1.

The centre of the circle given by `vecr.(hati+2hatj+2hatk)=15and|vecr-(hatj+2hatk)=4` isA. (0,1,2)B. (1,3,4)C. (-1,3,4)D. none of these

Answer» Correct Answer - b
The equation of the line through the center `hatj+2hatj` and normal to the given plane is
`vecr=hatj=2hatk+lamda(hati+2hatj+2hatk)" "(i)`
This meets the plane for which
`[hatj+2hatk+lamda(hati+2hatj+2hatk)].(hati+2hatj+2hatk)=15`
or `6+9lamda=15orlamda=1`
Putting in (i), we get
`vecr=hatj+2hatk+(hati+2hatj+2hatk)=hati+3hatj+4hatk`
Hence, centre is (1,3,4).


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