What is the equation of the plane passing through (- 1 3 2 and perpendicular to the planes x 2y 2z 5 and 3x 3y 2z 8?
The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is. 2x -4y +3z-8 =0. 2x-4y-3z+8 =0. 2x+4y+3z+8 =0.
What is the equation of a plane in 3D?
We also know it as the vector equation of a plane. In addition, the general equation of a plane in 3D space is A ∙ 0 + B ∙ 0 + C ∙ 0 + D = 0 => D = 0.
How to find the equation of a line passing through a point?
Vector equation of a line passing through a point and parallel to a vector is r = a+λb where λ ∈ R The equation of the plane containing the given point is A(x−1)+B(y−1)+C (z+1)= 0 ——– (1) Applying the condition of perpendicularity to the plane (1) with the given planes x+2y+3z−7= 0 and 2x−3y+4z = 0
Is the normal to the required plane perpendicular to the normals?
Since the required plane is perpendicular to the planes x + 2 y + 2 z = 5 and 3 x + 3 y + 2 z = 8, its normal would be perpendicular to the normals to the planes x + 2 y + 2 z = 5 and 3 x + 3 y + 2 z = 8. ⇒ The normal to the required plane is perpendicular to i ^ + 2 j ^ + 2 k ^ and 3 i ^ + 3 j ^ + 2 k ^.
Which plane Fullfill y=0 for each of its points?
That means we need (A,B,C)= (0,1,0), then the equation of the plane we are looking for must be something similar to y+D=0, we only have to determin If we are talking three dimensional geometry, then the xz plane is the plane that fullfills y=0 for each of its points.
What is the orthogonal plane to y=0?
If we are talking three dimensional geometry, then the xz plane is the plane that fullfills y=0 for each of its points. Notice that the vector (0,1,0) is orthogonal to the plane y=0. hence a parallel plane to y=0 must be orthogonal to this very same vector.