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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 21 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 21 Maths Textbook Solution.

Answers (1)

Answer:  < A=\frac{\pi }{4},< B=\frac{\pi }{2},< C=\frac{\pi }{4}

Hint: You must know the rules of finding angle of triangle

Given: Find angle of triangle A\left ( 0,-1,-2 \right ),B\left ( 3,1,4 \right ) and C\left ( 5,7,1 \right )

Solution: Given that,

                             \overrightarrow{O A}=0 \vec{\imath}-1 \hat{\jmath}-2 \hat{k}, \overrightarrow{O B}=3 \vec{\imath}+1 \hat{\jmath}+4 \hat{k}, \overrightarrow{O C}=5 \vec{\imath}+7 \hat{\jmath}+1 \hat{k}

\begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\ &\overrightarrow{A B}=3 \vec{\imath}+2 \hat{\jmath}+6 \hat{k} \\ &\overrightarrow{A B}=\sqrt{9+4+36} \\ &\overrightarrow{A B}=\sqrt{49} \\ &\overrightarrow{A B}=7 \end{aligned}

\begin{aligned} &\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B} \\ &\overrightarrow{B A}=-3 \vec{\imath}+2 \hat{\jmath}-6 \hat{k} \\ &\overrightarrow{B A}=\sqrt{9+4+36} \\ &\overrightarrow{B A}=\sqrt{49} \\ &\overrightarrow{B A}=7 \\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \end{aligned}

\begin{aligned} &\overrightarrow{B C}=2 \vec{\imath}+6 \hat{\jmath}-3 \hat{k} \\ &\overrightarrow{B C}=\sqrt{4+36+9} \\ &\overrightarrow{B C}=\sqrt{49} \\ &\overrightarrow{B C}=7 \end{aligned}

\begin{aligned} &\overrightarrow{C B}=\overrightarrow{O B}-\overrightarrow{O C} \\ &\overrightarrow{C B}=-2 \vec{\imath}-6 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{C B}=\sqrt{4+36+9} \\ &\overrightarrow{C B}=\sqrt{49} \\ &\overrightarrow{C B}=7 \end{aligned}

\begin{aligned} &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\ &\overrightarrow{C A}=-5 \vec{l}-8 \hat{j}-3 \hat{k} \\ &\overrightarrow{C A}=\sqrt{25+64+9} \\ &\overrightarrow{C A}=\sqrt{98} \\ &\overrightarrow{C A}=7 \sqrt{2} \end{aligned}

\begin{aligned} &\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A} \\ &\overrightarrow{A C}=5 \vec{\imath}+8 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{A C}=\sqrt{25+64+9} \\ &\overrightarrow{A C}=\sqrt{98} \\ &\overrightarrow{A C}=7 \sqrt{2} \end{aligned}

\begin{aligned} &\operatorname{Cos} A=\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}||\overrightarrow{A C}|} \\\\ &\operatorname{Cos} A=\frac{15+16+18}{(7)(7 \sqrt{2})} \\\\ &\operatorname{Cos} A=\frac{49}{49 \sqrt{2}} \\\\ &\operatorname{Cos} A=\frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} &A=\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ &A=\frac{\pi}{4} \\ &\operatorname{Cos} B=\frac{\overrightarrow{B A} \cdot \overrightarrow{B C}}{|\overrightarrow{B A}||\overrightarrow{B C}|} \end{aligned}

\begin{aligned} &\operatorname{Cos} B=\frac{-6-12+18}{(7)(7)} \\ &\operatorname{Cos} B=\frac{0}{49} \\ &\operatorname{Cos} B=0 \\ &B=\operatorname{Cos}^{-1}(0) \\ &B=\frac{\pi}{2} \end{aligned}

\begin{aligned} &\operatorname{Cos} C=\frac{\overrightarrow{C B} \cdot \overrightarrow{C A}}{|\overrightarrow{C B}||\overrightarrow{C A}|} \\ &\operatorname{Cos} C=\frac{10+48-9}{(7)(7 \sqrt{2})} \\ &\operatorname{Cos} C=\frac{49}{49 \sqrt{2}} \end{aligned}

\begin{aligned} &\operatorname{Cos} C=\frac{1}{\sqrt{2}} \\ &C=\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ &C=\frac{\pi}{4} \end{aligned}

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