- #1
Galadirith
- 109
- 0
Hi guys this problem should be simple enough but I just can't get the correct answer. Well the question is:
Show that the shortest distance between the line ([itex]l_1[/itex]) with equation
[tex]\frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}[/tex]
and the line ([itex]l_2[/itex]) with equations
i. [tex] x -2y -z =0[/tex]
ii. [tex]x -10y -3z =-7[/tex]
is [itex]\frac{1}{2}\sqrt{14}[/itex]
So my attempt
first find direction vector of [itex]l_1[/itex] so let's get parametric form:
[tex]\mu = \frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}[/tex]
[tex]x = 3\mu -4[/tex]
[tex]y = 2\mu +3[/tex]
[tex]z = 5\mu -6[/tex]
which gives the parametric equation
[tex]\mathbf{r} = \left( \begin{array}{c} -4 \\ 3 \\ -6 \end{array} \right) + \mu \left( \begin{array}{c} 3 \\ 2 \\ 5 \end{array} \right)[/tex]
so now to find the parametric equation of [itex]l_2[/itex]:
[tex]x = t[/tex]
using i.
[tex] t - 2y - z = 0[/tex]
[tex] z = t - 2y[/tex]
using ii.
[tex] t - 10y - 3(t - 2y) = -7 [/tex]
[tex] t - 10y -3t + 6y = -7 [/tex]
[tex] -2t - 4y = -7 [/tex]
[tex] y = \frac{7}{4} - \frac{1}{2}t [/tex]
using i.
[tex] z = t - 2(\frac{7}{4} - \frac{1}{2}t) [/tex]
[tex] z = t - \frac{7}{2} + t [/tex]
[tex] z = 2t - \frac{7}{2} [/tex]
so written in parametric vector form:
[tex] \mathbf{r} = \left( \begin{array}{c} 0 \\ \frac{7}{4} \\ - \frac{7}{2} \end{array} \right) + t \left( \begin{array}{c} 2 \\ -1 \\ 4 \end{array} \right) [/tex]
(obviously note the fact that I multiplied my direction vector x2 just to give integers)
so what is required is to project an arbitary vector from and point on [itex]l_1[/itex] to [itex]l_2[/itex] in the direction of the vector orthogonal to both the lines
so to find this vector will do the cross of the two direction vectors of the lines:
[tex] \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 5 \\ 2 & -1 & 4 \end{array} \right| =
\mathbf{i} \left| \begin{array}{cc} 2 & 5 \\ -1 & 4 \\ \end{array} \right|
- \mathbf{j} \left| \begin{array}{cc} 3 & 5 \\ 2 & 4 \\ \end{array} \right|
+ \mathbf{k} \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \\ \end{array} \right| =
\mathbf{i} (8 + 5) - \mathbf{j} (12 -10) + \mathbf{k} (-3 -4) =
13\mathbf{i} - 2\mathbf{j} -7\mathbf{k} = \mathbf{n}[/tex]
Now to find a random vector [itex]\mathbf{R}[/itex] between [itex]l_1[/itex] and [itex]l_2[/itex]
for this I will simply use their starting vectors when their respective parameters are each 0:
[tex] \mathbf{R} = l_2(0) - l_1(0) \left( \begin{array}{c} 0 + 4 \\ \frac{7}{4} - 3 \\ - \frac{7}{2} + 6 \end{array} \right) = \left( \begin{array}{c} 4 \\ -\frac{5}{4} \\ \frac{5}{2} \end{array} \right) [/tex]
so the final step should be to take the scalar product of R and [itex] \mathbf{\hat{n}}[/itex]:
[tex] \mathbf{R} \cdot \mathbf{\hat{n}} = \frac{1}{\sqrt{169 + 4 + 49}} \left( \begin{array}{c} 4 \\ -\frac{5}{4} \\ \frac{5}{2} \end{array} \right) \cdot
\left( \begin{array}{c} 13 \\ -2 \\ -7 \end{array} \right) =
\frac{52 + \frac{5}{2} - \frac{35}{2}}{\sqrt{169 + 4 + 49}} =
\frac{\left( \frac{104 + 5 - 35}{2} \right)}{\sqrt{222}} =
\frac{37}{\sqrt{222}} [/tex]
and that's it, and its not what they give that it should be, I have tried approaching it other ways but I always get the same answer, I am pretty sure I havnt made any numeral errors, and pretty sure i havnt made any conceptual errors, so I am really stumped, and help would be so appreciated guys, thanks allot :D
Show that the shortest distance between the line ([itex]l_1[/itex]) with equation
[tex]\frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}[/tex]
and the line ([itex]l_2[/itex]) with equations
i. [tex] x -2y -z =0[/tex]
ii. [tex]x -10y -3z =-7[/tex]
is [itex]\frac{1}{2}\sqrt{14}[/itex]
So my attempt
first find direction vector of [itex]l_1[/itex] so let's get parametric form:
[tex]\mu = \frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}[/tex]
[tex]x = 3\mu -4[/tex]
[tex]y = 2\mu +3[/tex]
[tex]z = 5\mu -6[/tex]
which gives the parametric equation
[tex]\mathbf{r} = \left( \begin{array}{c} -4 \\ 3 \\ -6 \end{array} \right) + \mu \left( \begin{array}{c} 3 \\ 2 \\ 5 \end{array} \right)[/tex]
so now to find the parametric equation of [itex]l_2[/itex]:
[tex]x = t[/tex]
using i.
