Questions: Equations of Lines and Planes in 3D

Question 1

Consider two planes: 2x + 3y − z = 5 and 4x + 6y − 2z = 11. What is their geometric relationship?

• A. They intersect in a line, since two planes generically intersect
• B. They are the same plane
• C. They are parallel but distinct
• D. They are perpendicular, since one equation is a multiple of the other

Answer: (C)

The normal vector of the first plane is ⟨2, 3, −1⟩; the normal of the second is ⟨4, 6, −2⟩ = 2⟨2, 3, −1⟩ — they are proportional, so the planes are parallel. But 5 ≠ 2·(11/2) when checked: the second equation 4x + 6y − 2z = 11 is not simply twice the first (which would give 2·5 = 10 ≠ 11), so the planes are distinct. Parallel normals mean parallel planes; the different constants confirm they don't coincide. The tempting wrong answer is 'they intersect' — two generic planes do intersect, but only when their normals are not proportional.

Question 2

What is the direction vector of the line of intersection of planes x + 2y + 3z = 4 and 2x − y + z = 5?

• A. ⟨1, 2, 3⟩ + ⟨2, −1, 1⟩ = ⟨3, 1, 4⟩ (sum of normal vectors)
• B. ⟨1, 2, 3⟩ · ⟨2, −1, 1⟩ (dot product of normal vectors)
• C. ⟨1, 2, 3⟩ × ⟨2, −1, 1⟩ = ⟨5, 5, −5⟩ (cross product of normal vectors)
• D. Any vector parallel to both planes, found by inspection

Answer: (C)

The line of intersection lies in both planes simultaneously, so it must be perpendicular to both normal vectors. The cross product of two vectors produces a vector perpendicular to both — so **n₁** × **n₂** gives the direction of the intersection line. In this case: ⟨1, 2, 3⟩ × ⟨2, −1, 1⟩ = ⟨(2)(1)−(3)(−1), (3)(2)−(1)(1), (1)(−1)−(2)(2)⟩ = ⟨5, 5, −5⟩. The dot product gives a scalar, not a vector; the sum of normals has no geometric meaning for this purpose.

Question 3

In the plane equation ax + by + cz = d, the coefficients a, b, c are the components of a normal vector to the plane.

• T. True
• F. False

Answer: True

This follows directly from how the equation is derived. Given normal **n** = ⟨a, b, c⟩ and a point (x₀, y₀, z₀), any other point (x, y, z) in the plane satisfies **n** · ((x, y, z) − (x₀, y₀, z₀)) = 0, which expands to ax + by + cz = ax₀ + by₀ + cz₀ = d. The normal vector components appear directly as the coefficients — reading off the normal from a plane equation is immediate.

Question 4

In 3D, a line can be fully described by specifying a slope and a point on the line, just as in 2D.

• T. True
• F. False

Answer: False

Slope is a single number describing steepness relative to a reference direction, and it only works in 2D where there is one relevant reference axis. In 3D, a line can tilt in two independent directions (toward x and toward y), so a single scalar slope cannot capture the full direction. Instead, a 3D line requires a direction vector ⟨a, b, c⟩ — three components — and a point. The parametric form r(t) = r₀ + t**d** replaces the slope-intercept form entirely in 3D.

Question 5

Why does describing a plane in 3D require a normal vector rather than a direction vector, and how do you find the normal vector when you know two vectors lying in the plane?

Model answer: A plane has infinitely many directions within it, so no single direction vector characterizes it uniquely. What uniquely specifies a plane's orientation is a vector perpendicular to all directions in it — the normal vector. If two vectors **u** and **v** lie in the plane, their cross product **n** = **u** × **v** is perpendicular to both and therefore to the entire plane. The normal vector then determines the plane equation directly: its components become the coefficients a, b, c.

The contrast with lines is instructive: a line has a unique direction (up to scaling), so a direction vector captures it. A plane has a unique *orientation* but not a unique direction — the perpendicular to that orientation is what captures it. The cross product is the natural tool because it is defined as 'perpendicular to both inputs,' which is exactly what finding a normal requires.