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  • Need solution for RD Sharma Maths Class 12 Chapter 27 Straight Line of Space Excercise Fill in the blanks Question 10

Need solution for RD Sharma Maths Class 12 Chapter 27 Straight Line of Space Excercise Fill in the blanks Question 10

Answers (1)

Answer :    -\frac{10}{7}

Hint :      Use vector dot product and vector equation of line.

Given :    \begin{aligned} &\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2} \text { and } \frac{x-1}{3 k}=\frac{5-y}{-1}=\frac{6-z}{5} \\ & \end{aligned}

              \text { let } \overrightarrow{\mathrm{b} 1}=-3 \hat{\imath}+2 \mathrm{k} \hat{\mathrm{j}}+2 \hat{k} \text { and } \overrightarrow{\mathrm{b} 2}=3 \mathrm{k} \hat{\imath}+\hat{\jmath}-5 \hat{k}

              \overrightarrow{\mathrm{b} 1} \cdot \overrightarrow{\mathrm{b} 2}=0   as the lines are at right angle

            \begin{aligned} &-9 k+2 k-10=0 \\ & \end{aligned}

            -7 k=10 \\

                 k=-\frac{10}{7} 

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