Find the length of the perpendicular drawn from the point (5, 4, –1) to the line
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