Answer :

Given: - Point (5, 4, – 1) and equation of line


Let, PQ be the perpendicular drawn from P to given line whose endpoint/ foot is Q point.


As we know position vector is given by



Therefore,


Position vector of point P is



and, from a given line, we get





On comparing both sides we get,


x = 1 + 2λ, y = 9λ, z = 5λ


; Equation of line


Thus, coordinates of Q i.e. General point on the given line


Q((1 + 2λ), 9λ, 5λ)


Now as we know (TIP) ‘if two points A(x1,y1,z1) and B(x2,y2,z2) on a line, then its direction ratios are proportional to (x2 – x1,y2 – y1,z2 – z1)’


Hence


Direction ratio of PQ is


= (2λ + 1 – 5), (9λ – 4), (5λ + 1)


= (2λ – 4), (9λ – 4), (5λ + 1)


and by comparing with line equation, direction ratios of the given line are


(hint: denominator terms of line equation)


= (2,9,5)


Since PQ is perpendicular to given line, therefore by “condition of perpendicularity.”


a1a2 + b1b2 + c1c2 = 0 ; where a terms and b terms are direction ratio of lines which are perpendicular to each other.


2(2λ – 4) + (9)(9λ – 4) + 5(5λ + 1) = 0


4λ – 8 + 81λ – 36 + 25λ + 5 = 0


110λ – 39 = 0



Therefore coordinates of Q


i.e. Foot of perpendicular


By putting the value of λ in Q coordinate equation, we get




Now,


Distance between PQ


Tip: - Distance between two points A(x1,y1,z1) and B(x2,y2,z2) is given by








units


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