The lines x = ay $$-$$ 1 = z $$-$$ 2 and x = 3y $$-$$ 2 = bz $$-$$ 2, (ab $$\ne$$ 0) are coplanar, if :
JEE · Math · previous-year question
- A.b = 1, a$$\in$$R $$-$$ {0}correct
- B.a = 1, b$$\in$$R $$-$$ {0}
- C.a = 2, b = 2
- D.a = 2, b = 3
Answer
A. b = 1, a$$\in$$R $$-$$ {0}
Explanation
Lines are $$x = ay - 1 = z - 2$$ $$\therefore$$ $${x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}$$ .... (i) and $$x = 3y - 2 = bz - 2$$ $$\therefore$$ $${x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}$$ .... (ii) $$\therefore$$ lines are co-planar $$\therefore$$ $$\left| {\begin{matrix} 0 & { - {1 \over a} + {2 \over 3}} & { - 2 + {2 \over b}} \\ 1 & {{1 \over a}} & 1 \\ 1 & {{1 \over 3}} & {{1 \over b}} \\ \end{matrix} } \right| = 0$$ $$\therefore$$ $$\left| {\begin{matrix} 0 & {{2 \over 3} - {1 \over a}} & {{2 \over b} - 2} \\ 0 & {{1 \over a} - {1 \over 3}} & {1 - {1 \over b}} \\ 1 & {{1 \over 3}} & {{1 \over b}} \\ \end{matrix} } \right| = 0$$ $$\therefore$$ $${1 \over a} - {1 \over {ab}} = 0$$ $$\Rightarrow$$ b = 1 and a $$\in$$ R $$-$$ {0}
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