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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 24 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 24 Maths Textbook Solution.

Answers (1)

Answer:  \cos ^{-1}\left ( \frac{10}{\sqrt{102}} \right )

Hint:  You must know the rules of solving vectors.

Given:  If vertices  A,Band C of \Delta ABC have position vectors \left ( 1,2,3, \right ) \left ( -1,0,0 \right )and \left ( 0,1,2 \right ) respectively. What is the magnitude of < ABC?

Solution: Given that,

\begin{aligned} &\overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{O B}=-\hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\ &\overrightarrow{O C}=0 \hat{\imath}+1 \hat{\jmath}+2 \hat{k} \\ &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \end{aligned}

\begin{aligned} &\overrightarrow{A B}=-2 \hat{\imath}-2 \hat{\jmath}-3 \hat{k} \\ & \mid \overrightarrow{A B} \mid=\sqrt{4+4+9} \\ & \mid \overrightarrow{A B} \mid=\sqrt{17} \\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \end{aligned}

\begin{aligned} &\overrightarrow{B C}=\hat{\imath}+\hat{\jmath}+2 \hat{k} \\ &\mid \overrightarrow{B C} \mid=\sqrt{1+1+4} \\ & \mid \overrightarrow{B C} \mid=\sqrt{6} \\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\ \end{aligned}

\begin{aligned} &\overrightarrow{C A}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ & \mid \overrightarrow{C A} \mid=\sqrt{1+1+1} \\ & \mid \overrightarrow{C A} \mid=\sqrt{3} \end{aligned}

\begin{aligned} &\operatorname{Cos}<A B C=\frac{|\overrightarrow{A B} \cdot \overrightarrow{B C}|}{|\overrightarrow{A B}||\overrightarrow{B C}|} \\ &\operatorname{Cos}<A B C=\frac{|-2-2-6|}{|\sqrt{17}||\sqrt{6}|} \\ &\operatorname{Cos}<A B C=\frac{10}{\sqrt{102}} \end{aligned}

< ABC=\cos ^{-1}\left ( \frac{10}{\sqrt{102}} \right )

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