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If the distance between the plane Ax-2y+...

If the distance between the plane `Ax-2y+z=d` and the plane containing the lines `(x-1)/(2)=(y-2)/(3)= (z-3)/(4) and (x-2)/(3)= (y-3)/(4)= (z-4)/(5)` is `sqrt6`, then find the value of `|d|`.

Text Solution

Verified by Experts

The correct Answer is:
`|d|=6`
Equation of the plane containing the lines
`(x-2)/(2)=(y-3)/(5)" and "(x-1)/(2)=(y-2)/(3)=(z-3)/(4)`
is `a(x-2)+b(gamma-3)+c(z-4)=0" "...(i)`
where, `3a+4b+5c=0" "…(ii)`
`2a+3b+4c=0" "(iii)`
and`a(1-2)+b(2-3)+c(2-3)=0`
i.e.`" "a+b+c=0" "...(iv)`
From Eqs. (ii) and (iii), `(a)/(1)=(b)/(-2)=(c)/(1),` which satisfy Eq. (iv).
Plane through lines is `x-2y+z=0.`
Given plane is `Ax-2y+z=d` is `sqrt(6).`
`:.` Planes must be parallel, so A=1 and then
`(|d |)/(sqrt(6))=sqrt(6)implies|d|=6`
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