Question

If the points (1, 1, λ) and (–3, 0, 1) are equidistant from the plane `barr*(3hati + 4hatj - 12hatk) + 13` = 0, find the value of λ.

Answer 1

The equation of the plane is `barr*(3hati + 4hatj - 12hatk) + 13` = 0

If `barr = xhati + yhatj + zhatk`, then the above equation becomes

`(xhati + yhatj + zhatk)*(3hati + 4hatj - 12hatk) + 13` = 0

∴ 3x + 4y – 12z + 13 = 0

Let p₁ and p₂ be the distances of the points (1, 1, λ) and (–3, 0, 1) from the plane respectively.

Then p₁ = `|(3(1) + 4(1) + (-12)λ + 13)/(sqrt(3^2 + 4^2 + (-12)^2))|`

= `|(3 + 4 - 12λ + 13)/sqrt(9 + 16 + 144)|`

= `|(20 - 12λ)/13|`

and p₂ = `|(3(-3) + 4(0) + (-12)(1) + 13)/sqrt(3^2 + 4^2 + (-12)^2)|`

= `|(-9 + 0 - 12 + 13)/sqrt(9 + 16 + 144)|`

= `8/13`

According to the given condition,

p₁ = p₂

∴ `|(20 - 12λ)/13| = 8/13`

∴ 20 – 12λ = ± 8

∴ 20 – 12λ = 8  or 20 – 12λ = – 8

∴ 12λ = 12 or 12λ = 28

∴ λ = 1 or λ = `7/3`