KCET · Maths · Three Dimensional Geometry
The angle between the lines \( 2 x=3 y=-z \) and \( 6 x=-y=-4 z \) is
- A \( 0^{\circ} \)
- B \( 45^{\circ} \)
- C \( 90^{\circ} \)
- D \( 30^{\circ} \)
Answer & Solution
Correct Answer
(C) \( 90^{\circ} \)
Step-by-step Solution
Detailed explanation
Given lines, \(2 x=3 y=-z\)
\(\Rightarrow \frac{x}{1 / 2}=\frac{y}{1 / 3}=\frac{z}{-1} \rightarrow(1)\)
Direction ratios are \((1 / 2,1 / 3,-1)\) and \(6 x=-y=-4 z\)
\(\Rightarrow \frac{x}{1 / 6}=\frac{y}{-1}=\frac{z}{-1 / 4} \rightarrow(2)\)
Direction ratios are \((1 / 6,-1,-1 / 4)\)
Now, \(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\)
\(\frac{1}{12}-\frac{1}{3}+\frac{1}{4}=\frac{1-4+3}{12}=0\)
Therefore, lines are perpendicular, that is, angle between two lines is \(90^{\circ}\).
\(\Rightarrow \frac{x}{1 / 2}=\frac{y}{1 / 3}=\frac{z}{-1} \rightarrow(1)\)
Direction ratios are \((1 / 2,1 / 3,-1)\) and \(6 x=-y=-4 z\)
\(\Rightarrow \frac{x}{1 / 6}=\frac{y}{-1}=\frac{z}{-1 / 4} \rightarrow(2)\)
Direction ratios are \((1 / 6,-1,-1 / 4)\)
Now, \(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\)
\(\frac{1}{12}-\frac{1}{3}+\frac{1}{4}=\frac{1-4+3}{12}=0\)
Therefore, lines are perpendicular, that is, angle between two lines is \(90^{\circ}\).
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