Line parallel to a plane

Question

Determine the value of $m$ that makes the three lines parallel to the same plane.

$s$: $s(1,m,1), s \in \mathbb{R} $

$t$: $(1,-1,0) + w(2,-1,1), w \in \mathbb{R} $

$r$: $(m-1,1,0) + t(m,1,1), t \in \mathbb{R} $

What did he mean by "parallel to the same plane"?

Answer 1

Depending on your level of math, consider the following descriptions. They are all equivalent i.e. you only have to work out one of them.

• The direction vectors span a plane or a line, but not the entire space
• There exists a plane which is parallel to the three lines. Note that this is not always true; in particular it is not so for the principal axes
• The determinant of the matrix containing the direction vectors is $0$.
• The direction vector of the third line is a linear combination of the direction vectors of the first and second line.

Technically having all lines parallel would also provide a solution, but this is likely not intended (nor possible!).

Answer 2

Determine the value of m that makes the three lines parallel to the same plane. s: s(1,m,1),s∈R t:(1,−1,0)+w(2,−1,1),w∈R r:(m−1,1,0)+t(m,1,1),t∈R What did he mean by "parallel to the same plane"?

It means that there exist a plane having such That each of these lines is parallel to that plane! That, in turn, means that a normal vector to the plane is perpendicular to all of the vectors in the direction of the line (the dot product of the direction vector with the normal vector is 0).

A vector in the direction of the first line is <1, m, 1>, for the second line, <2, 1, 1>, and to the third line . Any plane can be written in the form Ax+ By+ Cz= 1 so we can write the normal vector as .

Then we want A+ mB+ C= 0, 2A+ B+ C= 0, and mA+ B+ C= 0. You want t find values of m so that all three of those are true. One way to do that is to try to solve for A, B, and C. (One obvious solution is A= B= C= 0, the "trivial solution", but that does not define a plane.)

An obvious first step in solving for A, B, and C, is to eliminate C by subtracting the second equation from the first, -A+ (m-1)B= 0, and subtracting the first equation from the third, (m- 1)A+ (1- m)B= 0. If we subtract the first of those from the second we cancel B and get -mA= 0. As long as m is not 0, we can divide by m to get A= 0 and then B= C= 0. That is the "trivial solution". To get a non-trivial solution, we must have m= 0.