(a) Find the point at which the given lines intersect. r = 2, 3, 0 + t 3, βˆ’3, 3 r = 5, 0, 3 + s βˆ’3, 3, 0 (x, y, z) = (b) Find an equation of the plane that contains these lines.

Accepted Solution

Answer:(x,y,z)=(5,0,3)[tex]((x,y,z)-(2,3,0))*(-1,1,0)=0[/tex]Step-by-step explanation:a)The problem requires to find the intersection point of the lines, at that point the position 'r' of the lines is the same:[tex]r_{1} =r_{2} \\(2,3,0)+(3,-3,3)t=(5,0,3)+(-3,3,0)s\\[/tex]First, built the parametric equation system; this is just a simplification coordinate to coordinate of the vector equation:[tex]2+3t=5-3s\\3-3t=3s\\3t=3[/tex]From the last equation, [tex]t=1[/tex]And for whatever of the other two,[tex]s=0[/tex]You can check that replacing t=1 and s=0 the point gotten is (5,0,3), which is the intersection point (the point that belongs to both lines).b) The plane is defined by an orthogonal direction. The equation of the plane uses the fact that the dot product between two orthogonal vectors is always zero.The general equation of a plane is:[tex]((x,y,z)-(x_{0},y_{0},z_{0}))*(n)=0[/tex]Where (x,y,z) are the variables that may be part of the plane or not, [tex](x_{0},y_{0},z_{0})[/tex] is a point that belongs to the plane and n is a vector which is orthogonal to the plane.Due that both lines belong to the plane, the cross product between their direction vectors will give us the orthogonal vector.[tex]n=(3,-3,3)X(-3,3,0)=(-9,-9,0)[/tex]We can divide (-9,-9,0) by nine, because we only need the direction and the division does not affect it.[tex]n=(-1,-1,0)[/tex]Finally, we know that both lines are inside the plane, so any point that belong to a line, belong to the plane. For this reason, let's select any point, for example: (2,3,0) (It could be another). So, the equation of the plane is:[tex]((x,y,z)-(2,3,0))*(-1,-1,0)=0[/tex]