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Question: Answered & Verified by Expert
The point of intersection of the lines r=7i˙^+10j˙^+13k^+s(2i˙^+3j˙^+4k^)  and r=3i˙^+5j˙^+7k^+t(i˙^+2j˙^+3k^) is
MathematicsVector AlgebraJEE Main
Options:
  • A i˙^+j˙^-k^
  • B 2i˙^-j˙^+4k^
  • C i˙^-j˙^+k^
  • D i˙^+j˙^+k^
Solution:
2982 Upvotes Verified Answer
The correct answer is: i˙^+j˙^+k^
Two given lines intersect, if

7i˙^+10j˙^+13k^+s2i˙^+3j˙^+4k^=3i˙^+5j˙^+7k^+t(i˙^+2j˙^+3k^)

7+2si˙^+10+3sj˙^+(13+4s)k^=3+ti˙^+5+2tj˙^+(7+3t)k^

7+2s=3+t

2s-t=-4 ...(i)

10+3s=5+2t

3s-2t=-5 ...(ii)

and 13+4s=7+3t

4s-3t=-6 ...(iii)

On solving Eqs. (i) and (iii), we get

s=-3, t=-2

 Required point is

7i˙^+10j˙^+13k^+-3[2i˙^+3j˙^+4k^]

= i˙^+j˙^+k^ 

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