Given: - Point P(2, 4, – 1) and equation of 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 = λ – 5, y = 4λ – 3, z = – 9λ + 6

Therefore, coordinates of Q(λ – 5,4λ – 3, – 9λ + 6)

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

= (λ – 5 – 2), (4λ – 3 – 4), ( – 9λ + 6 + 1)

= (λ – 7), (4λ – 7), ( – 9λ + 7)

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

(hint: denominator terms of line equation)

= (1,4, – 9)

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.

⇒ 1(λ – 7) + (4)(4λ – 7) – 9( – 9λ + 7) = 0

⇒ λ – 7 + 16λ – 28 + 81λ – 63 = 0

⇒ 98λ – 98 = 0

⇒ λ = 1

Therefore coordinates of Q

i.e. Foot of perpendicular

By putting the value of λ in Q coordinate equation, we get

Now,

So, Equation of perpendicular PQ is

Tip: - Equation of a line joined by two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂) is given by