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

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Answer: Both the lines are perpendicular to each other

Hint: Angle between perpendicular line is 90^{\circ}

Given: A\left ( 0,0,0 \right ),B\left ( 2,1,1 \right ),C\left ( 3,5,-1 \right ) & B\left ( 4,3,-1 \right )

. Show AB is perpendicular to CD

Solution: we have A\left ( 0,0,0 \right ),B\left ( 2,1,1 \right ),C\left ( 3,5,-1 \right ) & B\left ( 4,3,-1 \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 )

Direction ratio of AB=\left ( 2-0,1-0,1-0 \right )

              \left ( a_{1}, b_{1}, c_{1} \right )=\left ( 2,1,1 \right )

Direction ratio of CD=\left ( 4-3,3-5,-1-\left ( -1 \right ) \right )

              \left ( a_{2}, b_{2}, c_{2} \right )==\left ( 1,-2,0 \right )

Now,

              \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{(2)(1)+(1)(-2)+(1)(0)}{\sqrt{2^{2}+(1)^{2}+(1)^{2}} \sqrt{1^{2}+(-2)^{2}+0^{2}}} \\ &\cos \theta=\frac{2-2}{\sqrt{6} \sqrt{5}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ &\theta=90^{\circ} \end{aligned}

Therefore, AB and CD are perpendicular to each other.

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