Home › JEE Main › Mathematics › Three Dimensional Geometry › Direction ratios of normal to plane 2x - y + 3z …
Direction ratios of normal to plane 2x - y + 3z + 7 = 0:
A(7, -7, 7)
B(2, -1, -3)
C(2, 1, 3)
D(2, -1, 3) (coefficients of x, y, z)
Answer & Solution
Correct answer: D. (2, -1, 3) (coefficients of x, y, z)
Coefficients of x, y, z directly give DR's of normal vector. Plane ax + by + cz + d = 0 has normal (a, b, c).
Related questions
The image of the point (1, 2, 3) in the plane x + y + z = 0 isThe foot of the perpendicular from origin to the plane 2x + 3y - 6z = 14 isThe line through the origin with direction ratios (1, 1, 1) meets the plane x + y + z = 6 A line has direction ratios (1, 1, 2) and meets the plane x - y + z = 0. The angle ϕ betweThe plane through the points (1,1,0), (1,0,1) and (0,1,1) has equationThe equation of a plane parallel to x + 2y - 3z = 5 and passing through (1, 1, 1) isThe shortest distance between r = (1,2,3) + λ(2,3,4) and r = (2,4,5) + μ(3,4,5) isThe line r = (2, -1, 4) + λ(3, 0, 2), expressed in symmetric Cartesian form, is