Question

Find the equation of the plane through the intersection of the planes `vec"r" * (hat"i" + 3hat"j") - 6` = 0 and `vec"r" * (3hat"i" - hat"j" - 4hat"k")` = 0, whose perpendicular distance from origin is unity.

Answer 1

Given planes are;

`vec"r" * (hat"i" + 3hat"j") - 6` = 0 

⇒ x + 3y – 6 = 0  .....(i)

And `vec"r" * (3hat"i" - hat"j" - 4hat"k")` = 0

⇒ 3x – y – 4z = 0   .....(ii)

Equation of the plane passing through the line of intersection of plane (i) and (ii) is

(x + 3y – 6) + k(3x – y – 4z) = 0  .....(iii)

(1 + 3k)x + (3 – k)y – 4kz – 6 = 0

Perpendicular distance from origin

= `|(-6)/sqrt((1 + 3"k")^2 + (3 - "k")^2 + (-4"k")^2)|` = 1

⇒ `36/(1 + 9"k"^2 + 6"k" + 9 + "k"^2 - 6"k" + 16"k"^2)` = 1  .....[Squaring both sides]

⇒ `36/(26"k"^2 + 10)` = 1

⇒ 26k² + 10 = 36

⇒ 26k² = 26

⇒ k² = 1

∴ k = `+- 1`

Putting the value of k in equation (iii) we get,

(x + 3y – 6) ± (3x – y – 4z) = 0

⇒ x + 3y – 6 + 3x – y – 4z = 0 and x + 3y – 6 – 3x + y + 4z = 0

⇒ 4x + 2y – 4z – 6 = 0 and – 2x + 4y + 4z – 6 = 0

Hence, the required equations are:

4x + 2y – 4z – 6 = 0 and – 2x + 4y + 4z – 6 = 0.