Question

The shortest distance between the lines $${\displaystyle \frac{x}{2}}={\displaystyle \frac{y}{2}}={\displaystyle \frac{z}{1}}$$ and $${\displaystyle \frac{x+2}{-1}}={\displaystyle \frac{y-4}{8}}={\displaystyle \frac{z-5}{4}}$$ lies in the interval:
a). $$[1,2)$$
b). $$(3,4]$$
c). $$[0,1)$$
d). $$(2,3]$$

Answer 1

Hint: Finding the space among parallel lines is to decide how ways aside the lines are. This may be executed by measuring the perpendicular distance between them. We might also derive a component for use of this technique and use this method at once to find the shortest distance between two parallel lines. For non-intersecting lines lying in the same plane, the shortest distance is the distance that is the shortest of all of the distances between points.
Formula Used:
Shortest distance: $${\displaystyle \frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\Vert \begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\Vert }}$$

Complete step-by-step solution:
Now let us compute the shortest distance between the pair of given lines $${\displaystyle \frac{x}{2}}$$ $$={\displaystyle \frac{y}{2}}$$ $$={\displaystyle \frac{z}{1}}$$and $${\displaystyle \frac{x+2}{-1}}={\displaystyle \frac{y-4}{8}}={\displaystyle \frac{z-5}{4}}$$.
Here we see that $$x_1$$ $$=$$ $$0$$ and $$x_2=-2$$, $$y_1=0$$ and $$y_2=4$$, $$z_1=0$$and $$z_2=5$$, $$a_1$$, $$b_1$$, $$c_1$$ is equal to
 $$2$$, $$2$$, $$1$$ respectively and $$a_2$$, $$b_2$$, $$c_2$$ is equal to $$-1$$, $$8$$, $$4$$ respectively.
Therefore we substitute the values inside the formula,
$${\displaystyle \frac{\left|\begin{array}{ccc}-2& 4& 5\\ 2& 2& 1\\ -1& 8& 4\end{array}\right|}{\Vert \begin{array}{ccc}i& j& k\\ 2& 2& 1\\ -1& 8& 4\end{array}\Vert }}$$
Then we compute the determinant of the numerator,
$$\begin{gathered} =-2(2.4-8.1)-4(2.4+1.1)+5(2.8+1.2) \\ =-2.0-4.9+5.18 \\ =-36+90 \\ =54 \end{gathered}$$
And then we calculate the denominator by first opening the determinant and then using the distance formula over the equation,
$$\begin{gathered} =i(2.4-8.1)-j(2.4+1.1)+k(2.8+1.2) \\ =i.0-j.9+k.18 \\ =0i-9j+18k \end{gathered}$$
Now we use the distance formula in the above-obtained equation i.e.
$$\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$$
$$x_1$$,$$y_1$$,$$z_1$$is equal to $$0$$so now we have,
$$\begin{gathered} \sqrt{0^2+{(-9)}^2+{18}^2} \\ =\sqrt{81+324} \\ =\sqrt{405} \\ =20.1 \end{gathered}$$
Hence, the shortest distance is $${\displaystyle \frac{54}{20.1}}$$ which is equal to $$2.68$$
Now we see that $$2.68$$ lies inside the interval $$(2,3]$$
Therefore the correct option is (d) $$(2,3]$$.
Additional information: In 3D space, two lines can either intersect each other at some point, parallel to each other or they could neither be intersecting nor parallel to every other, additionally referred to as skew lines. In the case of intersecting lines the shortest distance among them is $$0$$.

Note: One must be very fluent and careful while opening the determinants. The whole answer of the problem lies completely in how correctly the determinants are evaluated. Also one has to be very cautious while deciding the signs of terms of the lines while substituting them in the formulas.