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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.
Solution
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.
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