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  • Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise Very Short Answer Type Question 27 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise Very Short Answer Type Question 27 Maths Textbook Solution.

Answers (1)

Answer: \sqrt{398}

Hint: You must know the rules of vector functions

Given: If

\begin{aligned} &\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k} \\ &\vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}, \text { find }|3 \vec{a}-2 \vec{b}+4 \vec{c}| \\ &\vec{c}=\hat{i}+2 \hat{j}-\hat{k} \end{aligned}

Solution:

\begin{aligned} &\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k} \\ &\vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}, \\ &\vec{c}=\hat{i}+2 \hat{j}-\hat{k} \end{aligned}

Now,

\begin{aligned} &3 \vec{a}-2 \vec{b}+4 \vec{c}=3(3 \hat{i}-\hat{j}-4 \hat{k})-2(-2 \hat{i}+4 \hat{j}-3 \hat{k})+4(\hat{i}+2 \hat{j}-\hat{k}) \\ &=9 \hat{i}-3 \hat{j}-12 \hat{k}+4 \hat{i}-8 \hat{j}+6 \hat{k}+4 \hat{i}+8 \hat{j}-4 \hat{k} \\ &=17 \hat{i}-3 \hat{j}-10 \hat{k} \\ \end{aligned}

Hence, \begin{aligned} &\therefore|3 \vec{a}-2 \vec{b}+4 \vec{c}|=\sqrt{(17)^{2}-3^{2}-(10)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{289+9+100} \\ &=\sqrt{398} \end{aligned}

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