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Question
If →OA=i+j+k,→AB=3i−2j+k,→BC=i+2j−2k,→CD=2i+j+3k then the position vector of D is?
If →OA=i+j+k,→AB=3i−2j+k,→BC=i+2j−2k,→CD=2i+j+3k then the position vector of D is?
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Solution
The correct option is D −7i−4j−3k
−−→OA=→O−→A=i+j+kA=0i+0j+0k−i−j−kA=−i−j−k−−→AB=→A−→B=3i−2j+k→B=→A−3i+2j−k=−i−j−k−3i+2j−k→B=−4i+j−2k−−→BC=→B−→C=i+2j−2k−4i+j−2k=i+2j−2k+c→C=−5i−3j+0−−→CD=2i+j+3k→C−→D=2i+j+3k→D=→C−(2i+j+3k)=−5i−3j−2i−j−3k=−7i−4j−3k
−−→OA=→O−→A=i+j+kA=0i+0j+0k−i−j−kA=−i−j−k−−→AB=→A−→B=3i−2j+k→B=→A−3i+2j−k=−i−j−k−3i+2j−k→B=−4i+j−2k−−→BC=→B−→C=i+2j−2k−4i+j−2k=i+2j−2k+c→C=−5i−3j+0−−→CD=2i+j+3k→C−→D=2i+j+3k→D=→C−(2i+j+3k)=−5i−3j−2i−j−3k=−7i−4j−3k
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