Q2 of 136 Page 28

Find the perpendicular distance of the point (1, 0, 0) from the line Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Given: - Point P(1, 0, 0) and equation of line


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


Thus to find Distance PQ we have to first find coordinates of Q



x = 2λ + 1, y = – 3λ – 1, z = 8λ – 10


Therefore, coordinates of Q(2λ + 1, – 3λ – 1,8λ – 10)


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 – 1), ( – 3λ – 1 – 0), (8λ – 10 – 0)


= (2λ), ( – 3λ – 1), (8λ – 10)


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


(hint: denominator terms of line equation)


= (2, – 3,8)


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) + ( – 3)( – 3λ – 1) + 8(8λ – 10) = 0


4λ + 9λ + 3 + 64λ – 80 = 0


77λ – 77 = 0


λ = 1


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







= 2√6 unit


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