Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it will not count toward your grade.
how many solutions would a system of three linear equations in three variables have if it could be graphed by two overlapping planes intersected by a third plane? (1 point)
two solutions
one solution
no solutions
infinite number of solutions
check answer remaining attempts: 3

Explanation:

Step1: Recall plane intersection in 3D

In 3D, a linear equation in three variables represents a plane. Two overlapping planes mean they are the same plane (infinitely many points in common). A third plane intersecting this overlapping pair: the intersection of two overlapping planes (a line, since they are the same) and a third plane. The intersection of a line and a plane is either a point (if the line is not on the plane) or the line itself (if the line is on the plane). But wait, the two planes are overlapping (so their intersection is a plane, i.e., they are coincident), so the first two equations are the same plane. Then the third plane intersects this coincident plane (which is a plane) – the intersection of two planes is a line. Wait, no: two coincident planes (same plane) and a third plane. The intersection of the coincident plane (let's call it \( P_1 = P_2 \)) and \( P_3 \) is a line (since two planes intersect in a line, unless they are parallel or coincident). But wait, the system has three equations: two are the same plane (so equivalent equations), and the third is another plane. So the solution set is the intersection of the coincident plane (from the first two equations) and the third plane. The intersection of two planes is a line, which has infinitely many points (solutions). Wait, no: wait, the problem says "two overlapping planes intersected by a third plane". Overlapping planes are coincident (same plane), so the first two equations represent the same plane. Then the third plane intersects this plane (since it's "intersected by a third plane" – so the third plane is not parallel and not coincident with the first two). The intersection of two planes (one being the coincident pair, the other the third) is a line. A line has infinitely many points, so the system has infinitely many solutions.

Wait, let's re-express:

• Two overlapping planes: same plane (so equations \( A_1x + B_1y + C_1z = D_1 \) and \( k(A_1x + B_1y + C_1z) = kD_1 \) for some \( k

eq 0 \), so they are dependent, representing the same plane).

• Third plane: \( A_2x + B_2y + C_2z = D_2 \), not parallel to the first plane (since it intersects it).
• The intersection of the first plane (coincident pair) and the third plane is a line (since two non-parallel, non-coincident planes intersect in a line). A line has infinitely many points, so the system of equations (two same plane, one intersecting plane) has infinitely many solutions (all points on the line of intersection).

Step2: Evaluate options

• Two solutions: A line has more than two points, so no.
• One solution: A line has infinitely many, not one.
• No solutions: The planes intersect, so there are solutions.
• Infinite number of solutions: The intersection is a line, which has infinitely many points. So this is correct.

Answer:

infinite number of solutions