A plane is at a distance of 4 units from the origin and its normal vector is 2i − 3j + 6k. Find its vector equation.
A plane is at a distance of 4 units from the origin and its normal vector is 2i − 3j + 6k. Find its vector equation.
- A. r · (2i − 3j + 6k) = 4
- B. r · (2i − 3j + 6k) = 28
- C. r · (2i − 3j + 6k) = 14
- D. r · (i − j + k) = 4
Answer: B) r · (2i − 3j + 6k) = 28
Explanation: Normal n = 2i − 3j + 6k, magnitude = 7. Unit normal n̂ = (2i − 3j + 6k)/7. Equation is r · n̂ = d, so r · (2i − 3j + 6k)/7 = 4, meaning r · (2i − 3j + 6k) = 28.
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