The equation of the plane in cartesian form, which is at a distance of 6/\sqrt{29} from the origin and its normal vector drawn from the origin being 2i^{\^}-3j^{\^}+4k^{\^} , is

Correct Answer is : 2x - 3y + 4z = 6

TS EAMCET 04 May 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 61 of 160

Marks: +1, -0

The equation of the plane in cartesian form, which is at a distance of

√29

from the origin and its normal vector drawn from the origin being 2

−3

, is

Go to Question:

Prev Question

Next Question