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Let P=-(1,7,sqrt(2)) be a point and line L is 2sqrt(2)(x-1)=y-2,z=0. If PQ is the distance of plane sqrt(2)x+y-z=1 from point P measured along a line inclined at an angle of 45^(@) with the line L and is minimum then the value of PQis

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Solution :Line L is `(x-1)/1=(y-2)/2sqrt(2)=z/0`
This line L MAKES an ANGLE of `45^(@)` with the PLANE `SQRT(2)x+y-z=1`
`therefore` Required distance PQ is PREPENDICULAR distance of plane from P
I.e., `PQ=(|sqrt(2)+7-sqrt(2)-1|)/sqrt(2+1+1)=3`


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