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In higher mathematics, how to find the normal vector of a plane equation?
In the space coordinate system, the plane equation can be used as the cubic equation
The general equation AX + by + CZ + D = 0
So its normal vector is (a, B, c)
You can see from the point formula of the plane:
n·MM'=0,n=(A,B,C),MM'=(x-x0,y-y0,z-z0)
A(x-x0)+B(y-y0)+C(z-z0)=0
Finding the plane of three points can orient the product as normal
The graph of any cubic equation is always a plane, where the coefficients of X, y and Z are the coordinates of a normal vector of the plane
RELATED INFORMATIONS
- 1. What does the direction cosine of the normal vector of a plane equation mean?
- 2. Plane equation defined by normal vector and constant of plane The plane equation defined by normal vector N and constant D is x · n = D
- 3. If the intersection of the image of the first-order function y = KX + B and the inverse scale function y = KX is (2,3), then K=______ ，b=______ ．
- 4. XY + 2 = K + 8. When k is constant, X is proportional to y?
- 5. If xy = K + 1 / 4, is x proportional to y when k is constant? Such as the title
- 6. Xy = K + 3, when k is constant, X and y_ C = 3.14d, when C is constant, 3.14 and D_ Proportion
- 7. Given xy = K + 13, X and Y () A. Negative proportion B. positive proportion C. out of proportion D. not sure
- 8. Given that the set P = {(x, y) │ y = k, X ∈ R, K is constant}, q = {(x, y) │ y = (a ^ x) 1, a > 0, a ≠ 1, X ∈ r}, P ∩ q = has only one subset, then the value range of K?
- 9. Given the set P = {(x, y) | y = k}, q = {(x, y) | y = ax + 1}, and P ∩ q = ∞, then the value range of K is () A. （-∞，1）B. （-∞，1]C. （1，+∞）D. （-∞，+∞）
- 10. The solution of the equation 3x + YK + 2,2x-y = 5 is x > 0, y < 0, and the value range of K is obtained 3x+y=k+2
- 11. Given that point m (1,2, - 1) is in plane a and a normal vector of plane a is (- 1,1,3), the equation of plane a is obtained
- 12. [kneeling] given the normal vector of a plane and the two points it passes through, how to find the equation of the plane? Of course, it's a point method. In fact, you only need to know the point But my question is, if we bring in another known point, the result of the plane equation will be different. Isn't that contradictory?
- 13. Given the plane intercept equation x / A + Y / B + Z / C = 1, how to get the normal vector of the plane directly through the equation without converting it into a general equation
- 14. Given that the plane passes through point m (1, - 1,1) n (0,1, - 3) and is parallel to vector a (1,1,1), the plane equation is solved
- 15. If a plane passes through a point (1,0,1) and is parallel to the vectors a = {2,1,1} and B = {1, - 1,0}, the plane equation is solved?
- 16. A plane passing through point (1,0, - 1) and parallel to vectors a = (2,1,1) and B = (1, - 1,0), try to find the plane equation
- 17. Given the parallel vector of a plane and two points passing through the plane, how to solve the plane equation
- 18. In the geometric sense of vector, it is the distance (i.e. length) from the point of the vector on the coordinate plane to the origin / / must be to the origin (0,0)? Must go to the origin (0,0)? From point (4,5) to point (2,2), no?
- 19. It is known that I is an imaginary unit. In the complex plane, Z1 = 1 + I, Z2 = 2 + 3I correspond to points a and B, O is the origin, and vectors OP, OA and ob satisfy OP = OA + xob, If point P is in the fourth image term Then the value range of X is?
- 20. Points a and B (5,0) satisfy the vector OA * vector ob = vector OA * vector Ba, │ vector OA + vector ob = √ 185, and then calculate the a coordinate