MAT188H1 Chap.5 Planes and Normals
Planes and Normals
The normal constrains displacement within the plane
A plane in ordinary space can be described using a point on it and a nonzero normal vector. The normal points perpendicular to the plane. If p is the position vector of a known point and x represents another point on the plane, then x - p is a displacement within the plane.
Its dot product with the normal must be zero.
This leads to the point and normal equation n·(x - p) = 0. The subtraction is essential because the normal is perpendicular to displacements within the plane, not generally to the position vector of every point on it.
Only when the plane passes through the origin can an origin anchor reduce the equation to n·x = 0.
Coefficients reveal the perpendicular direction
Writing the vectors in components expands the equation into n₁x + n₂y + n₃z = D, where D = n·p. The coefficients on the left form a normal vector. The constant on the right controls which of the parallel planes with that normal is selected.
Keeping those roles distinct helps prevent a point on the plane from being mistaken for a normal vector.
To construct the equation, first calculate the constant from the given point. Then substitute the point back into the resulting equation as a check. If the equality fails, revisit the dot product and signs before attempting a sketch.
A correctly copied normal with an incorrectly computed constant gives a different plane, even though its orientation looks reasonable.
Membership and direction use related tests
A proposed point belongs to the plane when its coordinates satisfy the full equation, including the constant. A proposed direction within the plane instead needs dot product zero with the normal.
These are different tests because a direction describes displacement while a point describes location. A vector perpendicular to the normal can be translated into the plane; its tip in standard position need not already lie on the plane.
For a candidate line p + td, substitution into the plane equation produces n·p + t(n·d) = D. This is simply the line description inserted into the membership condition.
It separates the anchor contribution from the direction contribution. If the direction contribution vanishes and the anchor already satisfies the plane equation, every parameter value stays in the plane.
Intercepts help explain a sketch
To locate an axis intercept, set the other coordinates to zero and solve the remaining equation.
When the relevant coefficient is nonzero, this gives the coordinate where the plane meets that axis. These intercepts provide concrete points for a spatial drawing. Always verify them in the original equation rather than trusting a sketch whose perspective can conceal a sign error.
A zero coefficient needs a separate reading.
Setting the other coordinates to zero may produce an impossible equality or an equality true for every value of the remaining coordinate. Consequently, a missing intercept formula does not imply that the plane is undefined.
It indicates that dividing by that coefficient would be invalid and that the reduced equation should be interpreted directly.
Equivalent equations preserve the entire equality
Multiplying every term of a plane equation by the same nonzero scalar preserves its solution set. The normal then changes to a parallel vector, while the constant changes by the same factor.
Scaling only the coefficients without scaling the constant generally moves the plane. Comparison of equations must therefore include both sides, not merely a check that their normals are parallel.
The reading also introduces the vector form p + su + tv, with real parameters s and t and nonzero, nonparallel directions u and v. These directions run within the plane, unlike a normal.
The nonparallel condition is essential: scalar multiples would allow movement along only one direction and would not generate a plane. The point supplies location while the directions supply the allowed displacement.
For this introductory treatment, distinguish what each representation makes easiest. The point and normal equation is convenient for constructing a plane and testing membership in space.
The parameter form shows how points can be generated through two directions. A complete answer should state the role of each vector, preserve every nonzero or nonparallel condition, and check the resulting set against the given point.
What this chapter covers
- 01
Point and normal equation
- 02
General equation, membership and intercepts
- 03
In-plane directions and vector form
Worked example · free
Constructing a plane and checking its location
- 2Use the displacement from the known point: n·((x, y, z) - (2, -1, 3)) = 0. In components this becomes 2(x - 2) - (y + 1) + 4(z - 3) = 0.
- 2Expand and collect constants: 2x - y + 4z = 4 + 1 + 12 = 17. Independently checking the anchor gives 2(2) - (-1) + 4(3) = 17.
- 2At Q the left side is 2(0) - 1 + 4(4) = 15, which differs from 17. Q is therefore outside the plane, despite having coordinates close to those of points that might appear nearby in a sketch.
Key terms
- Normal Vector
- A nonzero vector perpendicular to a plane. Its components form the coefficients in a general equation of the plane, while its direction differs from the directions used to travel within the plane.
- Plane Equation
- An equation whose solution points form a plane. In space, a nonzero normal and a point determine the form n·(x - p) = 0 and its equivalent component equation.
- Axis Intercept
- A point where a geometric set meets a coordinate axis. For a plane, it is found by setting the other coordinates to zero and interpreting the resulting equation without dividing by zero.
- In-plane Direction
- A displacement parallel to a plane, characterized by a zero dot product with its normal. The displacement can be translated into the plane without requiring its standard-position endpoint to belong there.
- Vector Plane Form
- A description generating plane points from an anchor plus real scalar multiples of two nonzero, nonparallel directions. The directions must provide different ways to move within the plane rather than duplicating one another.
Planes and Normals FAQ
Why does the plane formula subtract the known point?
The subtraction creates a displacement joining two points of the plane. Perpendicularity applies to that displacement. Using a position vector alone would silently anchor the condition at the origin and could shift the intended plane. Preserve the subtraction until you have expanded all coordinates and collected the constant correctly.
Do parallel normals guarantee equal planes?
They establish matching orientation, but the equations may select different locations. Rescale the entire equation, including its constant, when comparing descriptions. An additional membership check using a known point distinguishes coincident planes from different planes that merely have parallel normals. Coefficient ratios alone leave that location question unanswered.
How can a direction lie within a plane without its endpoint belonging?
A direction vector represents a displacement that may be drawn from different starting points. Its perpendicularity to the normal permits motion along the plane once its tail is placed on the plane. Membership of the endpoint in standard position is a separate condition involving the constant in the plane equation.
Exam move
Place a point test and a direction test beside each plane equation and explain why their right sides differ. Rebuild the equation from its normal and a known point, then sketch any available intercepts. When comparing equations, scale the constant with the coefficients and verify a concrete point before concluding that the planes coincide.
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