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Question

# Column I Column IIa. Lines x−1−2=y+23=z−1 and →r=(3^i−^j+^k)+t(^i+^j+^k) are p. intersectingb. Lines x+51=y−37=z+33 and x−y+2z−4=0=2x+y−3z+5=0 are q. perpendicularc. Lines (x=t−3,y=−2t+1,z=−3t−2) and →r=(t+1)^i+(2t+3)^j+(−t−9)^k are r. paralleld. Lines →r=(^i+3^j−^k)+t(2^i−^j−^k) and →r=(−^i−2^j+5^k)+s(^i−2^j+34^k) are s. skew Then which of the following is correct ?

A
aq,s; br; cp,q; dp
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B
aq; br; cp,q; dp,q
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C
as; br,s; cp; dp,q
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D
aq,s; br,s; cp; dp
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Solution

## The correct option is A a→q,s; b→r; c→p,q; d→pa. Line x−1−2=y+23=z−1 is along the vector →a=−2^i+3^j−^k and line →r=(3^i−^j+^k)+t(^i+^j+^k) is along the vector →b=^i+^j+^k. Here →a⊥→b. Also, ∣∣ ∣∣3−1−1−(−2)1−0−23−1111∣∣ ∣∣≠0 b. The direction ratios of the line x−y+2z−4=0=2x+y−3z+5=0 are ∣∣ ∣ ∣∣^i^j^k1−1221−3∣∣ ∣ ∣∣=^i+7^j+3^k. Hence, the given two lines are parallel. c. The given lines are (x=t−3,y=−2t+1, z=−3t−2) and →r=(t+1)^i+(2t+3)^j+(−t−9)^k, ⇒x+31=y−1−2=z+2−3 and x−11=y−32=z+9−1 The lines are perpendicular as (1)(1)+(−2)(2)+(−3)(−1)=0. Also, ∣∣ ∣∣−3−11−3−2−(−9)1−2−312−1∣∣ ∣∣=0 Hence, the lines are intersecting. d. The given lines are →r=(^i+3^j−^k)+t(2^i−^j−^k) and →r=(−^i−2^j+5^k)+s(^i−2^j+34^k). ∣∣ ∣ ∣∣1−(−1)3−(−2)−1−52−1−11−23/4∣∣ ∣ ∣∣=0 Hence, the lines are coplanar and intersecting (as the lines are not parallel).

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