The normal vector to the plane 3x − 4y + 5z = 10 is:
The normal vector to the plane 3x − 4y + 5z = 10 is:
- A. 10i − 4j + 5k
- B. 3i − 4j + 5k
- C. 3i + 4j + 5k
- D. −3i + 4j − 5k
Answer: B) 3i − 4j + 5k
Explanation: In the Cartesian equation Ax + By + Cz = D, the vector n = Ai + Bj + Ck is normal to the plane. Hence, 3i − 4j + 5k.
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