When sailboats are racing, they sometimes get into a tacking duel. This occurs when the boats cross paths as they are travelling into the wind toward the same mark. A collision would be very expensive! GPS and RADAR technology can be used to predict the paths of the boats and determine if they will collide.
If the algebraic solution to a system of linear equations yields a unique pair of numbers, the lines intersect at exactly one point. If the algebraic solution gives an equation such as \(0x = 0\), the lines are coincident and have infinitely many solutions. If the system yields an impossible equation such as \(0x = 4\), the lines are parallel and have no solutions.
In three-space, lines can either intersect at a point, be parallel and distinct, be coincident, or be skew (not parallel and not intersecting).
Determine if the following lines intersect. If they do, find the coordinates of the point of intersection:
Line 1: \([x, y, z] = [7, 2, 6] + s[2, 1, 3]\)
Line 2: \([x, y, z] = [3, 9, 13] + t[1, 5, 5]\)
The direction vectors are not parallel since they aren't scalar multiples of each other.
First, we can write the corresonding parametric equations:
\(x_1 = 7 + 2s\)
\(y_1 = 2 + s\)
\(z_1 = 6 + 3s\)
\(x_2 = 3 + t\)
\(y_2 = 9 + 5t\)
\(z_2 = 13 + 5t\)
Next, we can set the equations to each other:
\(7 + 2s = 3 + t\)
\(2 + s = 9 + 5t\)
\(6 + 3s = 13 + 5t\)
Then, we can solve for \(t\). We can start by multiplying Equation 2 by 2:
\(2(2 + s = 9 + 5t)\)
\(4 + 2s = 18 + 10t\)
We can then solve for \(t\) by eliminating Equation 1 by Equation 2:
\(7 + 2s - (4 + 2s) = 3 + t - (18 + 10t)\)
\(7 + 2s -4 - 2s = 3 + t -18 - 10t\)
\(3 = -9t - 15\)
\(-9t = 18\)
\(t = -2\)
After, we can solve for \(s\) by substituting \(t\) into one of the equations. In this instance, we will substitute \(t\) into Equation 1:
\(7 + 2s = 3 + (-2)\)
\(2s = 3 -2 - 7\)
\(2s = -6\)
\(s = -3\)
By solving this system, we can determine that \(s =-3\) and \(t=-2\).
Finally, we can substitute \(s\) and \(t\) into Equation 3 to verify if our results are correct:
\(6 + 3(-3) = 13 + 5(-2)\)
\(6 - 9 = 13 -10\)
\(-3 = 3\)
Even though the first \(2\) lines meet, we can verify that there is no solution to the system of equations. This is because when we substituted the values for \(s\) and \(t\) into the third equation, the \(2\) sides weren't equal to each other.