JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(A\) be the point of intersection of the lines \(L_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}\) and \(L_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}\). Let \(B\) and \(C\) be the point on the lines \(L_1\) and \(L_2\) respectively such that \(\mathrm{AB}=\mathrm{AC}=\sqrt{15}\). Then the square of the area of the triangle ABC is :
- A \(54\)
- B \(63\)
- C \(57\)
- D \(60\)
Answer & Solution
Correct Answer
(A) \(54\)
Step-by-step Solution
Detailed explanation
Angle between both lines…
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