Question

If the two lines 
\(l1:\frac{(x−2)}{3}=\frac{(y+1)}{−2},z=2 \)
and
\( l2:\frac{(x−1)}{1}=\frac{(2y+3)}{α}=\frac{(z+5)}{2} \)
are perpendicular, then an angle between the lines l₂ and 
\(l3:\frac{(1−x)}{3}=\frac{(2y−1)}{−4}=\frac{z}{4} \)
is

• A. \(cos^{-1}(\frac{29}{4})\)
• B. \(sec^{-1}(\frac{29}{4})\)
• C. \(cos^{-1}(\frac{2}{29})\)
• D. \(cos^{-1}(\frac{2}{\sqrt29})\)

Answer 1

The Correct Option is B

The correct answer is (B) : \(sec^{-1}(\frac{29}{4})\)
∵ l₁ and l₂ are perpendicular, so
 \(3×1+(−2)(\frac{α}{2})+0×2=0\)
⇒ α = 3
Now angle between l₂ and l₃,
\(cos⁡θ=\frac{1(−3)+\frac{α}{2}(−2)+2(4)}{(\sqrt{1+\frac{α^2}{4}}+4\sqrt{9+4+16}}\)
\(⇒cos⁡θ=\frac{\frac{2}{29}}{2}\)
\(⇒θ=cos^{−1}⁡(\frac{4}{29})=sec−1⁡(\frac{29}{4}) \)