Home › JEE Main › Mathematics › Three Dimensional Geometry › Find equation of plane through (1, 0, 0), (0, 2,…
Find equation of plane through (1, 0, 0), (0, 2, 0), (0, 0, 3):
Ax - y - z = 0
BIntercept form: x/1 + y/2 + z/3 = 1
Cx + y + z = 1
Dx + y + z = 6
Answer & Solution
Correct answer: B. Intercept form: x/1 + y/2 + z/3 = 1
Intercept form: x/a + y/b + z/c = 1 with intercepts a = 1, b = 2, c = 3 (x-, y-, z-axis intercepts). Equivalent: 6x + 3y + 2z = 6.
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