KCET · Maths · Three Dimensional Geometry
The angle between the pair of lines \(\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}\) and \(\frac{x+1}{1}=\frac{y-4}{4}=\frac{z-5}{2}\) is
- A \(\theta=\cos ^{-1}\left[\frac{27}{5}\right]\)
- B \(\theta=\cos ^{-1}\left[\frac{8 \sqrt{3}}{15}\right]\)
- C \(\theta=\cos ^{-1}\left[\frac{19}{21}\right]\)
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
Given, equation of lines are
\[
\begin{aligned}
\frac{x+3}{3} & =\frac{y-1}{5}=\frac{z+3}{4} \\
\text { and } \quad \frac{x+1}{1} & =\frac{y-4}{4}=\frac{z-5}{2}
\end{aligned}
\]
Direction ratios of line (i) : \( < 3,5,4>\)
Direction ratios of line (ii) : \( < 1,4,2>\)
We know angle between pair of lines is given by
\[
\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|
\]
where, \( < a_1, b_1, c_1>\) and \( < a_2, b_2, c_2>\) are direction ratios of respectively lines.
\[
\begin{aligned}
& \cos \theta=\left|\frac{3 \times 1+5 \times 4+4 \times 2}{\sqrt{3^2+5^2+4^4} \sqrt{1^2+4^2+2^2}}\right| \\
& \quad=\left|\frac{3+20+8}{\sqrt{9+25+16} \sqrt{1+16+4}}\right|=\left|\frac{31}{5 \sqrt{2} \sqrt{21}}\right|=\frac{31}{5 \sqrt{42}} \\
& \theta=\cos ^{-1}\left(\frac{31}{5 \sqrt{42}}\right)
\end{aligned}
\]
\[
\begin{aligned}
\frac{x+3}{3} & =\frac{y-1}{5}=\frac{z+3}{4} \\
\text { and } \quad \frac{x+1}{1} & =\frac{y-4}{4}=\frac{z-5}{2}
\end{aligned}
\]
Direction ratios of line (i) : \( < 3,5,4>\)
Direction ratios of line (ii) : \( < 1,4,2>\)
We know angle between pair of lines is given by
\[
\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right|
\]
where, \( < a_1, b_1, c_1>\) and \( < a_2, b_2, c_2>\) are direction ratios of respectively lines.
\[
\begin{aligned}
& \cos \theta=\left|\frac{3 \times 1+5 \times 4+4 \times 2}{\sqrt{3^2+5^2+4^4} \sqrt{1^2+4^2+2^2}}\right| \\
& \quad=\left|\frac{3+20+8}{\sqrt{9+25+16} \sqrt{1+16+4}}\right|=\left|\frac{31}{5 \sqrt{2} \sqrt{21}}\right|=\frac{31}{5 \sqrt{42}} \\
& \theta=\cos ^{-1}\left(\frac{31}{5 \sqrt{42}}\right)
\end{aligned}
\]
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