The equation of a plane which passes through (2, –3, 1) and is normal to the line joining…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
The equation of a plane which passes through (2, –3, 1) and is normal to the line joining the points (3, 4, –1) and
(2, –1, 5) is given by :
(2, –1, 5) is given by :
- x + 5y – 6z + 19 = 0
- x – 5y + 6z – 19 = 0
- x + 5y + 6z + 19 = 0
- x – 5y – 6z – 19 = 0
Answer
(A) x + 5y – 6z + 19 = 0
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