MHT CET · Maths · Three Dimensional Geometry
The angle between the lines \(\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}\) and \(\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}\) is
- A \(\sin ^{-1}\left(\frac{2}{3}\right)\)
- B \(\cos ^{-1}\left(\frac{2}{3}\right)\)
- C \(\cos ^{-1}\left(\frac{1}{3}\right)\)
- D \(\sin ^{-1}\left(\frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(B) \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\cos \theta=\left|\frac{4(2)+1(2)+8(1)}{\sqrt{4^{2}+1^{2}+8^{2}} \sqrt{2^{2}+2^{2}+1^{2}}}\right|=\left|\frac{8+1+8}{\sqrt{81} \cdot \sqrt{9}}\right|=\frac{18}{27}\)
\(\therefore \cos \theta=\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
\(\therefore \cos \theta=\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
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