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Question
Find the equation of the plane passing through the line of intersection of the planes `vec"r"*(2hat"i" - 7hat"j" + 4hat"k")` = 3 and 3x – 5y + 4z + 11 = 0, and the point (– 2, 1, 3)
Solution
`vec"r"*(2hat"i" - 7hat"j" + 4hat"k")` = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 ........(1)
This passes through the point (– 2, 1, 3).
(1) ⇒ (– 4 – 7 + 12 – 3) + λ(– 6 – 5 + 12 + 11) = 0
– 2 + λ(12) = 0
⇒ 12λ = 2
λ = `2/12`
⇒ λ = `1/6`
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + `1/6` (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0
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Choose the correct alternative:
If `vec"a" = hat"i" + hat"j" + hat"k", vec"b" = hat"i" + hat"j", vec"c" = hat"i"` and `(vec"a" xx vec"b")vec"c" - lambdavec"a" + muvec"b"` then the value of λ + µ is
