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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 1 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2  Question 1 Maths Textbook Solution.

Answers (1)

Answer: three direction cosines i.e. a, b, c are perpendicular to each other

Given: show that three lines with directions cosines \frac{12}{13},\frac{-3}{13},\frac{-4}{13},\frac{4}{13},\frac{12}{13},\frac{3}{13},\frac{3}{13},\frac{-4}{13},\frac{12}{13}; are mutually perpendicular

Hint: for mutally perpendicular

                \vec{a}.\vec{b}=0,\vec{b}.\vec{c}=0,\vec{a}.\vec{c}=0

solution:

             \begin{aligned} &1^{\text {st }} \text { vector } \vec{a}=\frac{12}{13} \hat{i}-\frac{3}{13} \hat{j}-\frac{4}{13} \hat{k} \\ \end{aligned}

             \begin{aligned} &2^{\text {nd }} \text { vector } \vec{b}=\frac{4}{13} \hat{i}+\frac{12}{13} \hat{j}+\frac{3}{13} \hat{k} \\ \end{aligned}

             \begin{aligned} &3^{\text {rd }} \text { vector } \vec{c}=\frac{3}{13} \hat{i}-\frac{4}{13} \hat{j}+\frac{12}{13} \hat{k} \end{aligned}

\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{12}{13} \times \frac{4}{13}-\frac{3}{13} \times \frac{12}{13}-\frac{-4}{13} \times \frac{3}{13}=0 \\ \end{aligned}

\begin{aligned} &\vec{a} \cdot \vec{c}=\frac{36}{13 \times 13}+\frac{-12}{13 \times 13}-\frac{48}{13 \times 13}=0 \\ \end{aligned}

\begin{aligned} &\vec{b} \cdot \vec{c}=\frac{12}{13 \times 13}-\frac{48}{13 \times 13}+\frac{36}{13 \times 13}=0 \\\\ &=\frac{12-48+36}{13 \times 13}=\frac{0}{13 \times 13}=0 \end{aligned}

so all the three dimensions cosines are mutually perpendicular to each other .

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