Find the equation of the perpendicular drawn from the point P(–1, 3, 2) to the line Also, find the coordinates of the foot of the perpendicular from P.
Given: -
Point P( – 1, 3, 2) 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 = 2λ, y = λ + 2, z = 3λ + 3
⇒ ; Equation of line
Thus, coordinates of Q i.e. General point on the given line
⇒ Q(2λ, (λ + 2), (3λ + 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 + 1), (λ + 2 – 3), (3λ + 3 – 2)
= (2λ + 1), (λ – 1), (3λ + 1)
and by comparing with line equation, direction ratios of the given line are
(hint: denominator terms of line equation)
= (2,1,3)
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 + 1) + (λ – 1) + 3(3λ + 1) = 0
⇒ 4λ + 2 + λ – 1 + 9λ + 3 = 0
⇒ 14λ + 4 = 0
Therefore coordinates of Q
i.e. Foot of perpendicular
By putting the value of λ in Q coordinate equation, we get
2λ, (λ + 2), (3λ + 3)
Position Vector of Q
Now,
Equation of line passing through two points with position vectoris given by
Here,
⇒