posted by Derek .
I am so confused with this last part of the unit! What in the world is going on: Use normal vectors to determine the interaction, if any, for each of the following pairs of planes. Give a geometric interpretation in each case and the number of solutions for the corresponding system of linear equations. If the planes intersect in a line, determine a vector equation of the line. (Note, you don't have to answer all three, just one would be great, so I can see what you did).
a.) x + 2y + 3z = −4
2x + 4y + 6z = 10
b.)2x − y + 2z = −8
4x − 2y + 4z = −16
c.) x + 3y − 5z = −12
2x + 3y − 4z = −6
for the first one, did you notice that the normals are
(1,2,3) and (2,4,6) or 2(1,2,3) ?
so the normals are parallel, which means that the planes are parallel.
Does that make sense? OK!
But the constants are -4 and 10 which are not in the same ratio as 1:2:3 to 2:4:6
So the first question represents two parallel and distinct planes.
For the second one, did you notice that multiplying the first equation by 2 gives you the second equation?
So the two equations actually represent the same plane
For the third one, the two normals are not scalar multiples of each other, so the two planes are not parallel.
They must therefore intersect in a straight line.
(Think of two pages of an open book intersecting in the spine of the book)
subtract the first equation from the second to get
now let's pick any value for z
z=0, then x = 6 and from the first equation y=-6
z=6, then x=0 and from the first equation y = 6
So we now have two points (0,6,6) and (6,-6,0) which lie on the line of intersection.
notice the two points satisfy both plane equations.
That means we can find the direction vector of our line which is (6,-12,-6) or reduced to (1,-2,-1)
So now we have the direction of our line and a point on the line.
Then a possible equation of the line in vector form is
vector r = (0,6,6) + t(1,-2,-1)