Question

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:

• A. $1$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$\frac{1}{\sqrt{3}}$

Explanation for the correct option:

Given: the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0$, $2x-y+z+3=0$

Plane through the line of intersection of lines $x+y+z+1=0$ and $2x-y+z+3=0$, is $x+y+z+1+\lambda \left(2x-y+z+3\right)=0$ (plane passing through the intersection of two lines)

$x+y+z+1+2\lambda x-\lambda y+\lambda z+3\lambda =0$

$x(1+2\lambda )+y(1-\lambda )+z(1+\lambda )+1+3\lambda =0$ -------------$1$

It should be parallel to given line $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ -------------$2$

Therefore direction cosine and normal plane vector will be perpendicualr

$$\begin{gathered} \Rightarrow 0(1+2\lambda )+\left(-1\right)(1-\lambda )+1(1+\lambda )=0 \\ \Rightarrow \lambda -1+\lambda +1=0 \\ \Rightarrow \lambda =0 \end{gathered}$$ $(\because a_1{a}_2+b_1{b}_2+c_1{c}_2=0)$

And $1$will become

$$\begin{gathered} x(1+2(0\left)\right)+y(1-0)+z(1+0)+1+3\left(0\right)=0 \\ \Rightarrow x+y+z+1=0 \end{gathered}$$

And line pass through $(1,-1,0)$.

Shortest distance from a point $(x_1,y_1,z_1)$ to line $ax+by+cz+d=0$ is $\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}$

Shortest distance of $(1,-1,0)$ from this plane $=\frac{1-1+0+1}{\sqrt{1^2+1^2+1^2}}$

$=\frac{1}{\sqrt{3}}$

Thus, the shortest distance is $\frac{1}{\sqrt{3}}$.

Hence, option C is correct.