Question

Find the image A' of the point A(2, 1, 2) in the line `l : vecr = 4hati + 2hatj + 2hatk + λ(hati - hatj - hatk)`. Also, find the equation of line joining AA'. Find the foot of prependicular form A on the line l.

Answer 1

Given: `vecr= 4hati + 2hatj + 2hatk + λ(hati - hatj - hatk)`

Change above line equation in Cartesian form 

`(x - 4)/1 = (y - 2)/(-1) = (z - 2)/(-1) = λ("let")`   ...(i)

Let A' (p, q, r) be the image of A(2, 1, 2).

B → foot of perpendicular and arbitrary point on given line.

Let B(λ + 4, −λ + 2, −λ + 2)   ...(ii)

Now, the direction ratios of the line.

AB = (λ + 4 − 2, −λ + 2 − 1, − λ + 2 − 2)

AB = (λ + 2, −λ + 1, −λ)

Also AB ⊥ line.

∴ a₁a₂ + b₁b₂ + c₁c₂ = 0

Here, a₁ = λ + 2, b₁ = −λ + 1, c₁ = −λ

and a₂ = 1, b₂ = −1, c₂ = −1

∴ (λ + 2) (1) + (−λ + 1) (−1) − λ (–1) = 0

λ + 2 + λ − 1 + λ

= 0

λ = `-1/3`

Putting λ = `–1/3` in equation (ii), we get

∴ `B(-1/3 + 4, 1/3 + 2, 1/3 + 2) = B(11/3, 7/3, 7/3)`

Hence, the coordinates of the foot of the perpendicular from pointA on the line l is `B(11/3, 7/3, 7/3)`

Since B = midpoint of AA'.

`(11/3, 7/3, 7/3) = ((2 + p)/2, (1 + q)/2, (2 + r)/2)`

`11/3 = (2 + p)/2, 7/3 = (1 + q)/2, 7/3 = (2 + r)/2`

Solving,

p = `16/3, q = 11/3, r = 8/3`

Hence, the coordinates of the image of the point

A'`(16/3, 11/3, 8/3)`

Equation of line AA'

`(x - 2)/(16/3 - 2) = (y - 1)/(11/3 - 1) = (z - 2)/(8/3 - 2)`

⇒ `(3(x - 2))/10 = (3(y - 1))/8 = (3(z - 3))/2`

or `(x - 2)/5 = (y - 1)/4 = (z - 2)/1`