JEE Mains · Maths · STD 12 - 7.2 definite integral
Let a line \(L\) be perpendicular to both the lines \(L_1: \dfrac{x+1}{3} = \dfrac{y+3}{5} = \dfrac{z+5}{7}\) and \(L_2: \dfrac{x-2}{1} = \dfrac{y-4}{4} = \dfrac{z-6}{7}\). If \(\theta\) is the acute angle between the lines \(L\) and \(L_3: \dfrac{x - \dfrac{8}{7}}{2} = \dfrac{y - \dfrac{4}{7}}{1} = \dfrac{z}{2}\), then \(\tan\theta\) is equal to:
- A \(\dfrac{3}{2}\sqrt{2}\)
- B \(\dfrac{5}{2}\sqrt{2}\)
- C \(\dfrac{5}{3}\sqrt{2}\)
- D \(\dfrac{4}{3}\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{5}{2}\sqrt{2}\)
Step-by-step Solution
Detailed explanation
The direction vector of line \(L_1\) is \(\vec{d_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}\). The direction vector of line \(L_2\) is \(\vec{d_2} = \hat{i} + 4\hat{j} + 7\hat{k}\). Since line \(L\) is perpendicular to both \(L_1\) and \(L_2\), its direction vector \(\vec{d}\) is given…
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