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TS EAMCET · Maths · Three Dimensional Geometry

If a plane is at a distance of 6 units from the origin and the vector 2i^+6j^-3k^ is its normal, then the equation of the plane in Cartesian form is

  1. A 2 x+3 y-6 z-35=0
  2. B 2 x+6 y-3 z-42=0
  3. C 2 x+6 y-3 z-35=0
  4. D 2 x-6 y+3 z-42=0
Verified Solution

Answer & Solution

Correct Answer

(B) 2 x+6 y-3 z-42=0

Step-by-step Solution

Detailed explanation

Given, Distance from origin to the plane d=6 units And Normal vector N→=2i^+6j^-3k^ ∴Unit normal vector n^=N→N→=2i^+6j^-3k^22+62+32=2i^+6j^-3k^7 Equation of plane in normal form is r→·n^=d r→·2i^+6j^-3k^7=6 Put…