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JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(ABC\) be a triangle with vertices at points \(A (2, 3, 5), B (-1, 3, 2)\) and \(C\left( {\lambda ,5,\mu } \right)\) in three dimensional space. If the median through \(A\) is equally inclined with the axes, then \(\left( {\lambda ,\mu } \right)\) is equal to
- A \((10, 7)\)
- B \((7, 5)\)
- C \((7 , 10)\)
- D \((5, 7)\)
Answer & Solution
Correct Answer
(C) \((7 , 10)\)
Step-by-step Solution
Detailed explanation
Since \(AD\) is the median \(\therefore D=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)\) Now, \(dR's\) of \(AD\) is \(a=\left(\frac{\lambda-1}{2}-2\right)=\frac{\lambda-5}{2}\) \(b=4-3=1, c=\frac{\mu+2}{2}-5=\frac{\mu-8}{2}\) Also, \(a, b, c\) are \(dR's\)…
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