Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P (5, 4, 2) to the line . Also find the image of P in this line.

Given: a point P (5, 4, 2), line .

To find: the length and coordinates of the foot of the perpendicular from the point P to the given line. Also find the image of point P in the given line.

Given line is

So the Cartesian form of the line will be,

Let Q(α, β, γ) be the foot of perpendicular drawn from P(5, 4, 2) to the line (i) and P’ (x₁ , y₁ , z₁) be the image of P on the line (i)

Now as point Q lies on the given line, so let

So from above condition we get

Hence the coordinates of Q are (2k-1, 3k+3, 1-k)

Now

Parallel vector of the given line is

Obviously,

Substituting corresponding values, we get

2(α-5)+3(β-4)+(-1)(γ-2)=0

⇒ 2α-10+3β-12-γ+2=0

⇒ 2α+3β-γ-20=0

Now substituting the values of α, β, γ, we get

⇒ 2(2k-1)+3(3k+3)-(1-k)-20=0

⇒ 4k-2+9k+9-1+k-20=0

⇒ 14k-14=0

⇒ 14k=14

⇒ k=1

Substituting the value of k in coordinates of Q, we get

Q(2k-1, 3k+3, 1-k)

⇒ Q(2(1)-1, 3(1)+3, 1-(1))

⇒ Q(2-1, 3+3, 1-1)

⇒ Q(1, 6, 0)

Hence coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P (5, 4, 2) to the line is Q (1,6,0)

Now we will find the length of this perpendicular

So

Or the length of the perpendicular is √24 or 2√6 units.

Now Let P’(x₁, y₁, z₁) be the image of point P (5, 4, 2) in the line

Now from figure it is clear that M(1, 6, 0) is the midpoint of line PP’

Hence now applying the formula

Hence the image of the given point P(5, 4, 2) is P’(-3, 8, -2).