Find the equation of the line through the point (1,-1,1) and perpendicular to the lines joining the points (4,3,2), (1,-1,0) and (1,2,-1), (2,1,1).

OR
Find the position vector of the foot of perpendicular drawn from the point P(1,8,4) to the line joining A(O,-1,3) and B(5,4,4). Also find the length of this perpendicular.

Given: Equation of line passing through the point (1, -1, 1)

Firstly, we find the direction ratios of the line (L₁) joining the points

(4, 3, 2) and (1, -1, 0)

x₂ – x₁ = 1 – 4 = -3

y₂ – y₁ = -1 – 3 = -4

z₂ – z₁ = 0 – 2 = -2

So, DR’s of line (L₁) are (-3, -4, -2)

Now, the direction ratios of the line (L₂) joining the points

(1, 2, -1) and (2, 1, 1)

x₂ – x₁ = 2 – 1 = 1

y₂ – y₁ = 1 – 2 = -1

z₂ – z₁ = 1 – (-1) = 1 + 1 = 2

So, DR’s of line (L₂) are (1, -1, 2)

A vector perpendicular to L₁ and L₂ is

∴ Equation of line passing through (1, -1, 1) and perpendicular to L₁ and L₂ is

OR

Given: A(0, -1, 3) and B(5, 4, 4) are the two points

To find: position vector and length of PQ

Proof:

Vector equation of a line passing through two points with position vectors

Given two points are A(0, -1, 3) and B(5, 4, 4)

A(0, -1, 3) B(5, 4, 4)

So,

∴ Point Q is (5λ, -1 + 5λ, 3 + λ)

Now, we have to find the equation of

Two points are P(1, 8, 4) and Q(5λ, -1 + 5λ, 3 + λ)

P(1, 8, 4) Q(5λ, -1 + 5λ, 3 + λ)

So,

∵ PQ ⊥ AB

⇒ 5(5λ – 1) + 5(5λ – 9) + (λ – 1) = 0

⇒ 25λ – 5 + 25λ – 45 + λ – 1 = 0

⇒ 51λ – 51 = 0

⇒ 51 λ = 51

⇒ λ = 1

Putting the value of λ

∴ Foot of perpendicular Q is [5(1), -1 + 5(1), 3 + (1)]

= (5, 4, 4)

Now, we have to find the length of perpendicular PQ

= 4√2 units