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Please Solve RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Fill in the blanks Question 15 Maths Textbook Solution.

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Answer : The value of k is  2

Hint :   The condition of coplanar is

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

Given :  The lines are ,

            \overrightarrow{\mathrm{r}}=2 \overrightarrow{\mathrm{a} 1}-3 \overrightarrow{\mathrm{a} 2}+\lambda \overrightarrow{\mathrm{b} 1}       -------1

            \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a} 2}+\mu \overrightarrow{\mathrm{b} 2}     ------2

Solution :  From the condition of co planarity

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

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

          \Rightarrow (4 \vec{a} 2-2 \vec{a}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0

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

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

          \Rightarrow 2 \overrightarrow{\mathrm{a} 2} \cdot(\overrightarrow{\mathrm{b} 2} x \overrightarrow{\mathrm{b} 1})-\overrightarrow{\mathrm{a} 1} \cdot(\overrightarrow{\mathrm{b} 2} x \overrightarrow{\mathrm{b} 1})=0 \\

         \Rightarrow 2[\overrightarrow{\mathrm{a} 2 \mathrm{~b} 1 \mathrm{~b} 2}]=[\overrightarrow{\mathrm{a} 1 \mathrm{~b} 1 \mathrm{~b} 2}]

         a \cdot(b \times c)=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=0                    [  If a,b,c are coplanar  ]

         \begin{aligned} &\Rightarrow \quad[\overrightarrow{a 1}\overrightarrow{b1}\overrightarrow{b2}]=2[\overrightarrow{a 2} \overrightarrow{b1}\overrightarrow{b 2}] \\ & \end{aligned}

         \Rightarrow \quad k=2

           So , the value of  k= 2

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