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