The proof of the judgment theorem that the line and plane are vertical: The judgment theorem: If a straight line is perpendicular to two intersecting straight lines in a plane, then the straight line is perpendicular to the plane. Pay attention to the keyword "intersect". If it is a parallel straight line, it cannot be judged that the line and surface are perpendicular. Extended information
Proof of the judgment theorem that the line and plane are perpendicular
Judgment theorem:
If a straight line intersects two straight lines in the plane are perpendicular, then this line is perpendicular to this plane.
Pay attention to the keyword "intersect". If it is a parallel straight line, it cannot be judged that the line is perpendicular. The reasons for the need for intersection are explained below.
Proof by contradiction:
Suppose a straight line l is perpendicular to two intersecting straight lines AB and CD on the surface S, then l⊥surface S
Suppose l is not vertical On the surface S, either l∥S, or it is oblique to S and the angle is not equal to 90.
When l∥S, then l cannot be perpendicular to both AB and CD. This is because when l⊥AB, any plane R passing through l intersects S at m, then it can be known from the property of parallel lines and planes that m∥l
∴m⊥AB
Also ∵l⊥CD
∴m⊥CD
∴AB∥CD, contradicting the known conditions.
When l is obliquely intersecting S, the intersection point is a straight line n⊥l in S, then n and l form a new plane T, and T and S are obliquely intersecting (if T⊥S, then n is the intersection line of two planes. From the perpendicular nature of the surfaces, we can know that l⊥S is inconsistent with l skew S).
∵l⊥AB
∴AB∥n
∵l⊥CD
∴CD∥n
∴AB∥CD, contradicting the known conditions.
In summary, l⊥S
Algebraic method:
As shown in the figure, the two intersecting straight lines a and b in l and α are both perpendicular. Verify: l ⊥α
Proof: A straight line parallel to a or b must be perpendicular to l, so the following discussion will focus on straight lines that are not parallel to a or b.
First translate a, b, l until they intersect at point O, draw any straight line g through O, take a point G different from O on g, and draw GB∥a intersecting b at B through G , through G, let GA∥b intersect a with A. Connect AB, let the intersection point of AB and OG be C
∵OA∥GB,OB∥GA
∴The quadrilateral OAGB is a parallelogram
∴C is in AB Point
According to the center line theorem,
Take a point D different from O on l, connect DA and DB, and use the center line theorem
Subtraction of the two equations can be Got
Also noticed that OD⊥OA, OD⊥OB
∴ Got
That is
∴OD⊥OC
From the `arbitrariness of g, it can be seen that l is perpendicular to any straight line in α
∴l⊥α
Vector method:
Suppose straight line l It is a straight line perpendicular to both straight lines a and b intersecting in α. Proof: l⊥α
Proof: Let the direction vectors of a, b, l be a, b, l
∵a and b intersect, that is, a and b are not straight lines
∴According to the basic theorem of plane vectors, any vector c within α can be written in the form of c= λa+ μb
< p> ∵l⊥a, l⊥b∴l·a=0, l·b=0
l·c=l·(λa+ μb)=λl·a+ μl·b=0=0
∴l⊥c
Suppose c is the direction vector of any straight line c in α, then l⊥c
< p>According to the arbitrariness of c, l is perpendicular to any straight line in α∴l⊥α
Property theorem that line and surface are vertical
Property theorem 1: If a line is perpendicular to a plane, then the line is perpendicular to all lines in the plane.
Property Theorem 2: Passing through a point in space, there is and is only one straight line perpendicular to the known plane.
Property Theorem 3: If among two parallel straight lines, one line is perpendicular to a plane, then the other line is also perpendicular to the plane.
Property Theorem 4: Two straight lines perpendicular to the same plane are parallel.
Corollary: If two straight lines in space are parallel to a third straight line, then the two straight lines are parallel. (This corollary means that the transitivity of parallel lines holds not only in plane geometry but also in space geometry.)