Question

Let\({L_1}\)be the line through the points\((1,2,6)\)and\((2,4,8)\). Let\({L_2}\)be the line of intersection of the planes\({\pi _1}\)and\({\pi _2}\), where\({\pi _1}\)is the plane\(x - y + 2z + 1 = 0\)and\({\pi _2}\)is the plane through the points\((3,2, - 1),(0,0,1)\), and\((1,2,1)\). Calculate the distance between\({L_1}\)and\({L_2}\).

Short answer

The distance between the line \({L_1}\) which passes through the points \((1,2,6)\) and \((2,4,8)\), and the line \({L_2}\) of the intersects the planes \({\pi _1}\) and \({\pi _2}\) is \(6\).

Step-by-step solution

Formula used

The distance between two lines is given by \(D = \frac{{|n \cdot b|}}{{|n|}}\), where \(n\) is the normal vector, which is determined by direction vectors of two lines and \(b\) is the vector joined between the points on the two lines.

Calculate the direction vector for the given lines

The first line \({L_1}\) is drawn between the points \((1,2,6)\) and \((2,4,8)\). The second line \({L_2}\) is the intersection of planes \({\pi _1}\) and \({\pi _2}\).

Note that, the direction vector joined from the point \(P\left( {{x_1},{y_1},{z_1}} \right)\) to \(Q\left( {{x_2},{y_2},{z_2}} \right)\) is \(v = \left\langle {\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)} \right\rangle \).

The direction vector for the line \({L_1}\) is obtained by substituting \({x_1} = 1,{y_1} = 2,{z_1} = 6,{x_2} = 2,{y_2} = 4\), and \({z_2} = 8\) in the formula named as \({{\rm{v}}_1}\).

\(\begin{array}{l}{v_1} = \langle (2 - 1),(4 - 2),(8 - 6)\rangle \\ = \langle 1,2,2\rangle \end{array}\)

Since the equation of the plane \({\pi _1}\) is \(x - y + 2z = - 1\), then its normal vector is \({{\rm{n}}_1} = \langle 1, - 1,2\rangle \).

The plane \({\pi _2}\) is passing through the points \((3,2, - 1),(0,0,1)\) and \((1,2,1)\).

Consider the direction vector from the points \((0,0,1)\) to \((3,2, - 1)\) in the plane \({\pi _2}\) as \({\pi _{21}}\) and direction vector from the points \((0,0,1)\) to \((1,2,1)\) as \({\pi _{22}}\).

The direction vector \({\pi _{21}}\) is obtained by substituting \({x_1} = 0,{y_1} = 0,{z_1} = 1,{x_2} = 3,{y_2} = 2\), and \({z_2} = - 1\) ,

\(\begin{array}{l}{\pi _{21}} = \langle (3 - 0),(2 - 0),( - 1 - 1)\rangle \\ = \langle 3,2, - 2\rangle \end{array}\)

The direction vector \({\pi _{22}}\) is obtained by substituting \({x_1} = 0,{y_1} = 0,{z_1} = 1,{x_2} = 1,{y_2} = 2\), and \({z_2} = 1\),

\(\begin{array}{l}{\pi _{22}} = \langle (1 - 0),(2 - 0),(1 - 1)\rangle \\ = \langle 1,2,0\rangle \end{array}\)

The normal vector \({{\rm{n}}_2}\) to the plane \({\pi _2}\) is the cross product of direction vectors \({\pi _{21}}\) and \({\pi _{22}}\).

The normal vector \({{\rm{n}}_2}\) with \({\pi _{21}} = \langle 3,2, - 2\rangle \) and \({\pi _{22}} = \langle 1,2,0\rangle \) is

\(\begin{array}{l}{{\rm{n}}_2} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\3&2&{ - 2}\\1&2&0\end{array}} \right|\\ = {\rm{i}}((2)(0) - (2)( - 2)) - {\rm{j}}((3)(0) - (1)( - 2)) + {\rm{k}}((3)(2) - (1)(2))\\ = {\rm{i}}(0 + 4) - {\rm{j}}(0 + 2) + {\rm{k}}(6 - 2)\\ = 4{\rm{i}} - 2{\rm{j}} + 4{\rm{k}}\end{array}\)

Calculate the direction vector \({{\rm{v}}_2}\) of the line \({L_2}\):

It is the cross product of normal vectors of two planes.

