<p>I encountered this problem and felt that it was a little tricky. I am not sure what the answer is. Could anyone help me with the thinking process? Really appreciate it.</p>
<p>Two perpendicularly intersecting planes each contains two non-parallel lines. What is the least number of points at which the lines could intersect?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4</p>
<p>I guess we can make up a similar question along the same line (most number of points instead of least number of points):
Two perpendicularly intersecting planes each contains two non-parallel lines. What is the most number of points at which the lines could intersect?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4</p>
<p>2 non parallel lines will always intersect therefore the answer cannot be zero. I think that the answer for the least possible intersection is 1 as all of the lines could intersect at the same point.</p>
<p>I agree w/ Remi56783. Arrange two pencils to make an X. Then arrange two other pencils to also form an X thru the same common vertex, oriented w/ their plane perpendicular to the plane of the first X and you will see how it works.</p>
<p>BTW, the variation where you ask for the most number of possible intersection points is also kind of interesting. Assuming the four lines are distinct, I believe the answer is 4. But it’s not that easy for me to visualize. You have the two points where each of the pairs meet. And I believe that you can arrange for each of the lines in one of the planes to intersect with ONE of the lines in the other plane, but not both… I’m picturing a cube with an X drawn across the diagonals of the front face and another X drawn across the diagonals of the top face.</p>