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11.The unit vector perpendicular to the vectors 6i +2j+ 3kand Зі-6-2k is-2i-3j+6k2i +3j-6k2i-3j-6k2i+3j+6 |
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Answer» The unit vector is given by cross product of these vectors , divided by the magnitude of that vector. the cross product is 14i+21j-42j so, the magnitude is √14²+21²+42² = 49 so the unit vector is.(14i+21j-42j)/49 = (2i+3j-6k)/7 option C. |
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