1.

If the distance between the plane Ax 2y + z = d and the plane containing the lines2 1x=3 2y=4 3zand3 2x=4 3y=5 4zis 6 , then |d| is

Answer» Correct Answer - `|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`


Discussion

No Comment Found

Related InterviewSolutions