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Question
Find the Cartesian equation of the plane, passing through the line of intersection of the planes `vecr. (2hati + 3hatj - 4hatk) + 5 = 0`and `vecr. (hati - 5hatj + 7hatk) + 2 = 0` intersecting the y-axis at (0, 3).
Solution
The given planes are :
`vecr. (2hati + 3hatj - 4hatk) + 5 = 0`
`vecr. (hati - 5hatj + 7hatk) + 2 = 0`
Any plane through the line of intersection of the given planes is :
`vecr. { (2hati + 3hatj - 4hatk) + λ (hati - 5hatj + 7hatk) } = -5 - λ(2)` ...(1)
It passes through the point (0, 3, 0), i.e;
`(0hati + 3hatj + 0hatk) . { (2hati + 3hatj - 4hatk) + λ (hati - 5hatj + 7hatk) } = - 5 -2 λ`
`3hatj . { (2 + λ) hati + (3 - 5 λ) hatj + (-4 + 7 λ) hatk } = - 5 -2 λ`
⇒ 3(3 - 5λ) = -5 - 2λ
⇒ 9 - 15λ = -5 - 2λ
⇒ 14 = 15λ - 2λ
⇒ 13λ = 14
λ = `(14)/(13)`
Substituting this value of λ in (1), we have
`vecr. { (2hati + 3hatj - 4hatk) +(14)/(13) (hati - 5hatj + 7hatk )} = -5 - 2 (14/13)`
⇒ `vecr. (40hati - 31hatj + 46hatk) = -65 - 28`
⇒ `vecr. (40hati - 31hatj + 46hatk) = -93`
⇒ `vecr. (40hati - 31hatj + 46hatk) + 93 = 0`
Which is the required equation.
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