MHT CET · Maths · Three Dimensional Geometry
The angle between two lines \(\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}\) and \(\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}\) is
- A \(\cos ^{-1}\left(\frac{4}{9}\right)\)
- B \(\cos ^{-1}\left(\frac{2}{9}\right)\)
- C \(\cos ^{-1}\left(\frac{1}{9}\right)\)
- D \(\cos ^{-1}\left(\frac{5}{9}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{4}{9}\right)\)
Step-by-step Solution
Detailed explanation
The required angle
\(\begin{aligned} & \theta=\cos ^{-1}\left(\frac{2 \times 1+2 \times 2+(-1) \times 2}{\sqrt{2^2+2^2+1^2} \cdot \sqrt{1^2+2^2+2^2}}\right) \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{4}{3 \times 3}\right) \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{4}{9}\right)\end{aligned}\)
\(\begin{aligned} & \theta=\cos ^{-1}\left(\frac{2 \times 1+2 \times 2+(-1) \times 2}{\sqrt{2^2+2^2+1^2} \cdot \sqrt{1^2+2^2+2^2}}\right) \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{4}{3 \times 3}\right) \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{4}{9}\right)\end{aligned}\)
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