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Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise very Short Answer type Question 45 Maths Textbook Solution.

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Answer: \frac{4}{13}\hat{i}+\frac{3}{13}\hat{j}-\frac{12}{13}\hat{k}

Hint: You must know the rules of vector functions

Given: Find unit vector in direction of sum of vectors

\begin{aligned} &\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}, \\ &\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k} \end{aligned}

Solution: We have,

\begin{aligned} &\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}, \\ &\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k} \end{aligned}

Sum, \begin{aligned} &\vec{p}=\vec{a}+\vec{b}=2 \hat{i}+2 \hat{j}-5 \hat{k}+2 \hat{i}+\hat{j}-7 \hat{k} \\ \end{aligned}

\begin{aligned} &=4 \hat{i}+3 \hat{j}-12 \hat{k} \\ \end{aligned}

∴Required unit vector

\begin{aligned} &\frac{\vec{p}}{|\vec{p}|}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{(4)^{2}+(3)^{2}+(-12)^{2}}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{16+9+144}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{169}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{13} \\ &\therefore \frac{4}{13} \hat{i}+\frac{3}{13} \hat{j}-\frac{12}{13} \hat{k} \end{aligned}

 

 

 

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