Question

Consider the plane: 3x-y+4z=9 and the line given by the vector equation: r(t)=<1-2t, 7+6t, 4+3t>.

Show that they are parallel and find the distance between them.

Answer 1

@Algebraic! @amistre64 @eseidl

Answer 2

The line will be parallel to the plane if it is orthogonal to a vector normal to the plane. A vector normal to the plane is n=<3,-1,4> and a vector parallel to the line is v=<-2,6,3> (I just computed two points on the line, t=0 and t=1, and then used these to get a vector along the line). Is v perpendicular to n? We check their dot product:\[\vec n * \vec v=<3,-1,4>*<-2,6,3>=0\]Since these vectors are perpendicular, the plane and the line must be parallel.

Answer 3

Pick any point P on the line and any point Q on the plane. We compute the distance by taking the projection of the vector PQ onto the normal vector n:\[D=\frac{\left| \vec PQ * \vec n \right|}{\left| \vec n \right|}\]We'll take Q=(3,0,0) and P=(1,7,4), then PQ=<-2,7,4>. PQ*n=<-2,7,4>*<3,-1,4>=3 and the magnitude of the normal vector is\[\left| \vec n \right|=\sqrt {26}\]we have:\[D=\frac{3}{\sqrt {26}}=\frac{3\sqrt{26}}{26}\]Hope this is correct!