EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    Of the vectors given below, the parallel vectors are, \[{{A}^{\to }}=6\widehat{i}+8\widehat{j}\] \[{{B}^{\to }}=210\widehat{i}+280\widehat{k}\] \[{{C}^{\to }}=5.1\widehat{i}+6.8\widehat{k}\] \[{{D}^{\to }}=3.6\widehat{i}+6\widehat{j}+48\widehat{k}\]

    A)  \[{{A}^{\to }}\] and \[{{B}^{\to }}\]                        

    B)  \[{{A}^{\to }}\] and \[{{C}^{\to }}\]

    C)  \[{{A}^{\to }}\] and \[{{D}^{\to }}\]                        

    D)  \[{{C}^{\to }}\]and \[{{D}^{\to }}\]

    Correct Answer: C

    Solution :

                     \[\vec{A}=6\hat{i}+8\hat{j}\] \[\vec{B}=210\hat{i}+280\hat{k}\] \[\vec{C}=5.1\hat{i}+6.8\hat{j}=\frac{1}{20}(6\hat{i}+8\hat{j})\] \[\vec{D}=3.6\hat{i}+8\hat{j}+4.8\hat{k}\] Hence, it is clear that\[\text{\vec{A}}\] and\[\vec{C}\] are parallel and we can write as \[\vec{A}=\frac{1}{20}\vec{C}\] This implies that\[\text{\vec{A}}\]is parallel to\[\vec{C}\] and magnitude of \[\text{\vec{A}}\] is \[\frac{1}{20}.\]


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