Question

How many points are possible in the 3D plane that satisfy the given three statements?

• x = 2
• xi + yj + zk = 5
• y = mz + c, where m and c are constants

• A. 0
• B. 1
• C. 2
• D. 3
• E. None of the above

Answer 1

Answer: (E)

None of the above

Answer 2

We are given three conditions in ℝ³:

1. $x=2$ (a plane parallel to the $yz$–plane).
2. $x+y+z=5$ (interpreting $xi+yj+zk=5$ as the plane $x+y+z=5$).
3. $y=mz+c$ (a plane relating $y$ and $z$).

Step 1: Since $x=2$, substitute into $x+y+z=5$:

$$2+y+z=5 \quad\Rightarrow\quad y+z=3 \quad\Rightarrow\quad y=3-z.$$

Step 2: Equate the two expressions for $y$:

$$3-z= m\,z+c.$$

Rearrange to obtain:

$$(m+1)z = 3-c.$$

Cases:

• If $m \neq -1$:
  Then $z=\frac{3-c}{m+1}$ is unique. Hence, one point is obtained.

• If $m=-1$:

  • If additionally $c=3$ then the equation becomes an identity (i.e. $3-z=3-z$) and any $z$ satisfies it; thus, the intersection is the entire line of points in the plane $x=2$ satisfying $y+z=3$ (infinitely many solutions).
  • If $c\neq 3$ then the equation becomes inconsistent, yielding no solution.

Thus, the number of intersection points can be:

• One point,
• No point,
• or Infinitely many points.

Since the answer is not uniquely one of these fixed numbers but depends on the values of $m$ and $c$ (with the possibility of infinitely many points not listed), the correct answer is: None of the above