- #1
Galadirith
- 107
- 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 (l_1) with equation
\frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}
and the line (l_2) with equations
i. x -2y -z =0
ii. x -10y -3z =-7
is \frac{1}{2}\sqrt{14}
So my attempt
first find direction vector of l_1 so let's get parametric form:
\mu = \frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}
x = 3\mu -4
y = 2\mu +3
z = 5\mu -6
which gives the parametric equation
\mathbf{r} = \left( \begin{array}{c} -4 \\ 3 \\ -6 \end{array} \right) + \mu \left( \begin{array}{c} 3 \\ 2 \\ 5 \end{array} \right)
so now to find the parametric equation of l_2:
x = t
using i.
t - 2y - z = 0
z = t - 2y
using ii.
t - 10y - 3(t - 2y) = -7
t - 10y -3t + 6y = -7
-2t - 4y = -7
y = \frac{7}{4} - \frac{1}{2}t
using i.
z = t - 2(\frac{7}{4} - \frac{1}{2}t)
z = t - \frac{7}{2} + t
z = 2t - \frac{7}{2}
so written in parametric vector form:
\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)
(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 l_1 to l_2 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:
\left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 5 \\ 2 & -1 & 4 \end{array} \right| =<br /> \mathbf{i} \left| \begin{array}{cc} 2 & 5 \\ -1 & 4 \\ \end{array} \right|<br /> - \mathbf{j} \left| \begin{array}{cc} 3 & 5 \\ 2 & 4 \\ \end{array} \right|<br /> + \mathbf{k} \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \\ \end{array} \right| = <br /> \mathbf{i} (8 + 5) - \mathbf{j} (12 -10) + \mathbf{k} (-3 -4) = <br /> 13\mathbf{i} - 2\mathbf{j} -7\mathbf{k} = \mathbf{n}
Now to find a random vector \mathbf{R} between l_1 and l_2
for this I will simply use their starting vectors when their respective parameters are each 0:
\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)
so the final step should be to take the scalar product of R and \mathbf{\hat{n}}:
\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 <br /> \left( \begin{array}{c} 13 \\ -2 \\ -7 \end{array} \right) = <br /> \frac{52 + \frac{5}{2} - \frac{35}{2}}{\sqrt{169 + 4 + 49}} =<br /> \frac{\left( \frac{104 + 5 - 35}{2} \right)}{\sqrt{222}} = <br /> \frac{37}{\sqrt{222}}
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 (l_1) with equation
\frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}
and the line (l_2) with equations
i. x -2y -z =0
ii. x -10y -3z =-7
is \frac{1}{2}\sqrt{14}
So my attempt
first find direction vector of l_1 so let's get parametric form:
\mu = \frac{x+4}{3} = \frac{y-3}{2} = \frac{z+6}{5}
x = 3\mu -4
y = 2\mu +3
z = 5\mu -6
which gives the parametric equation
\mathbf{r} = \left( \begin{array}{c} -4 \\ 3 \\ -6 \end{array} \right) + \mu \left( \begin{array}{c} 3 \\ 2 \\ 5 \end{array} \right)
so now to find the parametric equation of l_2:
x = t
using i.
t - 2y - z = 0
z = t - 2y
using ii.
t - 10y - 3(t - 2y) = -7
t - 10y -3t + 6y = -7
-2t - 4y = -7
y = \frac{7}{4} - \frac{1}{2}t
using i.
z = t - 2(\frac{7}{4} - \frac{1}{2}t)
z = t - \frac{7}{2} + t
z = 2t - \frac{7}{2}
so written in parametric vector form:
\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)
(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 l_1 to l_2 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:
\left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 5 \\ 2 & -1 & 4 \end{array} \right| =<br /> \mathbf{i} \left| \begin{array}{cc} 2 & 5 \\ -1 & 4 \\ \end{array} \right|<br /> - \mathbf{j} \left| \begin{array}{cc} 3 & 5 \\ 2 & 4 \\ \end{array} \right|<br /> + \mathbf{k} \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \\ \end{array} \right| = <br /> \mathbf{i} (8 + 5) - \mathbf{j} (12 -10) + \mathbf{k} (-3 -4) = <br /> 13\mathbf{i} - 2\mathbf{j} -7\mathbf{k} = \mathbf{n}
Now to find a random vector \mathbf{R} between l_1 and l_2
for this I will simply use their starting vectors when their respective parameters are each 0:
\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)
so the final step should be to take the scalar product of R and \mathbf{\hat{n}}:
\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 <br /> \left( \begin{array}{c} 13 \\ -2 \\ -7 \end{array} \right) = <br /> \frac{52 + \frac{5}{2} - \frac{35}{2}}{\sqrt{169 + 4 + 49}} =<br /> \frac{\left( \frac{104 + 5 - 35}{2} \right)}{\sqrt{222}} = <br /> \frac{37}{\sqrt{222}}
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
… let's work this backwards 
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