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Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 11 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: \mathrm{A}(1,-1,2), \mathrm{B}(3,4,-2), \mathrm{C}(0,3,2) \& \mathrm{D}(3,5,6). Show AB is perpendicular to CD

Solution: we have \mathrm{A}(2,3,6), \mathrm{B}(1,2,2), \mathrm{C}(0,3,2) \& \mathrm{D}(3,5,6)

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 ( 3-1,4-\left ( -1 \right ),-2-2 \right )

              \left ( a_{1},b_{1},c_{1} \right )=\left ( 2,5,-4 \right )

Direction ratio of CD=\left ( 3-0,5-3,6-2 \right )

              \left ( a_{2},b_{2},c_{2} \right )=\left ( 3,2,4 \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)(3)+(5)(2)+(-4)(4)}{\sqrt{2^{2}+(5)^{2}+(-4)^{2}} \sqrt{3^{2}+(2)^{2}+4^{2}}} \\ &\cos \theta=\frac{6+10-16}{\sqrt{45} \sqrt{29}} \\ &\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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