Maths · Scalar and vector products
The points , B(2,2,1) \) and \) determine a plane. The distance of the plane fro
The points \( A(-1,3,0), B(2,2,1) \) and \( C(1,1,3) \) determine a plane. The distance of the plane \( A, B, C \) from the point \( D(5,7,8) \) is
- A. \( \sqrt{66} \)
- B. \( \sqrt{71} \)
- C. \( \sqrt{73} \)
- D. \( \sqrt{76} \)
Step-by-step solution
Vectors AB = (3,-1,1) and AC = (2,-2,3). Normal vector n = AB × AC = (-1,-7,-4). Plane equation: x+7y+4z=20. Distance from D: |5+49+32-20|/√(1+49+16)=66/√66=√66.
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