Question

A) determinine whether the plane 2x-y+z=2 and the line with parametric equation x=2-t, y= 1+3t, z=4t intersect or are parallel. B) if they intersect find the point of intersection. if they are parallel find their distance

Answer 1

The plane is the set of all points (x, y, z) such that\[2x - y + z =2 \]The line is the set of all points (x, y, z) such that\[x = 2 - t \]\[y=1+3t\]\[z=4t\]for all t which are real. If the line intersects the plane then there would be a point of intersection ie. a value of t such that the first equation is satisfied for some value of t. Put the values for the line's coordinates into that equation \[2(2-t) - (1+3t) + 4t = 2\]and simplify \[4-2t - 1 - 3t + 4t = 2\]\[-t=-1\]or\[t=1\]so we have a solution. Put that t value back into the (x, y, z) values for the line\[x = 2-t = 2-1 = 1\]\[y=1+3t = 1 + 3 =4\]\[z=4t=4\]thus (1,4, 4) is the point of intersection.
Double check to see if that actually is on the plane too ( it should ! )
\[2x-y+z = 2-4+4 = 2\]Yep.