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Question

A line l passing through the origin is perpendicular to the lines
l1:(3+t)^i+(1+2t)^j+(4+2t)^k,<t<l2:(3+2s)^i+(3+2s)^j+(2+s)^k,<s<
Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is (are)


A

(73,73,53)

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B

1,1,0

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C

(1,1,1)

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D

(79,79,89)

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Solution

The correct option is D

(79,79,89)


Equation of straight line is l:xx1a=yy1b=zz1c
which is perpendicular to l1 and l2
So, its DR's are cross-product of l1 and l2
Now, to find a point on l2 whose distance is given, assume a point and find its distance to obtian point
Let l:x0a=y0b=z0c
which is perpendicular to
l1:(3^i^j+4^k)+t(^i+2^j+2^k)l2:(3^i+3^j+2^k)+s(2^i+2^j+^k) DR's of l is∣ ∣ ∣^i^j^k122221∣ ∣ ∣=2^i+3^j2^kl:x2=y3=z2=k1,k2Now, A(2k1,3k1,2k1) and B(2k2,3k2,2k2).
Since, A lies on l1.
(2k1)^i+(3k1)^j(2k1)^k=(3+t)^i+(1+2t)^j+(4+2t)^k3+t=2k1,1+2t=3k1,4+2t=2k1 k1=1A(2,3,2)
Let any point on l2(3+2s,3+2s,2+s)
(232s)2+(332s)2+(22s)2=179s2+28s+37=179s2+28s+20=09s2+18s+10s+20=0(9s+10)(s+2)=0s=2,109.Hence,(1,1,0) and (79,79,89) are required points.


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