and are two vectors. The position vectors of the points A and C are and , respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that PQ is perpendicular to AB and CD both.


Given,




and Position Vectors (Vector from origin to a point is called a position vector)




line passing through A and along AB will have equation




and line passing through C and along CD will have equation




Now, PQ is a vector perpendicular to both AB and CD, such that P lies on AB and Q lies on CD


Therefore, coordinates of P and Q will have the form


P (6 + 3λ, 7 – λ, 4 + λ) (i)


Q (-3μ, 2μ – 9, 2 + 4μ) (ii)


Hence, vector PQ will be



Now, As PQ is perpendicular to both, therefore dot products


AB. PQ = 0 and CD. PQ = 0


AB. PQ = 3(-3μ – 6 – 3λ) – (2μ – 16 + λ) + (4μ – 2 – λ)


0 = -9μ – 18 – 9λ – 2μ + 16 - λ + 4μ – 2 - λ


-7μ – 11λ - 4 = 0 …. (iii)


CD.PQ = 3(-3μ – 6 – 3λ) + 2(2μ – 16 + λ) + 4(4μ – 2 – λ)


0 = -9μ – 18 – 9λ + 4μ - 32 + 2λ +16 μ – 8 - 4λ


29μ + 7λ - 22 = 0 …. (iv)


Solving (iii) and (iv), we get


λ = -1 and μ = 1


Put value of λ in (i) to get,


P (6 + 3(-1), 7 – (-1), 4 + (-1))


P (6-3,7+1,4-1)


P (3,8,3)


Put value of μ in (ii) to get,


Q (-3(1), 2(1) – 9, 2 + 4(1))


Q (-3, 2 – 9, 2 +4)


Q (3, -7, 6)


Hence, position vector of P and Q will be




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