TS EAMCET · Maths · Three Dimensional Geometry
If a plane is at a distance of units from the origin and the vector is its normal, then the equation of the plane in Cartesian form is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
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…
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