In the spatial rectangular coordinate system O-xyz, the normal vector of the plane OAB is A =(2,-2,1), if P (-1,3,2) is known, then the distance from P to the plane OAB is equal to () A.4 B.2 C.3 D.1

In the spatial rectangular coordinate system O-xyz, the normal vector of the plane OAB is A =(2,-2,1), if P (-1,3,2) is known, then the distance from P to the plane OAB is equal to () A.4 B.2 C.3 D.1

If the distance from point P to plane OAB is d, then d=|

OP•

A |
|

A|,
∵

A =(2,-2,1), P (-1,3,2),
D =|(−1,3,2)•(2,−2,1)|

4+4+1=2.
Therefore: B.

If the distance from point P to plane OAB is d, then d=|

OP•

A |
|

A|,
∵

A =(2,-2,1), P (-1,3,2),
D =|(−1,3,2)•(2,−2,1)|

4+4+1=2.
Selected from: B.

Given that a normal vector of the plane α is n=(1,1,1) the origin O=(0,0,0) is in plane A, then the distance from point P (4,5,3) to α I know the answer is four, three. Given that a normal vector of plane α is n=(1,1,1) and the origin O=(0,0,0) is in plane A, then the distance from point P (4,5,3) to α I know the answer is four, three.

The plane equation is x+y+z=0
Distance from point to plane formula: d=|4+5+3|/√(1+1+1)=12/√3=4√3

Method of finding known normal vector of plane Take x+2y+z=4 as an example.) I'm not good at math, even though I' m in college... Solution of known plane normal vector Take x+2y+z=4 as an example) I'm not good at math, even though I' m in college... Solution of known plane normal vector Take x+2y+z=4 for example,) I'm not good at math, even though I' m in college...

Ax+By+Cz+D=0+ D=0, the ternary primary equation is a general equation in a plane. The normal vector of a plane equation is the coefficient combination vector of x, y, z in the ternary primary equation, i.e. the vector n ={ A, B, C} is the normal vector of Ax + By + Cz + D=0. It can also be written as: normal vector n = A vector i + B vector j + C vector k, vector i, vector I.

If the angle between non-zero vector a and b is 120 degrees, a·a=-a·b, then |a|/|b|=? If the angle between nonzero vector a and b is 120 degrees, a·a=-a·b, then |a|/|b|=?

0.5
A·a=-a·b
|A a|=-|a b|*cos120 degrees=0.5*|a b|
|A |/|b |=0.5

If the nonzero vectors a, b satisfy |a|=2|b| and b⊥(a+b), then the angle between vectors a and b =?

B⊥(a+b)
Then b*(a+b)
=Ab+b^2
=Ab|^2=0
Ab=-|^b |^2
Cos (a, b)=ab/|a b||=-||b|^2/(2|b|)*|b||=-1/2
So it's 120 degrees.

B⊥(a+b)
Then b*(a+b)
=Ab+b^2
=Ab 2=0
Ab=-|^b |^2
Cos (a, b)=ab/|a b||=-||b|^2/(2|b|)*|b||=-1/2
So it's 120 degrees.

If vector a=(2,0) b is a non-zero vector, if the angle between vector a+b, b-a and positive direction of x is π/6 and 2π/3 respectively, then b=

If the vector a+b, b-a and the positive direction of the x-axis are π/6 and 2π/3, respectively, then the vector a+b and the vector b-a are perpendicular to each other. Let b=(x, y) a+b=(2+x, y) b-a=(x-2, y)(a+b)*(b-a)=x^2-4+y^2=0x^2+y^2=4 The vector a+b and the vector b-a are perpendicular to each other. Therefore, the parallelogram with a, b as the adjacent sides is rhombohedral.