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  • Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 5 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 5 maths textbook solution

Answers (1)

Answer: The direction cosine of the side of triangle ABC are

              \left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right),\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right),\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right)

Hint: Direction ratios are proportional to the length of side.

Given:  A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, 2). Find direction cosine of AB, BC and AC.

 

Solution:                                        A(3, 5, -4)

                           

                      B(-1, 1, 2)                                                    C(-5, -5, 2)          

Let us consider an \sqcup ABC

Direction ratio of AB = (-1-3, 1-5, 2+4)

                             = (-4, -4, 6)

 Direction cosines of AB are

              \begin{aligned} &=\left(\frac{-4}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}, \frac{-4}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}, \frac{6}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}\right) \\ &=\left(\frac{-4}{\sqrt{68}}, \frac{-4}{\sqrt{68}}, \frac{6}{\sqrt{68}}\right) \\ &=\left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right) \end{aligned}……………….. (1)

Direction ratio of BC = (-5+1, -5-1, -2-2)

                             = (-4, -6, -4)

Direction cosines of BC are

\begin{aligned} &=\left(\frac{-4}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}, \frac{-6}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}, \frac{-4}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}\right) \\ &=\left(\frac{-4}{\sqrt{68}}, \frac{-6}{\sqrt{68}}, \frac{-4}{\sqrt{68}}\right) \\ &=\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right) \end{aligned} ………………. (2)

Direction ratio of AC = (-5-3, -5-5, -2+4)

                             = (-8, -10, 2)

Direction cosines of AC are

  \begin{aligned} &=\left(\frac{-8}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}, \frac{-10}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}, \frac{2}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}\right) \\ &=\left(\frac{-8}{\sqrt{168}}, \frac{-10}{\sqrt{168}}, \frac{2}{\sqrt{168}}\right) \\ &=\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right) \end{aligned} ………………… (3)

By (1), (2) and (3) we get direction cosine of side AB, BC and AC are

              \left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right),\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right),\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right)

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