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

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

Answers (1)

Answer:  Proved

Hint:  You must know the rules of solving vectors

Given:  Show that the point whose position vectors are,

\vec{a}=4\hat{i}-3\hat{j}+\hat{k}, \ \ \ \vec{b}=2\hat{i}-4\hat{j}+5\hat{k}, \ \ \ \ \ \vec{c}=\hat{i}-\hat{j} from a right triangle.

Solution: Given that,

\begin{aligned} &\vec{a}=\overrightarrow{O A} \Rightarrow 4 \hat{\imath}-3 \hat{\jmath}+\hat{k} \\ &\vec{b}=\overrightarrow{O B} \Rightarrow 2 \hat{\imath}-4 \hat{\jmath}+5 \hat{k} \\ \end{aligned}

                                                                        \begin{aligned} &\qquad \vec{c}=\overrightarrow{O C} \Rightarrow \hat{\imath}-\hat{\jmath} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \Rightarrow-2 \hat{\imath}-\hat{\jmath}+4 \hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \Rightarrow-\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\\\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \Rightarrow 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A B} \cdot \overrightarrow{B C}=2-3-20 \Rightarrow-21 \neq 0 \\ &\overrightarrow{B C} \cdot \overrightarrow{C A}=-3-6-5 \Rightarrow-14 \neq 0 \\ &\overrightarrow{A B} \cdot \overrightarrow{C A}=-6+2+4 \Rightarrow 0 \end{aligned}

So, \overrightarrow{AB} is perpendicular to \overrightarrow{CA}

So, \Delta ABCis a right angle triangle.

Posted by

infoexpert21

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