MHT CET · Maths · Three Dimensional Geometry
The angle between the lines, whose direction cosines \(l, \mathrm{~m}, \mathrm{n}\) satisfy the equations \(l+\mathrm{m}+\mathrm{n}=0\) and \(2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0\), is
- A \(60^{\circ}\)
- B \(180^{\circ}\)
- C \(90^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(B) \(180^{\circ}\)
Step-by-step Solution
Detailed explanation
Substituting \(\mathrm{n}=-l-\mathrm{m}\) in
\(
\begin{aligned}
& 2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0, \text { we get } \\
& 2 l^2+2 \mathrm{~m}^2-(-l-\mathrm{m})^2=0 \\
& \Rightarrow l^2+\mathrm{m}^2-2 l \mathrm{~m}=0 \\
& \Rightarrow(l-\mathrm{m})^2=0 \\
& \Rightarrow l=\mathrm{m}
\end{aligned}
\)
If \(l=\mathrm{m}\), then \(\mathrm{n}=-2 \mathrm{~m}\)
\(
\Rightarrow \frac{l}{1}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{-2}
\)
The direction ratios of both the lines are same.
\(
\begin{array}{ll}
\therefore \quad & \cos \theta= \pm 1 \\
& \Rightarrow \theta=0^{\circ} \text { or } 180^{\circ}
\end{array}
\)
\(
\begin{aligned}
& 2 l^2+2 \mathrm{~m}^2-\mathrm{n}^2=0, \text { we get } \\
& 2 l^2+2 \mathrm{~m}^2-(-l-\mathrm{m})^2=0 \\
& \Rightarrow l^2+\mathrm{m}^2-2 l \mathrm{~m}=0 \\
& \Rightarrow(l-\mathrm{m})^2=0 \\
& \Rightarrow l=\mathrm{m}
\end{aligned}
\)
If \(l=\mathrm{m}\), then \(\mathrm{n}=-2 \mathrm{~m}\)
\(
\Rightarrow \frac{l}{1}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{-2}
\)
The direction ratios of both the lines are same.
\(
\begin{array}{ll}
\therefore \quad & \cos \theta= \pm 1 \\
& \Rightarrow \theta=0^{\circ} \text { or } 180^{\circ}
\end{array}
\)
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