Question

Show that the lines with symmetric equations \(x = y = z\) and \(x + 1 = y/2 = z/3\) are skew, and find the distance between these lines. [Hint: The skew lines lie in parallel planes.]

Short answer

The distance between skew lines with symmetric equations is\(\frac{1}{{\sqrt 6 }}\).

The lines with symmetric equations \(x = y = z\) and \(x + 1 = \frac{y}{2} = \frac{z}{3}\) are skew lines.

Step-by-step solution

Formula used

The separation between the parallel planes \(ax + by + cz + {d_1} = 0\) and \(ax + by + cz + {d_2} = 0\) is equal to:

\(D = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\)

Calculate the direction vectors of given line

The two lines must be parallel, skew or intersect lines. If the two lines are parallel, the direction vectors of both the lines are scalar multiples of each other.

The two lines \({L_1}\) and \({L_2}\) are in the form of symmetric equations.

The symmetric equation for the line \({L_1}\) is written as follows.

\(x = y = z\)

\(\frac{{x - 0}}{1} = \frac{{y - 0}}{1} = \frac{{z - 0}}{1}{\rm{ }}\) ……. (1)

The expressions for the symmetric equations for a line through the point \(\left( {{x_0},{y_0},{z_0}} \right)\) and parallel to the direction vector \(\langle a,b,c\rangle \).

\(\frac{{x - {x_0}}}{a} = \frac{{y - {y_0}}}{b} = \frac{{z - {z_0}}}{c}{\rm{ }}\) ……. (2)

Compare equation (1) and (2),

The direction vector of line \({L_1}\),

\({{\rm{v}}_1} = \langle 1,1,1\rangle \)

The symmetric equation for the line \({L_2}\) is

\(x + 1 = \frac{y}{2} = \frac{z}{3}\)

.

\(\frac{{x - ( - 1)}}{1} = \frac{{y - 0}}{2} = \frac{{z - 0}}{3}{\rm{ }}\) ……. (3)

Compare equation (2) and (3), we get the direction vector of line \({L_2}\),

\({v_2} = \langle 1,2,3\rangle \)

By observing the direction vectors of two lines \({L_1}\) and \({L_2}\), it is clear that the two lines are not scalar multiples of each other.

\({{\rm{v}}_1} \ne k{{\rm{v}}_2}\)

Therefore, the two line \({L_1}\) and \({L_2}\) are not parallel lines.

If the two lines have an intersection point, the parametric equations of the two lines must be equal.

The expressions for the parametric equations for a line through the point \(\left( {{x_0},{y_0},{z_0}} \right)\) and parallel to the direction vector \(\langle a,b,c\rangle \).

\(x = {x_0} + at,y = {y_0} + bt,z = {z_0} + ct\) ……. (4)

The point and the direction vector for line\({L_1}\) are \((0,0,0)\)and\(\left\langle {1,1,1} \right\rangle \)respectively.

Calculate the distance using distance formula

Let \(t = 0\) for line \({L_1}\), we get point \((0,0,0)\). Put into the distance from a point to a plane formula

\(\begin{array}{l}D = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\\ = \frac{{|1(0) - 2(0) + 1(0) + 1|}}{{\sqrt {{1^2} + {{( - 2)}^2} + {1^2}} }}\\ = \frac{1}{{\sqrt 6 }}\end{array}\)

Calculate the parametric equations

Calculation of parametric equations for the line \({L_1}\):

Substitute the values in equation (4),

\(\begin{array}{l}x = 0 + (1)t,y = 0 + (1)t,z = 0 + (1)t\\x = t,y = t,z = t\end{array}\)

The line \({L_1}\) parametric equations are \(x = t,y = t,z = t\).

From equation (3), it is clear that the point and the direction vector \(\left( {{v_2}} \right)\) for line \({L_2}\) are \(( - 1,0,0)\) and \(\langle 1,2,3\rangle \) respectively.

The lines \({L_2}\) parametric equations are:

Substitute the values in equation (4),

\(x = - 1 + (1)s,y = 0 + (2)s,z = 0 + (3)s\)

\(x = - 1 + s,y = 2s,z = 3s\)

The lines \({L_2}\) parametric equations are \(x = - 1 + s,y = 2s,z = 3s\).

From the parametric equations of the two lines, the parametric equations of the lines are not equal. Therefore, the two lines are not intersected lines.

As the two lines are neither parallel nor intersected lines, the two lines must be skew lines.

Hence, the lines with symmetric equations \(x = y = z\) and \(x + 1 = \frac{y}{2} = \frac{z}{3}\) are skew lines.