JEE Main & Advanced Mathematics Vector Algebra Question Bank Application of vectors in three dimensional geometry

  • question_answer
    The vector equation of the plane passing through the origin and the line of intersection of the plane \[\mathbf{r}.\mathbf{a}=\lambda \] and \[\mathbf{r}.\mathbf{b}=\mu \] is

    A)            \[\mathbf{r}.(\lambda \mathbf{a}-\mu \mathbf{b})=0\]

    B)            \[\mathbf{r}.\,(\lambda \mathbf{b}-\mu \mathbf{a})=0\]

    C)            \[\mathbf{r}.(\lambda \mathbf{a}+\mu \mathbf{b})=0\]

    D)            \[\mathbf{r}.(\lambda \mathbf{b}+\mu \mathbf{a})=0\]

    Correct Answer: B

    Solution :

               The equation of a plane through the line of intersection of the planes \[\mathbf{r}.\mathbf{a}=\lambda \] and \[\mathbf{r}.\,\mathbf{b}=\mu \] can be written as \[(\mathbf{r}.\mathbf{a}-\lambda )+k(\mathbf{r}.\mathbf{b}-\mu )=0\] or \[\mathbf{r}.(\mathbf{a}+k\mathbf{b})=\lambda +k\mu \]  .....(i)   This passes through the origin, therefore                    \[\mathbf{0}.(\mathbf{a}+k\mathbf{b})=\lambda +\mu k\Rightarrow k=\frac{-\lambda }{\mu }\]                    Putting the value of k in (i), we get the equation of the required plane as \[\mathbf{r}.(\mu \mathbf{a}-\lambda \mathbf{b})=0\ \Rightarrow \ \ \mathbf{r}\ .\ (\lambda \mathbf{b}-\mu \mathbf{a})=0\].


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