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Question
Find the equation of the plane passing through the intersection of the planes `vecr . (hati + hatj + hatk)` and `vecr.(2hati + 3hatj - hatk) + 4 = 0` and parallel to the x-axis. Hence, find the distance of the plane from the x-axis.
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Solution
We have,
`vecr.(hati + hatj + hatk) - 1 = 0` ....(1)
`vecr.(2hati + 3hatj - hatk) + 4 = 0` .....(2)
Equation of plane passing through the intersection of the planes (i) and (ii), is given by
`[vecr.(hati + hatj + hatk) - 1 ] = λ[vecr. (2hati + 3hatj - hatk) + 4 ] = 0`
⇒ `vecr.[(1 + 2λ)hati + (1 + 3λ)hatj + (1 - λ)hatk] - 1 + 4λ = 0` ....(iii)
If plane (iii) is parallel to x-axis, then
1 + 2λ = 0 ⇒ λ = `-1/2`
Therefore, the equation of the required plane is `vecr.(-hatj + 3hatk) = 6`
Distance of the plane `vecr.(-hatj + 3hatk) = 6` from x-axis is given by,
`d = ((1 xx 0 - 3 xx 0 + 6)/(sqrt(1^2 + 3^2))) = 6/sqrt10`
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