[tex] t - 2y - z = 0[/tex]
[tex] z = t - 2y[/tex]
using ii.
[tex] t - 10y - 3(t - 2y) = -7 [/tex]
[tex] t - 10y -3t + 6y = -7 [/tex]
[tex] -2t - 4y = -7 [/tex]
[tex] y = \frac{7}{4} - \frac{1}{2}t [/tex]
using i.
[tex] z = t - 2(\frac{7}{4} - \frac{1}{2}t) [/tex]
[tex] z = t - \frac{7}{2} + t [/tex]
[tex] z = 2t - \frac{7}{2} [/tex]
so written in parametric vector form:
[tex] \mathbf{r} = \left( \begin{array}{c} 0 \\ \frac{7}{4} \\ - \frac{7}{2} \end{array} \right) + t \left( \begin{array}{c} 2 \\ -1 \\ 4 \end{array} \right) [/tex]
(obviously note the fact that I multiplied my direction vector x2 just to give integers)
so what is required is to project an arbitary vector from and point on [itex]l_1[/itex] to [itex]l_2[/itex] in the direction of the vector orthogonal to both the lines
so to find this vector will do the cross of the two direction vectors of the lines:
[tex] \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 5 \\ 2 & -1 & 4 \end{array} \right| =
\mathbf{i} \left| \begin{array}{cc} 2 & 5 \\ -1 & 4 \\ \end{array} \right|
- \mathbf{j} \left| \begin{array}{cc} 3 & 5 \\ 2 & 4 \\ \end{array} \right|
+ \mathbf{k} \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \\ \end{array} \right| =
\mathbf{i} (8 + 5) - \mathbf{j} (12 -10) + \mathbf{k} (-3 -4) =
13\mathbf{i} - 2\mathbf{j} -7\mathbf{k} = \mathbf{n}[/tex]
Now to find a random vector [itex]\mathbf{R}[/itex] between [itex]l_1[/itex] and [itex]l_2[/itex]
for this I will simply use their starting vectors when their respective parameters are each 0:
[tex] \mathbf{R} = l_2(0) - l_1(0) \left( \begin{array}{c} 0 + 4 \\ \frac{7}{4} - 3 \\ - \frac{7}{2} + 6 \end{array} \right) = \left( \begin{array}{c} 4 \\ -\frac{5}{4} \\ \frac{5}{2} \end{array} \right) [/tex]
so the final step should be to take the scalar product of R and [itex] \mathbf{\hat{n}}[/itex]:
[tex] \mathbf{R} \cdot \mathbf{\hat{n}} = \frac{1}{\sqrt{169 + 4 + 49}} \left( \begin{array}{c} 4 \\ -\frac{5}{4} \\ \frac{5}{2} \end{array} \right) \cdot
\left( \begin{array}{c} 13 \\ -2 \\ -7 \end{array} \right) =
\frac{52 + \frac{5}{2} - \frac{35}{2}}{\sqrt{169 + 4 + 49}} =
\frac{\left( \frac{104 + 5 - 35}{2} \right)}{\sqrt{222}} =
\frac{37}{\sqrt{222}} [/tex]
and that's it, and its not what they give that it should be, I have tried approaching it other ways but I always get the same answer, I am pretty sure I havnt made any numeral errors, and pretty sure i havnt made any conceptual errors, so I am really stumped, and help would be so appreciated guys, thanks allot :D