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Question

# The position vectors of two points A and C are 9^i−^j+7^k and 7^i−2^j+7^k respectively. The point of intersection of the lines containing vectors −−→AB=4^i−^j+3^kand −−→CD=2^i−^j+2^k is P. If a vector −−→PQ is perpendicular to −−→AB~\text {and }\overrightarrow{CD} \text { and }PQ = 15\) units, the possible position vectors of Q are x1^i+x2^j+x3^k and y1^i+y2^j+y3^k. Then the value of 3∑i=1(xi+yi) is equal to

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Solution

## Cut equation of two lines AB, CD then their point of intersection P. Equations of lines AB and CD are: AB:→r=(9^i−^j+7^k)+t(4^i−^j+3^k) ...(1) CD:→r=(7^i−2^j+7^k)+s(2^i−^j+2^k) ...(2) Co-ordinates of point of intersection p of lines (1) and (2) x−94=y+1−1=z−73=t ⇒ Point on line AB is (9+4t,–1–t,7+3t) and point on line CD are (7+2s,–2–s,7+2s) ∴ 9+4t=7+2s ⇒ 2s–4t=2 ⇒ s–2t=1 ...(1) –1–t=–2–s ⇒ s–t=–1 ...(2) and 7+3t=7+2s ⇒ 2s–3t=0 ...(3) From (3), t=2s3 ∴ From (2), s−2s3=−1⇒s=−3 ∴ ⇒ t=–2 ∴ P≡(9–8),–1+2,7–6)P≡(1,1,1) Let Q≡(x,y,z) ∴ −−→PQ=(x−1)^i+(y+1)^j+(2−1)^k Given, ∣∣∣−−→PQ∣∣∣=15 ⇒√(x−1)2+(y−1)2+(2−1)2=15 ⇒ (x–1)2+(y–1)2+(z–1)2=225 ...(4) given, −−→PQ⊥−−→ABand−−→PQ⊥−−→CD ⇒ −−→PQ∥∥∥−−→AB×−−→CD Where −−→AB×−−→CD=^i^j^k4−132−12 =^i(−2+3)−^j(8−6)+^k(−4+2) =^i−2^j−2^k and so x−11=y−1−2=z−1−2=λ(say) ...(5) ⇒ x−1=λ, y−1=−2λ, z−1=−2λ Putting these values in (4) λ2+(−2λ)2+(−2λ)2=225 ⇒ 9λ2=225⇒λ=±5 when λ=5, then from (5) x=6,y=–9,z=–9 when λ=–5, then from (5), x=–4,y=11,z=11  ∴Coordinates of Q can be (6,–9,–9)and (–4,11,11) So that x1=6,x2=–6,x3=–9 and y1=–4,y2=11, y3=11 ∴ 3∑i=1(xi+yi)=(x1+y1)+(x2+y2)+(x3+y3) =(x1+x2+x3)+(y1+y2+y3) =(6–9–9)+(–4+11+11) =–12+18=6

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