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