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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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