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Let `A` and `B` be points with position vectors `veca` and `vecb` with respect to origin `O`. If the point `C` on `OA` is such that `2vec(AC)=vec(CO), vec(CD) ` is parallel to `vec(OB)` and `|vec(CD)|=3|vec(OB)|` then `vec(AD)` is (A) `vecb-veca/9` (B) `3vecb-veca/3` (C) `vecb-veca/3` (D) `vecb+veca/3`A. `3b-(a)/(2)`B. `3b+(a)/(2)`C. `3b-(a)/(3)`D. `3b+(a)/(3)` |
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Answer» Correct Answer - C Since, `OA=a,OB=b and 2AC=CO` by section formula, `OC=(2)/(3)a` Therefore, `|CD|=3|OB|` `impliesCD=3b` `impliesOD=OC+CD=(2)/(3)a+3b` Hence, `AD=OD-OA=(2)/(2)a+3b-a` `=3a-(1)/(3)a`. |
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