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  • Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 14 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 14 maths textbook solution

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Answer: Angle between AB and CD is 0^{\circ}

Given: A(1,2,3), B(4,5,7), C(-4,3,-6) \& D(2,9,2). Find angle between AB and CD

Solution: we have A(1,2,3), B(4,5,7), C(-4,3,-6) \& D(2,9,2)

Direction ratio of AB=\left ( 4-1,5-2,7-3 \right )

              \left ( a_{1},b_{1},c_{1} \right )=\left ( 3,3,4 \right )

Direction ratio of CD=\left ( 2-\left ( -4 \right ),9-3,2-\left ( -6 \right ) \right )

              \left ( a_{2},b_{2},c_{2} \right )=\left ( 6,6,8 \right )

Let \theta be the angle between two lines whose direction cosines are \left ( a_{1},b_{1},c_{1} \right ) &  \left ( a_{2},b_{2},c_{2} \right )

              \begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(3)(6)+(3)(6)+(4)(8)}{\sqrt{3^{2}+(3)^{2}+(4)^{2}} \sqrt{6^{2}+6^{2}+8^{2}}} \\ &\cos \theta=\frac{18+18+32}{\sqrt{9+9+16} \sqrt{36+36+64}} \\ &\cos \theta=\frac{68}{\sqrt{34} \sqrt{136}} \\ &\cos \theta=\frac{68}{34 \times 2} \\ \end{aligned}

               \theta=\cos ^{-1}(1) \\

             \theta=0^{\circ}

Therefore, angle between AB and CD is \theta=0^{\circ}

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