Line of intersection between 2 planes when there's unknowns

[Ttwnycorporation]

Hi, I would like to request for some help with my math. (sorry I'm not sure if this is the right topic for vectors)!!!

I am having trouble with part (i) of this question. There seems to be no way to do it because I can't find a common point for the two planes... Usually I just type the Cartesian equations into my calculator or Let a common point be (x,0,z), and then solve them, but this time the unknowns make it very complicated for me.

Could someone please help me out with this question? Thanks a lot in advance and also merry Christmas lol

 

[HallsofIvy]

Do you understand what the notation means? You are given the equation of a line, l, as \(\displaystyle r= \begin{pmatrix}1\\ 2 \\ 1\end{pmatrix}+ \lambda \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\). That is the same as \(\displaystyle x= 1+ \lambda\), \(\displaystyle y= 2\), \(\displaystyle z= 1- \lambda\). Do you see why "this line l lies in the plane y= 2"?

The plane \(\displaystyle \Pi_1\) has equation \(\displaystyle \Pi_1= r\cdot\begin{pmatrix}a \\ 1\\ b \end{pmatrix}= 5\). Since \(\displaystyle r= \begin{pmatrix} x \\ y \\ z\end{pmatrix}\) this is the same as ax+ y+ bz= 5.
(i) Find the equation of the line of intersection of \(\displaystyle \Pi_1\) and plane y= 2 In terms of a and b. The "plane y= 2" consists of all points (x, 2, z) so we set y= 2: ax+ 2+ bz= 5. ax+ bz= 3. bz= 3- ax, z= (3/b)- (a/b)x.
(ii) Hence or otherwise, determine the possible values of a and b if the plane \(\displaystyle \Pi_1\) intersects the plane y= 2 at l.
l can be written as \(\displaystyle x= 1+ \lambda\), \(\displaystyle y= 2\), \(\displaystyle z= 1-\lambda\) so in order for that to be the line in (i) we must have \(\displaystyle z= 1- \lambda= (3/b)- (a/b)(1+ \lambda)= (3/b)+ (a/b)x\). That reduces to \(\displaystyle 1+ (3/b)= -(a/b)+ (1- (a/b))\lambda\) and then \(\displaystyle (1- (a/b))\lambda)= 1+ (3/b)+ (a/b)\). That will be true for all \(\displaystyle \lambda\) if and only if \(\displaystyle 1- (a/b)= 0\) and \(\displaystyle 1+ (3/a)+ (a/b)= 0\). Solve those two equations for a and b.
(iii) Given that a= 1, b= 3 and \(\displaystyle \Pi_2\) is the reflection of \(\displaystyle \Pi_1\) about the plane y= 2, determine the equation of \(\displaystyle \Pi_2\).
Notice that a= 1, b= 3 do not satisfy the equations in (ii) so the line is not in plane \(\displaystyle \Pi_1\). A point (x, y, z) reflects in the plane y= 2 to (x, 4- y, z). The plane \(\displaystyle \Pi_1= x+ y+ 3z= 5\) reflects to \(\displaystyle \Pi_2= x+ 4- y+ 3z= 5\) so \(\displaystyle \Pi_2= x- y+ 3z= 1\).

 

[Ttwnycorporation]

Do you understand what the notation means? You are given the equation of a line, l, as \(\displaystyle r= \begin{pmatrix}1\\ 2 \\ 1\end{pmatrix}+ \lambda \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix}\). That is the same as \(\displaystyle x= 1+ \lambda\), \(\displaystyle y= 2\), \(\displaystyle z= 1- \lambda\). Do you see why "this line l lies in the plane y= 2"?
The plane \(\displaystyle \Pi_1\) has equation \(\displaystyle \Pi_1= r\cdot\begin{pmatrix}a \\ 1\\ b \end{pmatrix}= 5\). Since \(\displaystyle r= \begin{pmatrix} x \\ y \\ z\end{pmatrix}\) this is the same as ax+ y+ bz= 5.
(i) Find the equation of the line of intersection of \(\displaystyle \Pi_1\) and plane y= 2 In terms of a and b. The "plane y= 2" consists of all points (x, 2, z) so we set y= 2: ax+ 2+ bz= 5. ax+ bz= 3. bz= 3- ax, z= (3/b)- (a/b)x.
(ii) Hence or otherwise, determine the possible values of a and b if the plane \(\displaystyle \Pi_1\) intersects the plane y= 2 at l.
l can be written as \(\displaystyle x= 1+ \lambda\), \(\displaystyle y= 2\), \(\displaystyle z= 1-\lambda\) so in order for that to be the line in (i) we must have \(\displaystyle z= 1- \lambda= (3/b)- (a/b)(1+ \lambda)= (3/b)+ (a/b)x\). That reduces to \(\displaystyle 1+ (3/b)= -(a/b)+ (1- (a/b))\lambda\) and then \(\displaystyle (1- (a/b))\lambda)= 1+ (3/b)+ (a/b)\). That will be true for all \(\displaystyle \lambda\) if and only if \(\displaystyle 1- (a/b)= 0\) and \(\displaystyle 1+ (3/a)+ (a/b)= 0\). Solve those two equations for a and b.
(iii) Given that a= 1, b= 3 and \(\displaystyle \Pi_2\) is the reflection of \(\displaystyle \Pi_1\) about the plane y= 2, determine the equation of \(\displaystyle \Pi_2\).
Notice that a= 1, b= 3 do not satisfy the equations in (ii) so the line is not in plane \(\displaystyle \Pi_1\). A point (x, y, z) reflects in the plane y= 2 to (x, 4- y, z). The plane \(\displaystyle \Pi_1= x+ y+ 3z= 5\) reflects to \(\displaystyle \Pi_2= x+ 4- y+ 3z= 5\) so \(\displaystyle \Pi_2= x- y+ 3z= 1\).

Hi, thanks for the reply! For part (i), I still do not understand how you get the line of intersection. What does z= (3/b)- (a/b)x represent? So you substitute the plane y=2 into \(\displaystyle \Pi_1\), and then what happens??
Thanks a lot for the reply, I really appreciate it

 

[HallsofIvy]

Do you understand my statement that \(\displaystyle r\cdot \begin{pmatrix}a \\ 1 \\ b \end{pmatrix}= 5\) is the same thing as ax+ y+ bz= 5? And that the "plane y= 2" is the set of all points, (x, 2, z). The "line of intersection" of the two planes lies in both planes so any (x, y, z) must satisfy both ax+ y+ bz= 5 and y= 2 so ax+ 2+ bz= 5 or ax+ bz= 3. As I said, that can be written z= 3/b- (a/b)x. That is, any point, (x, y, z) on the line of intersection is of the form (x, 2, 3/b- (a/b)x).