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  • Provide Solution for RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 7

Provide Solution for RD Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.6 Question 7

Answers (1)

Answer:

Proved L H S = R H S

Hint:

Use vector magnitude formula for both L.H.S and R.H.S.

Given:

\hat{\imath}-\hat{j}, 4 \hat{\imath}-3 \hat{j}+\hat{k} \text { and } 2 \hat{i}-4 \hat{\jmath}+5 \hat{k}

Solution:

In a right triangle  C A^{2}=A B^{2}+B C^{2}  , where CA is the hypotenuse.

BC is the perpendicular and AB is the base.

Vertices of triangle are given below

A=(1,-1,0), B=(4,-3,1), C=(2,-4,5)

So,

 \begin{gathered} A \vec{B}=\vec{B}-\vec{A}=(4 \hat{\imath}-3 \hat{j}+\hat{k})-(\hat{i}-\hat{j}) \\ \\A \vec{B}=4 \hat{\imath}-3 \hat{\jmath}+\hat{k}-\hat{\imath}+\hat{\jmath}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\\\ |A \vec{B}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1} \\ \end{gathered}

|A \vec{B}|=\sqrt{14}                                                                 Eq.(i)

Similarly,

\begin{aligned} & B \vec{C}=\vec{C}-\vec{B}=(2 \hat{i}-4 \hat{\jmath}+5 \hat{k})-(4 \hat{\imath}-3 \hat{\jmath}+\hat{k}) \\\\ &B \vec{C}=-2 \hat{\imath}-\hat{\jmath}+4 \hat{k} \\ & \begin{array}{l} \end{array} \end{aligned}


|B \vec{C} |=\sqrt{(-2)^{2}+(-1)^{2}+4^{2}}=\sqrt{4+1+16} \\


|B \vec{C}|=\sqrt{21}                                                                 Eq.(ii)


\begin{aligned} &C \vec{A}=\vec{A}-\vec{C}=(\hat{\imath}-\hat{j})-(2 \hat{i}-4 \hat{\jmath}+5 \hat{k}) \\\\ &C \vec{A}=-\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\\\ &|C \vec{A}|=\sqrt{(-1)^{2}+3^{2}+(-5)^{2}}=\sqrt{1+9+25} \\\\ &|C \vec{A}|=\sqrt{35} \\ \end{aligned} Eq.(iii)

\begin{aligned} &C A^{2}=A B^{2}+B C^{2} \\\\ &(\sqrt{35})^{2}=(\sqrt{14})^{2}+(\sqrt{21})^{2} \\\\ &35=14+21 \\\\ &35=35 \end{aligned}

So, three points form a right angle triangle.

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infoexpert27

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