The vector proves that the plane is vertical There are two planes α, β. Now there is a straight line a in α which is perpendicular to the straight line B (B is in β) and a is perpendicular to the plane β. It is known that α intersects β and the straight line L, and it is proved that α is perpendicular to β Now I know that we only need to prove that the normal vector in two planes is perpendicular, but how to prove that B is the normal vector in α plane? That is, how to prove that B ⊥ l? (don't use any dihedral angle. If we use dihedral angle, we don't need to use so much effort to prove the perpendicularity directly.) You're using the method of spatial graphics My question clearly states that we should use space vector to prove it.

The vector proves that the plane is vertical There are two planes α, β. Now there is a straight line a in α which is perpendicular to the straight line B (B is in β) and a is perpendicular to the plane β. It is known that α intersects β and the straight line L, and it is proved that α is perpendicular to β Now I know that we only need to prove that the normal vector in two planes is perpendicular, but how to prove that B is the normal vector in α plane? That is, how to prove that B ⊥ l? (don't use any dihedral angle. If we use dihedral angle, we don't need to use so much effort to prove the perpendicularity directly.) You're using the method of spatial graphics My question clearly states that we should use space vector to prove it.

Note: B ⊥ L is not necessarily true. As shown in the figure, if a line corresponds to AB, then B corresponds to BF or be, and l corresponds to CD. It can be seen that B ⊥ L is not necessarily true. Proving that α is perpendicular to β is actually the theorem "if a plane passes through a vertical line of another plane, then the two planes are mutually

To find the scalar product of space vector, the formula of module length and the judging formula of perpendicularity
RT is used to solve solid geometry problems. Let a = (x1, Y1, z1) and B = (X2, Y2, Z2) be vectors, and find their coordinate operation formulas

Vector a = (x1, Y1, z1), B = (X2, Y2, Z2), and,
Quantity product AB = x1x2 + y1y2 + z1z2
The formula of module length of vector a = √ (x1 & # 178; + Y1 & # 178; + Z1 & # 178;)
a. B is equivalent to ab = 0, that is, x1x2 + y1y2 + z1z2 = 0

The condition of space vector perpendicularity
How can (A1, B1, C1) and (A2, B2, C2) be perpendicular?

a1a2+b1b2+c1c2=0

Two proofs of vector algorithm (λ a) · B = λ (a · b) = a · (λ b)
Two proofs of vector algorithm
(λa)·b=λ(a·b)=a·(λb)
a·b=a·ca⊥(b-c)

Let a = (x, y), B (m, n)
(λa)b(λx,λy)(m,n)(λxm+λyn)λ(xm+ym)λ(ab)
(xλm+yλn)(x,y)(λm,λn)a(λb)
ab=acab-ac=0a(b-c)=0a⊥(b-c)

Vector proving problem (λ a) · B = λ (a · b) = a · (λ b)
Two proofs of vector algorithm
(λa)·b=λ(a·b)=a·(λb)
a·b=a·ca⊥(b-c)

At this time, when the (λ a) · B \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= a · (λ b) when λ < 0 (λ a) · B = | λ a | B | cos = | λ | a | B | cos (π -...)

When the vector AB satisfies what condition | A-B | = | a | + | B|

When the direction of a and B is just opposite, and the value is equal, it meets the requirements of the topic
Because a = - B, La BL = l2al = LAL + LBL

When vectors a and B satisfy what conditions, | a + B | = | A-B |?

Ha ha, so easy when both sides square off a square and b square at the same time, that is, when A.B = 0

What conditions can vector b be linearly represented by vector group A

This condition is equivalent to AX = B. If you want to understand a problem, matrix A is actually composed of column vectors,
It is multiplied by an X vector to get another vector. That is to say, this vector can be expressed linearly by vector group A (some textbooks also call it linear)

Finding f (π / 4) with known function f (x) = sin2x-2sinx square
2: , find the minimum positive period of the function and the simple increasing interval 3, and find the set of X when finding the maximum value of the function and the maximum value of the area

f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1
f(π/4)=√2sin(3π/4)-1=0
The minimum positive period of a function T = π
Simple increasing interval 2K π - π / 2

Let x belong to (0, PI / 2), then y = (2sinx square + 1) / sin2x is the minimum

Y = (2sinx square + 1) / sin2x
=(3sin^2 x+cos^2 x)/2sinxcosx
=(3/2)tanx+(1/2)cotx
>=2 * radical (3 / 4)
=Root 3
If and only if TaNx = (radical 3) / 2, the equal sign holds
The minimum value of the function y = (2sinx square + 1) / sin2x is the root sign 3