Find the perpendicular distance of the point (3, –1, 11) from the line

Given: - Point P(3, – 1, 11) and the equation of the 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, y = – 3λ + 2, z = 4λ + 3


Therefore, coordinates of Q(2λ, – 3λ + 2,4λ + 3)


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 – 3), ( – 3λ + 2 + 1), (4λ + 3 – 11)


= (2λ – 3), ( – 3λ + 1), (4λ – 8)


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


(hint: denominator terms of line equation)


= (2, – 3,4)


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) + ( – 3)( – 3λ + 3) + 4(4λ – 8) = 0


4λ – 6 + 9λ – 9 + 16λ – 32 = 0


29λ – 47 = 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







unit


1