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Q.

If OA=i+2j+3k and OB=4i+k are the position vectors of the points A and B, then the position vector of a point on the line passing through B and parallel to the vector OA×OB which is at a distance of 189 units from B is

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a

6i+11j-7k

b

4i+11j-8k

c

2i-11j+8k

d

-2i-11j+8k

answer is A.

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Detailed Solution

BP=t(OA×OB) OA×OB=ijk123401                 =i(2-0)-j(-11)+k(-8) |BP|=t|OA×OB| 189=t 4+121+64    t=1 OP-(4i+k)=2i+11j-8k OP=6i+11j-7k

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