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The plane denoted by P(1) : 4x+7y+4z+81=...

The plane denoted by `P_(1) : 4x+7y+4z+81=0` is rotated through a right angle about its line of intersection with the plane `P_(2) : 5x+3y+ 10 z = 25`. If the plane in its new position is denoted by `P`, and the distance of this plane from the origin is d, then find the value of `[k//2]` (where `[*]` represents greatest integer less than or equal to k).

Text Solution

Verified by Experts

The correct Answer is:
`7`
`" "4x+7y+ 4z+ 81=0" "` (i)
`" "5x+3y+10z=25" "`(ii)
Equation of plane passing through their line of intersection is
`(4x+7y+4z+81)+ lamda(5x+3y+10z-25)=0`
or `" "(4+5lamda)x+ (7+3lamda)y+ (4+10lamda)z+ 81- 25lamda=0" "` (iii)
Plane (iii) `bot` to (i), so
`" "4(4+5lamda)+7(7+3lamda)+ 4(4+10lamda)=0`
`therefore" "lamda=-1`
From (iii), equation of plane is
`" "-x+4y-6z+106=0" "`(iv)
Distance of (iv) from (0, 0, 0)`=(106)/(sqrt(1+16+ 36))=(106)/(sqrt(53))`
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