Home › UP Board Class 12 › mathematics › Vectors & 3D Geometry › The Cartesian equation of a plane with intercept…
The Cartesian equation of a plane with intercepts 2, 3, 6 on the x, y, z axes is:
A$x/2 + y/3 + z/6 = 1$
B$2x + 3y + 6z = 1$
C$x + y + z = 11$
D$x/3 + y/2 + z/6 = 1$
Answer & Solution
Correct answer: A. $x/2 + y/3 + z/6 = 1$
Intercept form of plane: x/p + y/q + z/r = 1 where p, q, r are intercepts on x, y, z axes. So x/2 + y/3 + z/6 = 1.
Related questions
The angle θ between two PLANES with normals n1 and n2 satisfies:A point R divides the line joining P (position vector a) and Q (position vector b) INTERNAFor direction cosines l, m, n of any line in 3D:The vector triple product a × (b × c) simplifies via:The scalar triple product [a b c] geometrically represents:A UNIT VECTOR in the direction of v = 3i + 4j is:Two non-zero vectors a and b are PERPENDICULAR if and only if:If a = 2i + 3j + k and b = i − j + 2k, then a · b equals: