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  • Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 13 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 13 Maths Textbook Solution.

Answers (1)

Answer:

Hence proved

Hint:

Use distance formula   \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}

Given:

A(2 \hat{\imath}-\hat{\jmath}+\hat{k}), B(\hat{\imath}-3 \hat{\jmath}-5 \hat{k}), C(3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k})

Solution:

                                      
                                

Let BC length be a, AC length be b and AB length be c.

\begin{aligned} \text { Then, } a &=C-B=\sqrt{(3-1)^{2}+(-4+3)^{2}+(-4+5)^{2}} \\ &=\sqrt{4+1+1} \\ &=\sqrt{6} \end{aligned}       [ Use distance formula of two points ]

\begin{aligned} b &=C-A=\sqrt{(3-2)^{2}+(-4+1)^{2}+(-4-1)^{2}} \\ &=\sqrt{1+9+25} \\ &=\sqrt{35} \end{aligned}                    [ Use distance formula of two points ]

 
\begin{aligned} c &=A-B=\sqrt{(2-1)^{2}+(-1+3)^{2}+(1+5)^{2}} \\ &=\sqrt{1+4+36} \\ &=\sqrt{41} \end{aligned}                         [ Use distance formula of two points ] 

\begin{aligned} &a^{2}+b^{2}=(\sqrt{6})^{2}+(\sqrt{35})^{2} \\ &=6+35 \\ &=41 \\ &=(\sqrt{41})^{2}=c^{2} \\ &a^{2}+b^{2}=c^{2} \end{aligned}

ABC is right angle triangle.

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