# 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,

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