}{\sqrt{3}}$.