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JEE Main & Advanced Mathematics Vector Algebra Question Bank Self Evaluation Test - Vector Algebra

  • question_answer
    If  \[\vec{r}\cdot \vec{a}=\vec{r}\cdot b=\vec{r}\cdot \vec{c}=\frac{1}{2}\] for some non-zero vector \[\vec{r}\], then the area of the triangle whose vertices are \[A(\vec{a}),B(\vec{b})\] and \[C\left( {\vec{c}} \right)\] is (\[\vec{a},\text{ }\vec{b},\text{ }\vec{c}\] are non-coplanar)

    A) \[\left| [\vec{a}\,\vec{b}\,\vec{c}] \right|\]

    B) \[\left| {\vec{r}} \right|\]

    C) \[\left| [\vec{a}\,\vec{b}\,\vec{c}]\vec{r} \right|\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] Any vector \[\vec{r}\] can be represented in terms of three non-coplanar vectors \[\vec{a},\,\vec{b}\] and \[\vec{c}\] as \[\vec{r}=x(\vec{a}\times \vec{b})+y(\vec{b}\times \vec{c})+z(\vec{c}+\vec{a})\]                        ?(i) Taking dot product with \[\vec{a},\vec{b}\] and \[\vec{c}\], respectively we have \[x=\frac{\vec{r}\cdot \vec{c}}{[\vec{a}\,\vec{b}\,\vec{c}]},y=\frac{\vec{r}\cdot \vec{a}}{[\vec{a}\,\vec{b}\,\vec{c}]}\] and \[z=\frac{\vec{r}\cdot \vec{b}}{[\vec{a}\,\vec{b}\,\vec{c}]}\] From (i), we have \[[\vec{a}\vec{b}\vec{c}]\vec{r}=\frac{1}{2}(\vec{a}\times \vec{b}\times \vec{c}+\vec{c}\times \vec{a})\] \[\therefore \] Area of \[\Delta ABC=\frac{1}{2}|\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}|\] \[=|[\vec{a}\vec{b}\,\vec{c}]\vec{r}|\]


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