That is, calculate the \({n_1} \times {n_2}\) with \({n_1} = \langle 1, - 1,2\rangle \) and \({n_2} = \langle 4, - 2,4\rangle \)

\(\begin{array}{l}{v_2} = \left| {\begin{array}{*{20}{c}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&{ - 1}&2\\4&{ - 2}&4\end{array}} \right|\\ = {\rm{i}}(( - 1)(4) - ( - 2)(2)) - {\rm{j}}((1)(4) - (4)(2)) + {\rm{k}}((1)( - 2) - (4)( - 1))\\ = {\rm{i}}( - 4 + 4) - {\rm{j}}(4 - 8) + {\rm{k}}( - 2 + 4)\\ = 0{\rm{i}} + 4{\rm{j}} + 2{\rm{k}}\end{array}\)

Thus, the direction vectors of two lines are \({v_1} = \langle 1,2,2\rangle \) and \({v_2} = \langle 0,4,2\rangle \).

The normal vector is the cross product of direction vectors of both the lines.

Calculate the normal vector \(n = {v_1} \times {v_2}\) with \({v_1} = \langle 1,2,2\rangle \) and \({v_2} = \langle 0,4,2\rangle \)

\(\begin{array}{l}{\rm{n}} = \left| {\begin{array}{*{20}{l}}{\rm{i}}&{\rm{j}}&{\rm{k}}\\1&2&2\\0&4&2\end{array}} \right|\\ = {\rm{i}}((2)(2) - (4)(2)) - {\rm{j}}((1)(2) - (0)(2)) + {\rm{k}}((1)(4) - (0)(2))\\ = {\rm{i}}(4 - 8) - {\rm{j}}(2 - 0) + {\rm{k}}(4 - 0)\\ = - 4{\rm{i}} - 2{\rm{j}} + 4{\rm{k}}\end{array}\)

Consider one of the points in the line \({L_1}\) is \((1,2,6)\) and one of the points in the line \({L_2}\) is \((3,2, - 1)\).

Then, the vector \(b\) is obtain by joined from the points \((1,2,6)\) to \((3,2, - 1)\) as follows.

\(\begin{array}{l}{\rm{b}} = \langle (3 - 1),(2 - 2),( - 1 - 6)\rangle \\ = \langle 2,0, - 7\rangle \\ = 2{\rm{i}} + 0{\rm{j}} - 7{\rm{k}}\end{array}\)

Calculate the distance using distance formula

Substituting \(( - 4i - 2j + 4k)\) for \(n\) and ( \(2i + 0j - 7k)\) for \(b\) to obtain the distance.

\(D = \frac{{|( - 4i - 2j + 4k) \cdot (2i + 0j - 7k)|}}{{|( - 4i - 2j + 4k)|}}\)

\(\begin{array}{l} = \frac{{\left| {\begin{array}{*{20}{l}}{1( - 4i) \cdot - 2i + + (( - 4i) \cdot 0j) + \mid ( - 4i) \cdot ( - 7k)) + (( - 2j) \cdot 2i) + (( - 2j) \cdot 0j)}\\{ + (( - 2j) \cdot ( - 7k)) + (4k \cdot 2i) + (4k \cdot 0j) + (4k \cdot ( - 7k))}\end{array}} \right|}}{{\sqrt {{{( - 4)}^2} + {{( - 2)}^2} + {4^2}} }}\\ = \frac{{|( - 8) + 0 + 0 + 0 + 0 + 0 + 0 + 0 + ( - 28)|}}{{\sqrt {16 + 4 + 16} }}\\ = \frac{{\left| { - 36} \right|}}{{\sqrt {36} }}\end{array}\)

\(\begin{array}{l} = \frac{{36}}{6}\\ = 6\end{array}\)

Thus, the distance between the line \({L_1}\) which passes through the points \((1,2,6)\) and \((2,4,8)\), and the line \({L_2}\) of the intersects the planes \({\pi _1}\) and \({\pi _2}\) is \(6.\)