JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the lines \(x\,=\,ay\,+\,b,\,\,z\,=\,cy\,+\,d\) and \(x\, = \,a\,'z + \,b\,',\,\,y = \,c\,'z\, + \,d\,'\) are perpendicular, then
- A \(cc\,'\, + a + a\,' = 0\)
- B \(aa\,'\, + c + c\,' = 0\)
- C \(ab\,'\, + bc\,' + 1 = 0\)
- D \(bb\,'\, + cc\,' + 1 = 0\)
Answer & Solution
Correct Answer
(B) \(aa\,'\, + c + c\,' = 0\)
Step-by-step Solution
Detailed explanation
Line \(x=a y+b, z=c y+d\) \(\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}\) Line \(x=a^{\prime} z+b^{\prime}, y=c^{\prime} z+d^{\prime}\) \(\Rightarrow \frac{x-b^{\prime}}{a^{\prime}}=\frac{y-d^{\prime}}{c^{\prime}}=\frac{z}{1}\) Given both the lines are perpendicular…
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