Given l, and the direction vector of l is (2, m,1), the normal vector of plane α is (1,1 2,2), Then m=______.

Given l, and the direction vector of l is (2, m,1), the normal vector of plane α is (1,1 2,2), Then m=______.

L, and the direction vector of l is (2, m,1), and the normal vector of plane α is (1,1
2,2),
The vector is (2, m,1) and the normal vector of the plane α(1,1
2,2) Vertical
Then (2, m,1)(1,1
2,2)=2+1
2M+2=0
Solution m=-8
Therefore, the answer is:-8

Given that the direction vector of the line l is v =(1,-1,-2), the normal vector of the plane αμ=(-2,-2,1), then the angle between l and α is

V.μ=1*(-2)+(-1)*(-2)+(-2)*1.=-2+2-2.=-2.|v|=√6,|μ|=3.cos=v.μ/|v'=(-2)/(3√6).=-√6/9.["-" The angle between the positive direction vector of straight line l and the positive direction vector of normal vector is obtuse, but the angle between the two straight lines should be less than or equal to 90°, in fact The two...

Known vector M, N is the direction vector and the normal vector of the line l and the plane α, respectively, if cos < M, N >=-1 2, Then the angle formed by l and α is ______. Known vector M, N is the direction vector and the normal vector of the line l and the plane α, respectively, if cos < M, N >=-1 2, The angle formed by l and α is ______.

Vector

M,

N is the direction vector and normal vector of the line l and the plane α, respectively, cos <

M,

N >=-1
2,
∴<

M,

N >=120°
The angle between l and α is π
6
Therefore, the answer is:π
6

Vector

M,

N is the direction vector and normal vector of line l and plane α, respectively, cos <

M,

N >=-1
2,
∴<

M,

N >=120°
The angle between l and α is π
6
Therefore, the answer is:π
6

If the direction vector of the line l is a =(1,0,2) the normal vector of the plane α is n (-2,0,4) If the direction vector of the line l is a =(1,0,2) the normal vector of the plane α is n (-2,0,4), then

L.

How to find the normal vector of a plane equation? Detailed explanation How to find the normal vector of a given plane equation? Ask for detailed explanation How to find the normal vector of a given plane equation? Detailed Description

Let the plane equation be Ax+By+Cz+D=0, then its normal vector is (A/√(A ²+B2+C2), B/√(A2+B2+C2), C/√(A2+B2+C2).

What is a plane equation? How to find its normal vector?

An equation in space such as Ax+By+Cz+D=0 determines a plane.
His normal vector is, the vector (A, B, C)