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Please Solve RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Fill in the blanks Question 14 Maths Textbook Solution.

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Answer : The required answer is 2

Hint :   The condition of coplanar is

            (\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0    -----1

Given :  \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a} 1}+\lambda \overrightarrow{\mathrm{b} 1} \text { and } \overrightarrow{\mathrm{r}}=2 \overrightarrow{\mathrm{a} 2}+\lambda \overrightarrow{\mathrm{b} 2}        ------2

Solution :  From Eq.1 and Eq.2

                 \begin{aligned} &(\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0 \\ & \end{aligned}

                 (2 \overrightarrow{a 2}-\overrightarrow{a 1}) \cdot\lfloor{b 2} \times \overrightarrow{b 1})=0

                 \begin{gathered} 2(\overrightarrow{a 2} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1}))-\overrightarrow{a 1} \cdot(\vec{b} 2 \times \overrightarrow{b 1})=0 \\ \end{gathered}   

                 2 \overrightarrow{a 2} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})-\overrightarrow{a 1} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0      -------3

                    Now  \frac {[\overrightarrow{a 1}\overrightarrow{b 1}\overrightarrow{b2} ]}{[\overrightarrow{a 2}\overrightarrow {b1}\overrightarrow{b2} ]} = \frac{\overrightarrow{a1}.\left ( \overrightarrow{b1}\times \overrightarrow{b2}\right )}{\overrightarrow{a2}.\left ( \overrightarrow{b1}\times \overrightarrow{b2}\right )}                    [Using scalar triple formula]

                                             = \frac{2\overrightarrow {a2}\cdot \left ( \overrightarrow{b1}\times\overrightarrow{b2} \right )}{\overrightarrow {a2}\cdot \left ( \overrightarrow{b1}\times\overrightarrow{b2} \right )}                  Substituting the values in Numerator

                                             = 2

                                            

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