MHT CET · Maths · Three Dimensional Geometry
The acute angle between the lines \(x=-y, z=0\) and \(x=0, z=0\) is
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{18}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Given lines are
\(\frac{x-0}{1}=\frac{y-0}{-1}=\frac{z-0}{0} \text { and } \frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0}\)
\(\text { Now, the angle }=\cos ^{-1}\left(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}\right)\)
\(=\cos ^{-1}\left(\frac{1 \times 0+(-1) \times 1+0 \times 0}{\sqrt{1^2+(-1)^2+0^2} \cdot \sqrt{0^2+1^2+0^2}}\right)\)
\( =\cos ^{-1}\left|\frac{-1}{\sqrt{2}}\right|[\text { for acute angle }]\)
\(=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)
\(\frac{x-0}{1}=\frac{y-0}{-1}=\frac{z-0}{0} \text { and } \frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0}\)
\(\text { Now, the angle }=\cos ^{-1}\left(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b_2^2+c_2^2}}\right)\)
\(=\cos ^{-1}\left(\frac{1 \times 0+(-1) \times 1+0 \times 0}{\sqrt{1^2+(-1)^2+0^2} \cdot \sqrt{0^2+1^2+0^2}}\right)\)
\( =\cos ^{-1}\left|\frac{-1}{\sqrt{2}}\right|[\text { for acute angle }]\)
\(=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)
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